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\begin{document}


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\begin{center}
{\Large\bf Real Analysis} \\
{Dr Nikolai Chernov}
\end{center}

\section{Lebesgue measure in $\IR^2$}
\label{SecLMR2}

\begin{definition} An \textbf{interval} $I \subset \IR$ is a set of the form
$[a,b]$ or $[a,b)$ or $(a,b]$ or $(a,b)$, where $a \leq b$ are real
numbers. Its length is $|I| = b-a$.
\end{definition}

\begin{definition} A \textbf{rectangle} $R \subset \IR^2$ is a set of the form
$R = I_1 \times I_2$, where $I_1, I_2$ are intervals. The area of a
rectangle is $\Area(R) = |I_1| \times |I_2|$.
\end{definition}

\begin{definition} An \textbf{elementary set} is a finite union of disjoint
rectangles, i.e.\ $B = \cup_{i=1}^n R_i$, where $R_i \cap R_j =
\emptyset$ for $i \neq j$. The area of an elementary set is $\Area
(B) = \sum_{i=1}^n \Area (R_i)$.
\end{definition}

\begin{theorem}
The area of an elementary set does not depend on how it is
partitioned into disjoint rectangles.
\end{theorem}

\begin{theorem}
Finite unions, intersections and differences of elementary sets are
elementary sets. Thus, elementary sets make a ring, see below.
\end{theorem}

\begin{definition} A \textbf{ring} is a nonempty collection of subsets of a
set $X$ closed under finite unions, intersections and differences.
\end{definition}

\begin{definition} An \textbf{algebra} is a ring containing $X$ itself.
\end{definition}

From now on we will work with sets $A \subset X$ within a fixed
rectangle $X \subset \IR^2$. Elementary sets in $X$ make an algebra.

\begin{definition} The \textbf{outer measure} of a set $A \subset X$ is
$\muo(A) = \inf \sum_{i=1}^{\infty} \Area (R_i)$ where the infimum
is taken over all countable covers of $A$ by rectangles, $A \subset
\cup_{i=1}^{\infty} R_i$.
\end{definition}

\begin{definition} The \textbf{inner measure} of a set $A \subset X$ is
$\mui(A) = \Area (X) - \muo(A^c)$ where $A^c = X \setminus A$ is the
complement of $A$.
\end{definition}

\begin{lemma0} For any elementary set $B$ we have $\muo(B) = \Area
(B)$.
\end{lemma0}

Proof: BWOC, suppose $\muo(B) < \Area (B)$. If $B$ is a closed set
covered by open rectangles, then due to compactness there is a
finite subcover. If $B$ is not closed, shrink it a bit to make it
closed. If the rectangles are not open, expand them a bit to make
them open.

\begin{theorem}
For any set $A\subset X$ we have $0 \leq \mui(A) \leq \muo(A) \leq
\Area (X)$.
\end{theorem}

Note: $\mui(A) \leq \muo(A)$ is equivalent to $\muo(A) + \muo(A^c)
\geq \Area (X)$, which follows from the previous lemma.

\begin{definition}  \label{Defmeasuable}
A set $A \subset X$ is \textbf{measurable} if $\mui(A) = \muo(A)$;
then its \textbf{Lebesgue measure} is defined by $\bm(A) = \mui(A) =
\muo(A)$.
\end{definition}

Note: a set $A \subset X$ is measurable if and only if $\muo(A) +
\muo(A^c) = \Area (X)$. This implies that if $A$ measurable, then
$A^c$ is measurable, too.\medskip

Note:  Let $\cup_i R_i = C_1\supset A$ and $\cup_j R_j' = C_2\supset
A^c$ be countable covers of $A$ and $A^c$ by rectangles,
respectively. We can choose them so that $\sum_i \Area(R_i) + \sum_j
\Area(R_j') < \Area (X) + \varepsilon$. Then $\sum_{i,j} \Area
(R_i\cap R_j) < \varepsilon$.\medskip

Note: an empty set is measurable and $\bm (\emptyset) =0$.

\begin{exercise}
Prove that every countable set $A \subset X$ is measurable and $\bm
(A)=0$.
\end{exercise}

\begin{exercise}[Bonus]
Let $A \subset X$ consist of points $(x,y)$ such that either $x$ or
$y$ is a rational number. Is $A$ measurable? What is its Lebesgue
measure?
\end{exercise}

\begin{theorem}
The Lebesgue measure is translationally invariant, i.e.\ if $A
\subset X$ is measurable and $a \in \IR^2$, then the set
$$
    A+a = \{x+a \colon x\in A\}
$$
is measurable and $\bm (A+a) = \bm (A)$.
\end{theorem}

The Lebesgue measure can be constructed in $\IR$ by using intervals
(and their length) instead of rectangles (and area). The Lebesgue
measure in $\IR$ is translationally invariant, too, i.e.\ if $A$ is
measurable and $a \in \IR$, then the set $A+a = \{x+a \colon x\in
A\}$ is also measurable and $\bm (A+a) = \bm (A)$.

The Lebesgue measure can be extended to unbounded sets $A \subset
\IR^2$ as follows: $A$ is measurable if and only if $A \cap X$ is
measurable for every rectangle $X \subset \IR^2$. Its measure is
$$
  \bm (A) = \lim_{n\to \infty} \bm (A\cap X_n),
$$
where $X_n = [-n,n] \times [-n,n]$ is a growing sequence of
rectangles. The value of $\bm (A)$ does not depend on the particular
growing sequence of rectangles.

\begin{theorem}[$\sigma$-subadditivity]  \label{Tmsigadd}
If $A \subset \cup_{i=1}^{\infty} A_i$, then $\muo(A) \leq
\sum_{i=1}^{\infty} \muo(A_i)$.
\end{theorem}

\begin{corollary}
If $A_1 \subset A_2$, then $\muo(A_1) \leq \muo(A_2)$.
\end{corollary}

\begin{theorem}
Every elementary set $B \subset X$ is measurable and its Lebesgue
measure is equal to its area, i.e.\ $\bm(B) = \Area (B)$.
\end{theorem}

\begin{lemma}
For any two sets $A_1, A_2 \subset X$ we have $|\muo(A_1) -
\muo(A_2)| \leq \muo(A_1 \Delta A_2)$.
\end{lemma}

Proof: BWOC, suppose $\muo(A_1)>\muo(A_2)+\muo(A_1\Delta A_2)$, then
$\muo(A_1)>\muo(A_2)+\muo(A_1\setminus A_2)$, which contradicts
Theorem~\ref{Tmsigadd}

\begin{theorem}[Approximation]
A set $A \subset X$ is measurable if and only if for any
$\varepsilon>0$ there exists an elementary set $B \subset X$ such
that $\muo(A \Delta B) < \varepsilon$.
\end{theorem}

Note that, in the context of the above theorem, $|\bm(A) - \Area(B)|
\leq \varepsilon$. \medskip

Proof: ``$\Rightarrow$'': Let $\cup_i R_i = C_1\supset A$ and
$\cup_j R_j' = C_2\supset A^c$ be countable covers of $A$ and $A^c$
by rectangles, respectively, such that $\sum_i \Area(R_i) + \sum_j
\Area(R_j') < \Area (X) + \varepsilon$. Define $B = \cup_{i=1}^N
R_i$ so that $\sum_{i=N+1}^{\infty} \Area(R_i) < \varepsilon$. Then
$A \Delta B \subset (C_1\setminus B) \cup (B\cap C_2)$, and
$\muo(C_1\setminus B)<\varepsilon$ by construction, while
$\muo(B\cap C_2)<\varepsilon$ according to the second remark after
Definition~\ref{Defmeasuable}.

``$\Leftarrow$'': Note that $\muo(A^c \Delta B^c) = \muo(A \Delta B)
< \varepsilon$, hence by the previous lemma $\muo(A) \approx
\muo(B)=\Area(B)$ and $\muo(A^c) \approx \muo(B^c)=\Area(B^c)$ (here
$\approx$ means that the two measures are $\varepsilon$-close).
Lastly, note that $\Area (B) + \Area (B^c) = \Area (X)$, hence
$\muo(A) + \muo(A^c) \approx \Area (X)$.

\begin{theorem}
Finite unions, intersections and differences of measurable sets are
measurable.
\end{theorem}

\begin{theorem}[Additivity]
If $A_1, \ldots, A_n$ are disjoint measurable sets, then $\bm \bigl(
\cup_{i=1}^n A_i \bigr) = \sum_{i=1}^n \bm(A_i)$.
\end{theorem}

\begin{theorem}
Countable unions and intersections of measurable sets are
measurable. Thus measurable sets make a $\sigma$-algebra, see below.
\end{theorem}

\begin{definition} A \textbf{$\sigma$-algebra} is an algebra closed
under countable unions and intersections. (It is enough to mentions
countable unions.)
\end{definition}

\begin{theorem}[$\sigma$-additivity]
If $\{A_i\}_{i=1}^\infty$ is a countable collection of pairwise
disjoint measurable sets, then $\bm \bigl( \cup_{i=1}^{\infty} A_i
\bigr) = \sum_{i=1}^{\infty} \bm(A_i)$.
\end{theorem}

\begin{exercise}
Prove that every open set $A \subset X$ is measurable (hint:
represent it by a countable union of rectangles). Prove that every
closed set $A \subset X$ is measurable.
\end{exercise}

Note: every $G_\delta$ set is measurable and every $F_\sigma$ set is
measurable. (Recall: $G_\delta$ set is a countable intersection of
open sets; $F_\sigma$ set is a countable union of closed sets.)

Note: every set obtained from open or closed sets by a sequence
countable unions, intersections, differences, etc., is a measurable
set.

\begin{theorem}
If $A \subset X$ is measurable and $\bm(A) = 0$, then every subset
$A' \subset A$ is measurable and $\bm(A')=0$.
\end{theorem}

\begin{theorem}
If $A \subset X$ is measurable and $\bm(A) > 0$, then there is a
non-measurable subset $A' \subset A$.
\end{theorem}

\begin{theorem}[Continuity - I]
Let $A_1 \supset A_2 \supset \cdots \supset A_n \supset \cdots$ be a
sequence of measurable sets (called \emph{monotonically decreasing
sequence}). Then $\lim_{n \to \infty} \bm(A_n) = \bm(A)$, where $A =
\cap_{n=1}^{\infty} A_n$.
\end{theorem}

\begin{theorem}[Continuity - II]
Let $A_1 \subset A_2 \subset \cdots \subset A_n \subset \cdots$ be a
sequence of measurable sets (called \emph{monotonically increasing
sequence}). Then $\lim_{n \to \infty} \bm(A_n) = \bm(A)$, where $A =
\cup_{n=1}^{\infty} A_n$.
\end{theorem}

Example: the middle-third Cantor set $F \subset [0,1]$ is a closed
uncountable set of zero measure. Its cardinality is the same as that
of $\IR$.

\begin{definition}
Two sets $A$ and $B$ have the same \textbf{cardinality},
$\card(A)=\card(B)$, if there is a bijection $\varphi \colon A
\leftrightarrow B$. The set $A$ has cardinality smaller than that of
$B$, $\card(A)<\card(B)$, if there is an injection $\varphi \colon A
\to B$ but not vice versa.
\end{definition}

Note: `same cardinality' is an equivalence relations. Thus all sets
of the same cardinality make an (equivalence) class. For any two
sets $A$ and $B$ there is either an injection $A \to B$ or an
injection $B\to A$ (this follows from a `well ordering theorem',
which we do not quote here).

\begin{theorem}[Cantor-Bernstein-Schroeder]
If there is an injection $A \to B$ and an injection $B \to A$, then
there is a bijection $A \leftrightarrow B$. Hence, for any two sets
$A$ and $B$ there are three possibilities: $\card(A)=\card(B)$ or
$\card(A)>\card(B)$ or $\card(A)<\card(B)$.
\end{theorem}

For a finite set $A$ of $n$ elements, we simply put $\card(A)=n$.
For the set of integers $\IN$, we put $\card(\IN) = \aleph_0$
(`aleph-null'). Then any countable set has cardinality $\aleph_0$,
in particular, $\card(\IZ) = \card(\IQ) = \aleph_0$.

The cardinality of $\IR$ is denoted by $\fC$ and is called
\textbf{continuum}. We have $\card(I) = \fC$ for any interval $I
\subset \IR$ of positive length. We have $\card(\IR^n) = \fC$ for
every $n \geq 1$. We also have $\aleph_0 < \fC$.

The \textbf{continuum hypothesis} states that there is no set $A$
such that $\aleph_0 < \card(A) < \fC$. In particular, every subset
of $\IR$ is either finite or countable or has the same cardinality
as $\IR$. The continuum hypothesis cannot be proved or disproved if
one uses standard mathematical axioms. Thus it may (or may not) be
adopted as an independent axiom.

\begin{definition}
For any set $A$, the \textbf{power set} (or \textbf{powerset}) of
$A$, denoted by $2^A$, is the set of all subsets of $A$.
\end{definition}

Note: if $\card(A)=n < \infty$, then $\card(2^A) = 2^n$.

\begin{theorem}[Cantor]
For any set $A$ we have $\card(A) < \card(2^A)$.
\end{theorem}

The set $2^{\IN}$ can be identified with the set of all infinite
sequences of zeros and ones. This set has cardinality $\fC$.

\begin{exercise}
Let $\fL$ denote the set of all Lebesgue measurable sets (in $\IR$).
Prove that $\card(\fL) > \fC$.
\end{exercise}

\begin{exercise}
Let $\fU$ denote the set of all open subsets $U \subset \IR$. Prove
that $\card(\fU) = \fC$. Do the same for open sets in $\IR^2$.
\end{exercise}

\section{General measures}

General measures are defined on sets with $\sigma$-algebras.

\begin{definition}
A set $X$ with a $\sigma$-algebra $\fM$ of its subsets is called a
\textbf{measurable space}. Sets $A \in \fM$ are said to be
\textbf{measurable}.
\end{definition}

\begin{definition}
Let $(X,\fM)$ be a measurable space. A \textbf{measure} is a
function $\mu$, defined on $\fM$, whose range is $[0,\infty]$ and
which is $\sigma$-additive. To avoid trivialities, we shall also
assume that $\mu(A)<\infty$ for at least one $A \in \fM$. A
measurable space with a measure is called a \textbf{measure space}.
\end{definition}

For any set $X$, there are two trivial $\sigma$-algebras. One is
\emph{minimal}, it consists of the sets $X$ and $\emptyset$ only.
The other is \emph{maximal}, it contains all the subsets of $X$. The
latter one is denoted by $2^X$.

\begin{exercise}
Let $X = \{1,2,3\}$. Construct all $\sigma$-algebras of $X$.
\end{exercise}

If $X$ is finite or countable, we will always use the maximal
$\sigma$-algebra $2^X$. In that case every measure $\mu$ on $(X,
2^X)$ is determined by its values $p_x = \mu(\{x\})$ on one-point
sets $\{x\}, x\in X$.

\begin{theorem}
Let $\{\fM_\alpha\}$ be an arbitrary collection of $\sigma$-algebras
of a set $X$. Then their intersection $\cap_\alpha \fM_\alpha$ is a
$\sigma$-algebra of $X$ as well.
\end{theorem}

\begin{theorem}
Let $\fG$ be any collection of subsets of $X$. Then there exists a
unique $\sigma$-algebra $\fM^\ast \supset \fG$ such that for any
other $\sigma$-algebra $\fM \supset \fG$ we have $\fM^\ast \subset
\fM$. (In other words, $\fM^\ast$ is the minimal $\sigma$-algebra
containing $\fG$.)
\end{theorem}

\begin{definition}
We say that the minimal $\sigma$-algebra $\fM^\ast$ containing the
given collection $\fG$ is \textbf{generated} by $\fG$. We also
denote it by $\fM(\fG)$.
\end{definition}

\begin{exercise}
Let $X = [0,1]$ and $\fG$ consist of all one-point sets, i.e.\ $\fG
= \bigl\{ \{x\}, x\in X \bigr\}$. Describe the $\sigma$-algebra
$\fM(\fG)$.
\end{exercise}

It is interesting to compare the notion of $\sigma$-algebra with
topology. The standard topology in $\IR$ is not a $\sigma$-algebra.
On the other hand, the $\sigma$-algebra in the previous exercise is
not a topology.

\begin{definition}
Let $X$ be a topological space. The $\sigma$-algebra $\fM$ generated
by the collection of all open subsets $U \subset X$ is called the
\textbf{Borel} $\sigma$-algebra. Its members are called
\textbf{Borel sets}.
\end{definition}

Note: all closed sets, all $F_\sigma$ sets and all $G_\delta$ sets
are Borel sets. The cardinality of the Borel $\sigma$-algebra of
$\IR$ is $\fC$, i.e.\ continuum.

\begin{exercise}
Show that every Borel set in $\IR$ is Lebesgue measurable, but not
vice versa.
\end{exercise}

\begin{exercise}[Bonus]
Does there exist an infinite $\sigma$-algebra which has only
countably many members?
\end{exercise}

\begin{definition}
Let $(X,\fM)$ be a measurable space. For any $X \in \fM$, define
$\mu(A)=\infty$ if $A$ is an infinite set, and $\mu(A)=\,{\rm
card}(A)$ if $A$ is finite. Then $\mu$ is called \textbf{counting
measure}.
\end{definition}

\begin{definition}
Let $(X,\fM)$ be a measurable space and $x\in X$. The measure
$\delta_x$ defined by
$$
     \delta_x(A)=\left\{\begin{array}{ll} 1 & {\rm if}\ \ x\in A\\
     0 & {\rm otherwise}\end{array}\right .
$$
is called the \textbf{delta-measure} or the \textbf{Dirac measure}
(concentrated at $x$).
\end{definition}

\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space. Then $\mu(\emptyset)=0$ and
for every measurable sets $A \subset B$ we have $\mu(A) \leq
\mu(B)$.
\end{theorem}

\begin{theorem}[Continuity - I] \label{tmCIa}
Let $A_1 \supset A_2 \supset \cdots \supset A_n \supset \cdots$ be a
sequence of measurable sets, and $\mu(A_1)<\infty$. Then $\lim_{n
\to \infty} \mu(A_n) = \mu(A)$, where $A = \cap_{n=1}^{\infty} A_n$.
\end{theorem}

\begin{theorem}[Continuity - II]
Let $A_1 \subset A_2 \subset \cdots \subset A_n \subset \cdots$ be a
sequence of measurable sets. Then $\lim_{n \to \infty} \mu(A_n) =
\mu(A)$, where $A = \cup_{n=1}^{\infty} A_n$.
\end{theorem}

\begin{exercise}
Show that the assumption $\mu(A_1) <\infty$ in Theorem~\ref{tmCIa}
is indispensable. Hint: consider the counting measure on $\IN$ and
take sets $A_n = \{n,n+1,\ldots\}$.
\end{exercise}

In many cases $\sigma$-algebras tend to be rather large, an explicit
definition (or description) of $\mu(A)$ for all $A \in \fM$ is often
an impossible task. It is common to define $\mu(A)$ on a smaller
collection of sets, $\fE$, and extend it to $\fM(\fE)$.

\begin{definition}
A \textbf{semi-algebra} is a nonempty collection $\fE$ of subsets of
$X$ with two properties: (i) it is closed under intersections; i.e.\
if $A,B \in \fE$, then $A \cap B \in \fE$; and (ii) if $A \in \fE$,
then $A^c = \cup_{i=1}^n A_i$, where each $A_i \in \fE$ and $A_1,
\ldots, A_n$ are pairwise disjoint subsets of $X$.
\end{definition}

\begin{exercise}
Let $X \subset \IR^2$ be a rectangle. Verify that the collection of
all subrectangles $R \subset X$ is a semi-algebra.
\end{exercise}

\begin{theorem}[Extension]
Let $\fE$ be a semi-algebra of $X$. Let $\nu$ be a function on
$\fE$, whose range is $[0,\infty)$ and which is $\sigma$-additive,
i.e.\ for any $A \in \fE$ such that $A = \cup_{i=1}^\infty A_i$ and
$A_i \in \fE$ are pairwise disjoint sets, we have $\mu(A) =
\sum_{i=1}^\infty \mu(A_i)$. Then there is a unique measure $\mu$ on
the $\sigma$-algebra $\fM (\fE)$ that agrees with $\nu$ on $\fE$,
i.e.\ $\mu(A) = \nu(A)$ for all $A\in \fE$.
\end{theorem}

We accept this theorem without proof. Its proof is basically the
repetition of our construction of the Lebesgue measure in $\IR^2$.

\begin{corollary} \label{crLLL}
Let $X \subset \IR^2$ be a rectangle. There is a unique measure
$\mu$ on the Borel $\sigma$-algebra of $X$ such that for every
rectangle $R \subset X$ we have $\mu(R)=m(R)$, where $m(R)$ is the
the area of $R$.
\end{corollary}

The following useful theorem is also given without proof:

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $\fE$ a collection of
subsets of $X$ that generates $\fM$, i.e. such that $\fM (\fE)=
\fM$. Suppose two measures, $\mu_1$ and $\mu_2$, agree on $\fE$,
i.e.\ $\mu_1(A)=\mu_2(A)$ for all $A\in \fE$, and
$\mu_1(X)=\mu_2(X)$. Then $\mu_1=\mu_2$.
\end{theorem}

The extension theorem provides a measure on the minimal
$\sigma$-algebra containing the given semi-algebra. It is often
convenient to complete it in the way we constructed the Lebesgue
measure.

\begin{definition}
A measure $\mu$ on a measurable space $(X,\fM)$ is said to be
\textbf{complete} if every subset of any set of measure zero is
measurable, i.e.\ if $B \subset A$ and $\mu(A)=0$ imply $B\in \fM$
(and then obviously $\mu(B)=0$).
\end{definition}

Sets of zero measure are called \textbf{null sets}. Their
complements are called \textbf{full measure sets}.

\begin{theorem}[Completion]
Let $(X, \fM, \mu)$ be a measure space. Then
$$
   \bar{\fM} = \{A\cup E\colon A\in
   \fM,\ \ E \subset N
   \ \text{for some null set}\ N\}
$$
is a $\sigma$-algebra. For every $B = A \cup E$ as above define
$\bar{\mu} (B) = \mu(A)$. Then $\bar{\mu}$ is a complete measure on
$(X, \bar{\fM})$. Moreover, if $\mu^\ast$ is another complete
measure that coincides with $\mu$ on $\fM$, then $\mu^\ast$
coincides with $\bar{\mu}$ on $\bar{\fM}$.
\end{theorem}

$\bar{\mu}$ is called the \textbf{completion} of $\mu$. The
completion of the measure constructed in Corollary~\ref{crLLL} is
exactly the Lebesgue measure on $X \subset \IR^2$.

\section{Measurable functions}

Recall: given two topological spaces $X$ and $Y$, a function $f
\colon X \to Y$ is continuous iff for any open set $V \subset Y$ its
preimage $f^{-1} (V)$ is open, too.

\begin{definition}
Let $(X,\fM)$ be a measurable space and $Y$ a topological space. A
function $f\colon X \to Y$ is a \textbf{measurable function} iff for
any open set $V \subset Y$ its preimage is measurable, i.e.\ $f^{-1}
(V) \in \fM$.
\end{definition}

The most interesting functions for us are real-valued ($Y=\IR$) and
complex-valued ($Y=\IC$). We also consider the \textbf{extended real
line} $[-\infty, \infty]$ with topology in which $(a, \infty]$ and
$[-\infty, b)$ are open sets. Then $f \colon X \to [-\infty,
\infty]$ is an \textbf{extended real-valued function}.

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $Y,Z$ topological spaces. If
$f\colon X \to Y$ is measurable and $g \colon Y \to Z$ is
continuous, then their composition $g \circ f \colon X \to Z$ is
measurable.
\end{theorem}

Recall that the Borel $\sigma$-algebra is generated by open sets.

\begin{definition}
Let $X,Y$ be topological spaces. A function $f\colon X \to Y$ is a
\textbf{Borel function} iff for any open set $V \subset Y$ its
preimage is a Borel set.
\end{definition}

\begin{theorem}
Continuous functions are Borel functions.
\end{theorem}

There are many more Borel functions than continuous functions.

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $f\colon X \to Y$ a
function. Then  \begin{itemize} \item[\rm (a)] $\fG = \{E\subset
Y\colon f^{-1}(Y) \in \fM\}$ is a $\sigma$-algebra in $Y$;
\item[\rm (b)] if $Y$ is a topological space and $f$ measurable, then
$f^{-1}(E) \in \fM$ for any Borel set $E \subset Y$; \item[\rm (c)]
If $Y$ and $Z$ are topological spaces, $f\colon X \to Y$ is
measurable and $g \colon Y \to Z$ is a Borel function, then $g \circ
f \colon X \to Z$ is measurable.
\end{itemize}
\end{theorem}

Given a subset $A \subset X$ the function
$$
   \chi_A (x) = \left\{ \begin{array}{cc}
   1 & \text{if}\ \ x \in A\\
   0 & \text{if} \ \ x \notin A \end{array}\right .
$$
is called the \textbf{characteristic function} of $A$ (or the
\textbf{indicator} of $A$).

\begin{exercise}
Prove that $A \in \fM$ if and only if $\chi_A$ is measurable.
\end{exercise}

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $f\colon X \to [-\infty,
\infty]$. Then $f$ is measurable iff $f^{-1} \bigl( [-\infty, x)
\bigr)$ is a measurable set for every $x \in \IR$.
\end{theorem}

\begin{exercise}
In the context of the previous theorem, prove that $f$ is measurable
iff $f^{-1} \bigl( [-\infty, x] \bigr)$ is a measurable set for
every $x \in \IR$.
\end{exercise}

\begin{exercise}
In the context of the previous theorem, prove that $f$ is measurable
iff $f^{-1} \bigl( [-\infty, x) \bigr)$ is a measurable set for
every rational $x \in \IQ$.
\end{exercise}

\begin{exercise}
Let $(X,\fM)$ be a measurable space and $f\colon X \to [-\infty,
\infty]$ and $g\colon X \to [-\infty, \infty]$ two measurable
functions. Prove that the sets
$$
   \{x\colon f(x)<g(x)\}
   \qquad\text{and}\qquad
   \{x\colon f(x)=g(x)\}
$$
are measurable.
\end{exercise}

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $u\colon X \to \IR$ and
$v\colon X \to \IR$ two measurable functions. Let $Y$ be a
topological space and $\Phi \colon \IR^2 \to Y$ a continuous
function. Then $h(x) = \Phi (u(x), v(x))$ is a measurable function
from $X$ to $Y$.
\end{theorem}

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $u\colon X \to \IR$ and
$v\colon X \to \IR$ two functions. Then $f(x) = u(x) + \bi v(x)$ is
a measurable function from $X$ to $\IC$ if and only if both $u(x)$
and $v(x)$ are measurable functions.
\end{theorem}

\begin{theorem} \label{tm|f|}
Let $(X,\fM)$ be a measurable space and $f\colon X \to \IC$ and
$g\colon X \to \IC$ two measurable functions. Then $|f|$, $f+g$,
$f-g$, and $fg$ are measurable functions.
\end{theorem}

\begin{theorem}[Polar factorization]
Let $(X,\fM)$ be a measurable space and $f\colon X \to \IC$ a
measurable function. Then $f=g|f|$, where $g \colon X \to \IC$ is a
measurable function such that $|g|=1$.
\end{theorem}

Note: arithmetic operations in $[-\infty,\infty]$ are defined in an
obvious way (for example, $a+\infty = \infty$ for any $a\in
(-\infty,\infty]$, $a+(-\infty) = -\infty$ for any $a\in
[-\infty,\infty)$, $a\cdot \infty = \infty$ for every $a\in
(0,\infty]$, etc.; we also put $0\cdot\infty=0$, a bit arbitrarily;
the sum $\infty + (-\infty)$ is not defined). The set
$[-\infty,\infty]$ is naturally ordered, any subset $A \subset
[-\infty,\infty]$ obviously has $\inf A$ a $\sup A$. The convergence
of sequences in $[-\infty,\infty]$ is defined by using its topology.
For any sequence $\{a_n\}$ in $[-\infty,\infty]$ we naturally define
$\liminf a_n$ and $\limsup a_n$:
$$
  \limsup_{n\to\infty} a_n = \inf \{b_1,b_2,\ldots\},\qquad
  b_n = \sup\{a_n,a_{n+1},\ldots\}.
$$

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $f_n\colon X \to
[-\infty,\infty]$ measurable functions. Then
$$
  g=\sup_{n\geq 1} f_n
  \qquad{\rm and}\qquad
  h=\limsup_{n\to\infty} f_n
$$
are measurable functions. (Similarly for $\inf f_n$ and $\liminf
f_n$.)
\end{theorem}

\begin{corollary}
Let $(X,\fM)$ be a measurable space and $f_n\colon X \to
[-\infty,\infty]$ measurable functions. If the limit
$$
  g(x)=\lim_{n\to\infty} f_n(x)
$$
exists for every $x\in X$, then $g$ is a measurable function.
\end{corollary}

\begin{corollary} \label{crf+-}
Let $(X,\fM)$ be a measurable space and $f,g\colon X \to
[-\infty,\infty]$ are measurable functions. Then
$$
  \max\{ f(x),g(x)\}
  \qquad{\rm and}\qquad
  \min\{ f(x),g(x)\}
$$
are measurable functions. Also,
$$
  f^+(x) = \max\{ f(x),0\}
  \qquad{\rm and}\qquad
  f^-(x) = -\min\{ f(x),0\}
$$
are measurable functions.
\end{corollary}

The above functions $f^+$ and $f^-$ are called positive and negative
parts of $f$, respectively. Note that $f=f^+-f^-$ and $|f|=f^++f^-$.

\begin{theorem}
If $f=g-h$ and $g\geq 0$, $h\geq 0$, then $f^+\leq g$ and $f^-\leq
h$.
\end{theorem}

\begin{exercise}
Show that the function $f\colon \IR \to \IR$ defined by
$$
   f (x) = \left\{ \begin{array}{cc}
   x & \text{if}\ \ x \in \IQ\\
   0 & \text{if} \ \ x \notin \IQ \end{array}\right .
$$
is a Borel function.
\end{exercise}

\begin{exercise}
Show that the function $f\colon \IR \to \IR$ defined by
$$
   f (x) = \left\{ \begin{array}{cc}
   \sin (1/x) & \text{if}\ \ x \neq 0 \\
   0 & \text{if} \ \ x =0 \end{array}\right .
$$
is a Borel function.
\end{exercise}

\begin{exercise}[Bonus]
Let $f\colon \IR \to \IR$ be a monotonically increasing function,
i.e.\ $f(x_1) \leq f(x_2)$ for $x_1 \leq x_2$. Show that $f$ is a
Borel function.
\end{exercise}

\begin{definition}
A function $f\colon X \to Y$ is a \textbf{simple function} iff its
range $f(X)$ is finite.
\end{definition}

If $s \colon X \to \IR$ is a measurable simple function whose
(distinct) values are $\alpha_1, \ldots, \alpha_n$, then it can be
represented by
\beq \label{simple}
  s = \sum_{i=1}^n \alpha_i \, \chi_{A_i}
\eeq
where $A_i = s^{-1} (\{\alpha_i\})$ are disjoint measurable sets
such that $X = A_1 \cup \cdots \cup A_n$.

\begin{theorem}
Let $(X,\fM)$ be a measurable space and $f\colon X \to [0,\infty]$ a
measurable function. Then there exist simple functions $s_n \colon X
\to \IR$ such that
$$
   0\leq s_1\leq s_2\leq\cdots \leq f
$$
and $s_n(x) \to f(x)$ as $n\to \infty$ for every $x \in X$.
\end{theorem}


\section{Lebesgue integration}

\begin{definition}
Let $(X,\fM,\mu)$ be a measure space and $s \colon X \to [0,
\infty)$ a nonnegative measurable simple function represented by
(\ref{simple}). Then we define the \textbf{Lebesgue integral}
$$
   \int_X s \, d\mu = \sum_{i=1}^n \alpha_i \, \mu(A_i)
$$
and for any $E \in \fM$ we define the \textbf{Lebesgue integral}
$$
   \int_E s \, d\mu = \sum_{i=1}^n \alpha_i \, \mu(A_i \cap E).
$$
It is possible that for some $i$ we have $\alpha_i =0$ and $\mu(A_i
\cap E)=\infty$, then we use the rule $0 \cdot \infty =0$.
\end{definition}

Note: $\int_E s\, d\mu \in [0,\infty]$ for any such $s$. If
$\mu(E)=0$, then $\int_E s\, d\mu =0$.

If $s\equiv c \geq 0$ is a constant function, then $\int_E c\, d\mu
= c\, \mu(E)$.

Example: $\int_X \chi_A\, d\mu = \mu(A)$. Note: $\int_E 0\, d\mu =
0$, even if $\mu(E)=\infty$.

\begin{definition}
Let $(X,\fM,\mu)$ be a measure space and $f \colon X \to [0,
\infty]$ a nonnegative measurable function (possibly taking value
$+\infty$). Then for any $E \in \fM$ we define the \textbf{Lebesgue
integral}
$$
   \int_E f \, d\mu = \sup_{s \in L_f} \int_E s\, d\mu,
$$
where
$$
   L_f = \big\{s\colon X \to [0,\infty)\ \ \text{measurable simple},
   \ \ 0\leq s \leq f\big\}.
$$
Note that $L_f \neq \emptyset$ as it always contains the function
$s\equiv 0$.
\end{definition}

For simple functions, the above two definitions are equivalent.

\begin{theorem}[Basic properties of the Lebesgue integral]
Let $f,g \colon X \to [0,\infty]$ be measurable functions and $A,B,E
\in \fM$ measurable sets. Then
\begin{itemize}
\item[\rm (a)] if $f \leq g$, then $\int_E f\, d\mu \leq \int_E g\,
d\mu$;
\item[\rm (b)] if $A \subset B$, then $\int_A f\, d\mu \leq \int_B f\,
d\mu$;
\item[\rm (c)] for any constant $c \geq 0$ we have $\int_A cf\, d\mu
= c\int_A f\, d\mu$;
\item[\rm (d)] if $\mu(E) = 0$, then $\int_E f\, d\mu
= 0$, even if $f \equiv \infty$;
\item[\rm (e)] $\int_E f\, d\mu = \int_X \chi_E f \, d\mu$.
\end{itemize}
\end{theorem}

\begin{exercise}
Let $x_0 \in X$ and $\mu = \delta_{x_0}$ the $\delta$-measure.
Assume that $\{x_0\} \in \fM$. Show that for every measurable
function $f \colon X \to [0, \infty]$ we have
$$
   \int_X f \, d\mu = f(x_0).
$$
\end{exercise}

\begin{exercise}
Let $X = \IN$ and $\mu$ the counting measure on the $\sigma$-algebra
$\fM = 2^{\IN}$. Show that for every function $f \colon X \to [0,
\infty]$ we have
$$
   \int_X f \, d\mu = \sum_{n=1}^\infty f(n).
$$
\end{exercise}


\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space and $s \colon X \to [0,
\infty)$ a nonnegative measurable simple function. Then
$$
   \varphi(E) = \int_E s\, d\mu
$$
is a measure on $\fM$.
\end{theorem}


\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space and $s,t \colon X \to [0,
\infty)$ two nonnegative measurable simple functions. Then
$$
   \int_X (s+t) \, d\mu = \int_X s \, d\mu + \int_X t \, d\mu.
$$
\end{theorem}

\begin{theorem}[Lebesgue's Monotone Convergence]
Let $(X,\fM,\mu)$ be a measure space and $\{f_n\}$ a sequence of
measurable functions on $X$. Suppose that\\
{\rm (a)} $0\leq f_1(x) \leq f_2(x) \leq \cdots \leq \infty$ for
every $x
\in X$;\\
{\rm (b)} $f_n(x) \to f(x)$ as $n \to \infty$ for every $x \in X$.\\
Then $f$ is measurable and
$$
   \int_X f_n \, d\mu \to \int_X f \, d\mu
   \qquad {\rm as}\ \ n \to \infty.
$$
\end{theorem}

This can be written as
$$
  \lim_{n\to\infty} \int_X f_n \, d\mu
  = \int_X \lim_{n\to\infty}f_n \, d\mu.
$$


\begin{exercise}
Let $f \colon X \to [0, \infty]$ and $\int_X f \, d\mu =0$. Show
that $\mu\{x\colon f(x)\neq 0\}=0$.
\end{exercise}

\begin{lemma0}
Let $(X,\fM,\mu)$ be a measure space and $f,g \colon X \to [0,
\infty]$ two nonnegative measurable functions. Then
$$
   \int_X (f+g) \, d\mu = \int_X f \, d\mu + \int_X g \, d\mu.
$$
\end{lemma0}

\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space and $f_n \colon X \to [0,
\infty]$ a sequence of nonnegative measurable functions on $X$ and
$$
   f(x) = \sum_{n=1}^\infty f_n(x)
   \quad {\rm for}\ {\rm every}\ \ x \in X.
$$
Then $f$ is measurable and
$$
   \int_X f\, d\mu = \sum_{n=1}^{\infty}
   \int_X f_n \, d\mu.
$$
\end{theorem}

That is, summation and integration ``commute'' for positive
functions.

\begin{corollary}
If $a_{ij} \geq 0$ for all $i,j =1,2,\ldots$, then
$$
   \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{ij} =
   \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_{ij}.
$$
\end{corollary}

\begin{theorem}[Fatou's Lemma]
Let $(X,\fM,\mu)$ be a measure space and $f_n \colon X \to [0,
\infty]$ a sequence of nonnegative measurable functions on $X$ .
Then
$$
   \int_X \Bigl(\liminf_{n\to\infty}
   f_n \Bigr)\, d\mu \leq \liminf_{n\to\infty}
   \int_X f_n \, d\mu.
$$
\end{theorem}

\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space and $f \colon X \to [0,
\infty]$ a nonnegative measurable function on $X$ . Then
$$
   \varphi(E) = \int_E f\, d\mu
$$
is a measure on $\fM$. Furthermore,
$$
   \int_X g\, d\varphi = \int_X gf\, d\mu
$$
for every nonnegative measurable function $g$ on $X$. {\rm [}We
write $d\varphi = f\, d\mu$.{\rm ]}
\end{theorem}


\begin{corollary}
Let $(X,\fM,\mu)$ be a measure space and $f \colon X \to [0,
\infty]$ a nonnegative measurable function on $X$ . Then for every
$A,B \in \fM$, $A\cap B = \emptyset$
$$
   \int_{A\cup B} f\, d\mu =
   \int_A f\, d\mu + \int_B f\, d\mu.
$$
\end{corollary}

\begin{exercise}
Let $E \subset X$ be such that $\mu(E)>0$ and $\mu(E^c)>0$. Put $f_n
=\chi_E$ if $n$ is odd and $f_n = 1-\chi_E$ if $n$ is even. What is
the relevance of this example to Fatou's lemma?
\end{exercise}

\begin{exercise}
Construct an example of a sequence of nonnegative measurable
functions $f_n \colon X \to [0,\infty)$ such that
$f(x)=\lim_{n\to\infty} f_n(x)$ exists pointwise, but
$$
   \int_X f\, d\mu < \liminf_{n\to\infty} \int_X f_n \, d\mu.
$$
\end{exercise}

Next we define Lebesgue integral for general complex-valued
functions on a measure space $(X,\fM,\mu)$.

\begin{definition}
We say that a measurable function $f\colon X\to\IC$ is
\textbf{Lebesgue integrable} if
$$
    \int_X |f|\, d\mu <\infty.
$$
The set of all integrable functions is denoted by $L^1(\mu)$.
\end{definition}

Note: $|f|$ is measurable due to Theorem~\ref{tm|f|}. The above
integral is defined since $|f| \geq 0$.

\begin{definition}
Let $f \in L^1(\mu)$ and $u=\,{\rm Re}\, f$ and $v=\,{\rm Im}\, f$
are its real and imaginary parts, so  that $f = u+\bi v$. Let
$u=u^+-u^-$ and $v=v^+-v^-$ be the decomposition of $u$ and $v$ into
their positive and negative parts. Then for any $E \in \fM$ we
define the \textbf{Lebesgue integral} by
$$
    \int_E f\, d\mu = \int_E u^+\,d\mu-\int_Eu^-\,d\mu
    + \bi \int_E v^+\,d\mu-\bi\int_Ev^-\,d\mu.
$$
\end{definition}

Note: $u^{\pm},v^{\pm} \geq 0$ and measurable by
Corollary~\ref{crf+-}, so all the integrals are defined.

Note: $|u^{\pm}|\leq |u|\leq|f|$ and $|v^{\pm}|\leq |v|\leq|f|$, so
all the integrals are finite, hence $\int_E f\, d\mu \in \IC$.

For a real-valued function $f$ we simply have $\int_E f\, d\mu =
\int_E f^+\,d\mu-\int_Ef^-\,d\mu$. This implies that
$$
  {\rm Re}\int_E f\,d\mu = \int_E {\rm Re}\, f \, d\mu
  \quad\text{and}\quad
  {\rm Im}\int_E f\,d\mu = \int_E {\rm Im}\, f \, d\mu.
$$

Note: if $A\cap B=\emptyset$, then $\int_{A\cup B}f\,d\mu =
\int_Af\, d\mu +\int_Bf\,d\mu$.

\begin{theorem}
Let $f,g \in L^1(\mu)$ and $\alpha,\beta \in \IC$. Then
$$
   \int_X (\alpha f+\beta g)\, d\mu = \alpha\int_Xf\,d\mu
   +\beta\int_Xg\,d\mu.
$$
\end{theorem}

\begin{exercise}
Let $f,g \in L^1(\mu)$ be real-valued functions and $f\leq g$. Show
that
$$
   \int_X f\, d\mu \leq \int_X g\, d\mu.
$$
\end{exercise}

\begin{exercise}
Let $f_n \colon X\to [0,\infty]$ be a sequence of measurable
functions such that $f_1 \geq f_2 \geq \cdots \geq 0$ and
$\lim_{n\to\infty} f_n(x) = f(x)$ for every $x\in X$. Suppose
$f_1\in L^1(\mu)$. Show that
$$
   \lim_{n\to\infty}
   \int_X f_n \, d\mu=\int_X f \, d\mu.
$$
\end{exercise}

\begin{theorem}
If $f \in L^1(\mu)$, then
$$
   \Bigl|\int_X f\, d\mu\Bigr|\leq \int_X|f|\,d\mu.
$$
\end{theorem}

\begin{theorem}[Lebesgue's Dominated Convergence]
Let $f_n \colon X\to \IC$ be a sequence of measurable functions such
that $f(x) = \lim_{n\to\infty} f_n(x)$ exists for every $x\in X$.
Suppose there exists $g\in L^1(\mu)$ such that
$$
   |f_n(x)|\leq g(x)\qquad
   \forall x\in X,\ \  \forall n=1,2,\ldots
$$
Then $f\in L^1(\mu)$,
$$
   \lim_{n\to\infty} \int_X|f_n-f|\,d\mu =0
$$
and
$$
   \lim_{n\to\infty} \int_X f_n \, d\mu=\int_X f \, d\mu.
$$
\end{theorem}

\begin{corollary}
If $f\in L^1(\mu)$ and $E_1\supset E_2\supset \cdots$ is a sequence
of measurable sets such that $\mu(E_n)\to 0$ as $n\to\infty$, then
$\int_{E_n} f\, d\mu \to 0$ as $n\to\infty$.
\end{corollary}

Next we discuss the role played by sets of measure zero (null sets).

\begin{definition}
Let $P$ be a property which a point $x \in X$ may or may not have.
We say that $P$ holds \textbf{almost everywhere (a.e.)} on a set $E
\subset X$ if there exists $N \subset X$, $\mu(N)=0$, such that $P$
holds at every $x \in E \setminus N$.
\end{definition}

Example: we say that $f=g$ a.e.\ if $\mu\{x\in X\colon f(x)\neq
g(x)\}=0$ (this is an equivalence relation). We say that a sequence
of functions $f_n$ converges a.e.\ to a limit function $f(x)$ if
$\mu\{x\in X\colon f_n(x) \nrightarrow f(x)\}=0$.

\begin{theorem}
If $f = g$ a.e., then $\int_E f\, d\mu = \int_E g\,d\mu$ for any $E
\in \fM$.
\end{theorem}

\begin{definition}
Let $(X,\fM,\mu)$ be a measure space and $Y$ a topological space. We
say that a function $f$ with values in $Y$ is \textbf{measurable and
a.e.\ defined} on $X$ if there exists $N\subset X$, $\mu(N)=0$, such
that $f$ is defined on $X \setminus N$ and for every open set $V
\subset Y$ we have $f^{-1}(V) \setminus N \in \fM$.
\end{definition}

\begin{exercise}
Let $f$ be a function as above. Fix a $y\in Y$ and define
$$
   \tilde{f}(x) = \left\{ \begin{array}{cc}
   f(x) & \text{if}\ \ x \in X\setminus N\\
   y & \text{if} \ \ x \in N \end{array}\right ..
$$
Show that $\tilde{f} \colon X \to Y$ is measurable.
\end{exercise}

\begin{exercise}
Let $f$ be a function as above and $\mu$ a complete measure. Show
that
$$
   \tilde{f}(x) = \left\{ \begin{array}{cc}
   f(x) & \text{if}\ \ x \in X\setminus N\\
   \text{any point of }Y & \text{if} \ \ x \in N \end{array}\right ..
$$
Show that $\tilde{f} \colon X \to Y$ is measurable.
\end{exercise}

\begin{definition}
A sequence $f_n$ of functions $X \to \IC$ converges to a function
$f\colon X\to \IC$ \textbf{uniformly on} $X$ if for any
$\varepsilon>0$ there exists $N>0$ such that for all $n>N$
$$
   \sup_{x\in X}|f_n(x)-f(x)|<\varepsilon.
$$
Equivalently, $ \sup_{x\in X}|f_n(x)-f(x)|\to 0$ as $n\to \infty$.
\end{definition}

\begin{exercise}
Suppose $\mu(X) < \infty$. Let a sequence $\{f_n\}$ of bounded
complex measurable functions uniformly converge to $f$ on $X$. Prove
that $\lim_{n\to\infty} \int_Xf_n\,d\mu = \int_Xf\,d\mu$. Show that
the assumption $\mu(X) < \infty$ cannot be omitted.
\end{exercise}

\begin{theorem}
Let $\{f_n\}$ be a sequence of complex measurable functions defined
a.e.\ on $X$ and
$$
   \sum_{n=1}^{\infty} \int_X |f_n|\, d\mu < \infty.
$$
Then the series $f(x) =\sum_{n=1}^{\infty} f_n$ converges for a.e.\
$x\in X$ and
$$
   \sum_{n=1}^{\infty} \int_X f_n\, d\mu = \int_X f\, d\mu.
$$
\end{theorem}

\begin{exercise}
Give an example of a sequence of complex measurable functions
$f_n\colon X \to \IC$ (i.e., defined on the entire $X$) such that
$$
   \sum_{n=1}^{\infty} \int_X |f_n|\, d\mu < \infty
$$
but the series $f(x) =\sum_{n=1}^{\infty} f_n$ diverges for some
$x\in X$.
\end{exercise}

\begin{definition}
A sequence $f_n$ of functions $X \to \IC$ converges to a function
$f\colon X\to \IC$ \textbf{in measure} if for any $\varepsilon>0$
and $\delta>0$ there exists $N>0$ such that for all $n>N$
$$
   \mu\{x\in X\colon |f_n(x)-f(x)|>\varepsilon\}<\delta.
$$
Equivalently, $\mu\{x\in X\colon |f_n(x)-f(x)|>\varepsilon\}\to 0$
as $n\to \infty$.
\end{definition}

\begin{exercise}
Suppose $\mu(X) < \infty$ and $f_n$ are measurable functions defined
a.e.\ on $X$. Prove that if $f_n \to f$ a.e.\ on $X$, then $f_n\to
f$ in measure. What happens if $\mu(X) = \infty$?
\end{exercise}

\begin{theorem}
Let $(X,\fM,\mu)$ be a measure space. \begin{itemize} \item[\rm (a)]
If $f\colon X \to [0,\infty]$ is measurable, $E\in \fM$, and $\int_E
f\, d\mu=0$, then $f=0$ a.e.\ on $E$; \item[\rm (b)] If $f\in
L^1(\mu)$ and $\int_E f\,d\mu=0$ for every $E \in \fM$, then $f=0$
a.e.\ on $X$; \item[\rm (c)] If $f\in L^1(\mu)$ and
$$
   \Bigl|\int_X f\, d\mu\Bigr| = \int_X|f|\,d\mu,
$$
then there exists a constant $c\in \IC$ such that $cf=|f|$ a.e.\ on
$X$.
\end{itemize}
\end{theorem}

\begin{theorem}
If $\mu(X) < \infty$, $f\in L^1(\mu)$, $S\subset \IC$ closed, and
for every $E \in \fM$ with $\mu(E)>0$ we have $A_E(f) \in S$, where
$$
   A_E(f)=\frac{1}{\mu(E)}\int_E f\, d\mu
$$
is the average value of $f$ on the set $E$, then $f(x) \in S$ a.e.\
on $X$.
\end{theorem}

\begin{corollary}
In particular, if $\int_E f\, d\mu$ is real-valued for every $E$,
then $f$ is real-valued a.e.\ on $X$.
\end{corollary}

\begin{theorem}
If $E_k \in \fM$ and $\sum_{k=1}^{\infty} \mu(E_k)<\infty$, then
almost every $x \in X$ belongs to finitely many of the sets $E_k$.
\end{theorem}

\begin{exercise}
In the above theorem, let $A$ be the set of points which belong to
infinitely many of the sets $E_k$. Show that
$$
     A = \cap_{n=1}^{\infty} \cup_{k=n}^{\infty} E_k.
$$
Use this fact to prove the theorem without any reference to
integration. Hint: recall that every measure $\mu$ is countably
subbaditive, i.e.\ for any $E_n \in \fM$ (not necessarily disjoint)
$\mu(\cup E_n) \leq \sum \mu(E_n)$.
\end{exercise}

\begin{exercise}
Prove that if $f_n \to f$ in measure, then there is a subsequence
$\{f_{n_k}\}$ of $\{f_n\}$ such that $f_{n_k}\to f$ a.e. on $X$.
Hint: use the previous theorem.
\end{exercise}

\begin{exercise}
Suppose $f\in L^1(\mu)$. Prove that $\forall \varepsilon>0$ $\exists
\delta>0$ such that $\int_E|f|\, d\mu<\varepsilon$ whenever
$\mu(E)<\delta$. \label{Einteps}
\end{exercise}

\section{Regular measures in $\IR^k$}

In Section~\ref{SecLMR2} we constructed the Lebesgue measure on
$\IR^2$ (as well as on $\IR$, and our construction easily extends to
$\IR^k$ for $k \geq 3$). Here we continue the study of the Lebesgue
measure on $\IR^k$.

\begin{definition}
A measure $\mu$ defined on a $\sigma$-algebra $\fM$ in $\IR^k$ is
called a Borel measure if it is defined on all Borel sets (i.e.\
$\fM$ contains all Borel sets).
\end{definition}

A measure $\mu$ is Borel if $\fM$ contains all open intervals.

\begin{definition}
A Borel measure $\mu$ is said to be \textbf{outer regular} if
$$
   \mu(E) = \inf\{\mu(V)\colon E\subset V,\ \ V\text{ open}\}
   \quad \forall E\in\fM.
$$
A Borel measure $\mu$ is said to be \textbf{inner regular} if
$$
   \mu(E) = \sup\{\mu(K)\colon K\subset E,\ \ K\text{ compact}\}
   \quad \forall E\in\fM.
$$
A Borel measure $\mu$ is said to be \textbf{regular} if it is both
outer regular and inner regular.
\end{definition}

\begin{theorem}
If $\mu(E)<\infty$ for any bounded set $E \subset \IR^k$, then the
outer regularity and inner regularity are equivalent.
\end{theorem}

\begin{exercise}
Show that the counting measure on $\IR^k$ is inner regular, but not
outer regular.
\end{exercise}

\begin{theorem}
The Lebesgue measure on $\IR^k$ is regular.
\end{theorem}

\begin{exercise}
Find examples of Lebesgue measurable sets $E_1,E_2 \subset \IR$ such
that
$$
   \bm(E_1) < \inf\{\bm(A)\colon E\subset A,\ \ A\text{ closed}\}
$$
$$
   \bm(E_2) > \sup\{\bm(V)\colon V\subset E,\ \ V\text{ open}\}.
$$
where $\bm$ is the Lebesgue measure.
\end{exercise}

\begin{theorem}
Let $\mu$ be a regular Borel measure. Then for every $E \in \fM$
\begin{itemize} \item[\rm (a)] there exists a $G_{\delta}$-set $G$ such
that $E \subset G$ and $\mu(E) = \mu(G)$; \item[\rm (b)] there
exists an $F_{\sigma}$-set $F$ such that $F \subset E$ and $\mu(E) =
\mu(F)$.
\end{itemize}
\end{theorem}

So every measurable set is `almost' $G_\delta$ (and `almost'
$F_\sigma$), up to a null set.

\begin{theorem}
There exists a unique measure $\mu$ defined on the Borel
$\sigma$-algebra in $\IR$ such that $\mu(I) = |I|$ (the length of
$I$) for every open interval $I \subset \IR$.
\end{theorem}

This theorem can be extended to $\IR^2$ as follows: there exists a
unique measure $\mu$ defined on the Borel $\sigma$-algebra in
$\IR^2$ such that $\mu(R) = m(R)$ (the area of $R$) for every open
rectangle $R \subset \IR^2$. But the proof becomes more complicated.

Next recall that the Lebesgue measure $\bm$ on $\IR^2$ is complete.

\begin{theorem}
If $\mu$ is another complete measure on $\IR^2$ defined (at least)
on all Borel sets such that $\mu (R) = \Area(R)$ for every rectangle
$R$, then $\mu(E) = \bm(E)$ for every Lebesgue measurable set $E$.
\end{theorem}

Recall that a measure $\mu$ on $\IR$ (or $\IR^2$) is called
translation invariant if $\mu(E) = \mu(E+a)$ for every measurable
set $E$ and $a\in\IR$ (respectively, $a\in \IR^2$).

\begin{theorem}
Let $\mu$ be a translation invariant Borel measure on $\IR$ such
that $\mu(I) <\infty$ for at least one open interval
$I\neq\emptyset$. Then there exists a constant $c$ such that $\mu(E)
= c\bm(E)$ for every Borel set $E$; here $\bm$ is the Lebesgue
measure.
\end{theorem}

\begin{exercise}
Extend this theorem to $\IR^2$: show that if $\mu$ is a translation
invariant Borel measure on $\IR^2$ such that $\mu(R) <\infty$ for at
least one open rectangle $R\neq\emptyset$, then there exists a
constant $c$ such that $\mu(E) = c\bm(E)$ for every Borel set $E$.
\end{exercise}

\begin{exercise}
Let $f \colon \IR \to [0,\infty]$ be a Borel function, $f \in
L^1(\bm)$, and consider the measure $\rho(E) = \int_E f\, d\bm$,
where $\bm$ is the Lebesgue measure. Prove that $\rho$ is regular.
Hint: use the result of Exercise~\ref{Einteps}.

\end{exercise}

\section{Approximation of Lebesgue measurable functions}

In the previous section we have seen that Lebesgue measurable sets
can be arbitrarily well approximated by open sets (`from outside')
and by compact sets (`from inside'). Here we approximate Lebesgue
measurable functions by step functions and by continuous functions.

\begin{definition}
A function $\varphi \colon \IR \to \IC$ is called a \textbf{step
function} if $\varphi = \sum_{i=1}^n \alpha_i \chi_{I_i}$ for some
$\alpha_i \in \IC$ and disjoint bounded intervals $I_i \in \IR$.
\end{definition}

\begin{exercise}
Let $s \colon [a,b] \to \IR$ be a simple Lebesgue measurable
function. Show that for every $\varepsilon>0$ there is a step
function $\varphi \colon [a,b] \to \IR$ and a Lebesgue measurable
set $E \subset [a,b]$ such that $s(x) = \varphi(x)$ on $E$ and $\mu
\bigl( [a,b]\setminus E \bigr) <\varepsilon$. Hint: use the
regularity of $\mu$.
\end{exercise}

\begin{exercise}
Let $f \colon [a,b] \to \IR$ be a Lebesgue measurable function. Show
that for every $\varepsilon>0$ there is a step function $g \colon
[a,b] \to \IR$ such that
$$
   \mu \bigl\{ x\in [a,b]\colon |f(x)-g(x)|\geq\varepsilon \bigr\}
    <\varepsilon.
$$
Hint: first show that $|f|\leq M$ except for a set of small
measure,  then use the previous exercise.
\end{exercise}

\begin{exercise}
Let $f \in L^1$ with respect to the Lebesgue measure $\bm$ on $\IR$.
Prove that there is a sequence $\{g_n\}$ of step functions such that
$$
   \lim_{n\to\infty} \int_{\IR}|f-g_n|\,d\bm = 0.
$$
Hint: use the previous exercise.
\end{exercise}

\begin{definition}
Let $X$ be a topological space and $f \colon X \to \IC$. The
\textbf{support} of $f$ is defined by
$$
  {\rm supp}\, f = \overline{\{x\in X\colon f(x)\neq 0\}}.
$$
Also, $C_c(X)$ denotes the set of all continuous functions with
compact support.
\end{definition}

\begin{theorem}[Lusin]
Let $f \colon \IR \to \IC$ be a Lebesgue measurable function, $A
\subset \IR$ a Lebesgue measurable set with $\mu(A)<\infty$ (here
$\mu$ is the Lebesgue measure), such that $f(x)=0$ for all $x\notin
A$. Then for every $\varepsilon>0$ there exists $g \in C_c(\IR)$
such that
$$
  \mu\bigl\{x\colon f(x)\neq g(x)\bigr\}<\varepsilon.
$$
Furthermore, we may arrange it so that $\sup|g(x)|\leq \sup|f(x)|$.
\end{theorem}

\begin{corollary}
Assume that the hypotheses of Lusin's theorem are satisfied and that
$|f|\leq M$. Then there is a sequence $\{g_n\}$ such that $g_n\in
C_c(\IR)$, $|g_n|\leq M$, and $g_n(x) \to f(x)$ a.e.
\end{corollary}

\begin{definition}
Let $X$ be a topological space and $f\colon X \to \IR$ (or
$[-\infty,\infty]$). We say that $f$ is \textbf{lower
semicontinuous} if $\{x\colon f(x)>a\}$ is open for every $a\in\IR$.
We say that $f$ is \textbf{upper semicontinuous} if $\{x\colon
f(x)<a\}$ is open for every $a\in\IR$.
\end{definition}

\begin{proposition}
Simple properties of semicontinuous functions:
\begin{itemize}
\item[\rm (a)] $f$ is continuous $\Leftrightarrow$ $f$ is both upper and lower
semicontinuous; \item[\rm (b)] if $f$ is semicontinuous, then it is
Borel; \item[\rm (c)] $f$ is upper semicontinuous $\Leftrightarrow$
$-f$ is lower semicontinuous; \item[\rm (d)] if $f$ is upper (lower)
semicontinuous and $c>0$, then $cf$ is upper (lower) semicontinuous;
\item[\rm (e)] if $f,g$ are upper (lower) semicontinuous, then $f+g$
is upper (lower) semicontinuous; \item[\rm (f)] if $\{f_\gamma\}$ is
a family of upper (lower) semicontinuous functions, then $\inf
f_\gamma$ (resp., $\sup f_\gamma$)is upper (lower) semicontinuous;
\item[\rm (g)] $V \subset X$ is open $\Leftrightarrow$ $\chi_V$ is
lower semicontinuous; \item[\rm (h)] $A \subset X$ is closed
$\Leftrightarrow$ $\chi_A$ is upper semicontinuous; \item[\rm (i)]
if $f_n\geq 0$ are lower semicontinuous, then $\sum f_n$ is lower
semicontinuous.
\end{itemize}
\end{proposition}

\begin{exercise}
Prove that if $f\colon X \to \IR$ is upper (lower) semicontinuous
and $X$ is compact, then $f$ is bounded above (below) and attains
its maximum (minimum).
\end{exercise}

\begin{theorem}[Vitali-Caratheodory]
Let $f \colon \IR \to \IR$ be Lebesgue integrable function (i.e.\ $f
\in L^1$). Then for every $\varepsilon>0$ there exist functions $u,v
\colon \IR \to \IR$, $u \leq f \leq v$, where $u$ is upper
semicontinuous and bounded above, $v$ is lower semicontinuous and
bounded below, and
$$
  \int_{\IR} (v-u)\, d\mu < \varepsilon.
$$
\end{theorem}

\begin{theorem}
Let $f \colon \IR \to \IR$ be Lebesgue measurable function. Then
there exist Borel functions $g,h \colon \IR \to \IR$ such that $g(x)
= h(x)$ a.e.\ and $g(x) \leq f(x) \leq h(x)$ for every $x\in X$.
\end{theorem}

\begin{theorem}[Egorov]
Let $\mu(X)<\infty$ and $\{f_n\}$ a sequence of complex measurable
functions on $X$ such that $\lim_{n\to \infty} f_n(x) = f(x)$ for
every $x \in X$. Then for every $\varepsilon>0$ there is $E \subset
X$ such that $\mu(X\setminus E) <\varepsilon$ and $f_n\to f$
uniformly on $E$.
\end{theorem}

\section{Lebesgue integral versus Riemannian integral}

\begin{theorem}
Let $[a,b] \subset \IR$ be bounded. If the Riemannian integral
$$
   I = \int_a^b f(x)\, dx
$$
exists, then $f$ is Lebesgue integrable on $[a,b]$ and its Lebesgue
integral equals
$$
   \int_{[a,b]} f \, d\bm = I.
$$
\end{theorem}

\begin{theorem}
The Riemannian integral
$$
   \int_a^b f(x)\, dx
$$
exists if and only if $f$ is bounded and almost everywhere
continuous.
\end{theorem}

Note: the existence of improper Riemannian integrals do not imply
Lebesgue integrability. More precisely, if $f$ has a finite improper
Riemannian integral, then it is Lebesgue integrable if and only if
$|f|$ has a finite improper Riemannian integral. In that case the
Lebesgue integral is equal to the improper Riemannian integral.
(Note: if $f\geq 0$, then the existence of improper Riemannian
integral is equivalent to Lebesgue integrability.)

\begin{exercise}
Find a function $f(x)$ on $[0,\infty)$ such that the improper
Riemannian integral $\int_0^{\infty} f(x)\, dx = \lim_{A\to\infty}
\int_0^A f(x) \, dx$ exists (is finite), but $f$ is not Lebesgue
integrable.
\end{exercise}

\section{$L^p$ spaces}

\begin{definition}
A function $\varphi \colon (a,b) \to \IR$, where $-\infty \leq a < b
\leq \infty$ is said to be \textbf{convex} if
$$
  \varphi( px+qy)
  \leq p\varphi(x)+q\varphi(y)
$$
for all $a<x\leq y<b$ and $p,q>0$, $p+q=1$.
\end{definition}

\begin{lemma}
A function $f \colon (a,b) \to \IR$ is convex if and only if for any
$a<s<t<u<b$
$$
   \frac{\varphi(t)-\varphi(s)}{t-s} \leq
   \frac{\varphi(u)-\varphi(t)}{u-t}.
$$
\end{lemma}

\begin{lemma}
If $\varphi$ is differentiable, then it is convex if and only if
$\varphi'$ is monotonically increasing, i.e.\ $\varphi'(x) \leq
\varphi'(y)$ for all $x<y$.
\end{lemma}

\begin{theorem}
If $\varphi$ is convex on $(a,b)$, then it is continuous on $(a,b)$.
\end{theorem}

Note: this is not true on closed intervals $[a,b]$.

\begin{theorem}[Jensen's inequality]
Let $(\Omega,\fM,\mu)$ be a measure space and $\mu(\Omega)=1$. If
$f\colon \Omega \to (a,b) \subset \IR$ is Lebesgue integrable ($f
\in L^1_{\mu}$) and $\varphi$ is a convex function on $(a,b)$, then
$$
   \varphi\Bigl(\int_{\Omega}f\, d\mu\Bigr)
   \leq \int_{\Omega} (\varphi\circ f)\, d\mu.
$$
\end{theorem}

Note: the cases $a=-\infty$ and $b=\infty$ are not excluded.

\begin{exercise}
Prove that the supremum of any collection of convex functions on
$(a,b)$ is convex on $(a,b)$ (assume that the supremum is finite).
Prove that a pointwise limit of a sequence of convex functions is
convex.
\end{exercise}

\begin{exercise}
Let $\varphi$ be convex on $(a,b)$ and $\psi$ convex and
nondecreasing on the range of $\varphi$. Prove that $\psi \circ
\varphi$ is convex on $(a,b)$. For $\varphi > 0$, show that the
convexity of $\log \varphi$ implies the convexity of $\varphi$, but
not vice versa.
\end{exercise}

\begin{exercise}
Assume that $\varphi$ is a continuous real function on $(a,b)$ such
that
$$
 \varphi\Bigl(\frac{x+y}{2}\Bigr) \leq \frac{1}{2}\,\varphi(x)+
 \frac{1}{2}\,\varphi(y)
$$
for all $x,y \in (a,b)$. Prove that $\varphi$ is convex. (The
conclusion does not follow if continuity is omitted from the
hypotheses.)
\end{exercise}

Example: let $\Omega = \{1,2,\ldots,n\}$ and $\mu(\{i\})=1/n$ for
every $i=1,\ldots,n$; let $f(i) = \ln a_i$ for some $a_1, \ldots,
a_n>0$ and $\varphi(x)=e^x$ (this is a convex function!). Then
Jensen's inequality implies
$$
  (a_1a_2\cdots a_n)^{1/n} \leq \frac{a_1+a_2+\cdots+a_n}{n}
$$
(geometric mean is $\leq$ arithmetic mean). More general: if
$\mu(\{i\})=p_i$ and $p_1+\cdots+p_n=1$, then
$$
  a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n} \leq
  p_1a_1+p_2a_2+\cdots+p_na_n
$$

\begin{definition}
If $p,q>0$ and
$$
  \frac 1p + \frac 1q = 1,
$$
then we say that $p$ and $q$ are \textbf{conjugate exponents}.
\end{definition}

Note: we must have $p,q>1$. If $p\to 1$, then $q\to\infty$; thus
$p=1$ and $q=\infty$ are sometimes considered as conjugate
exponents, too.

An important special case is $p=q=2$.

\begin{theorem}
Let $p$ and $q$ be conjugate exponents, $(X,\fM,\mu)$ a measure
space, and $f,g \colon X \to [0,\infty]$ measurable functions. Then
$$
  \int_X fg\, d\mu \leq \biggl[\int_X f^p\,d\mu\biggr]^{1/p}
  \biggl[\int_X g^q\,d\mu\biggr]^{1/q}
$$
{\rm (\textbf{H\"older inequality}, in the case $p=q=2$ it is
\textbf{Schwarz inequality})} and
$$
  \biggl[\int_X (f+g)^p\, d\mu\biggr]^{1/p} \leq
  \biggl[\int_X f^p\,d\mu\biggr]^{1/p}+
  \biggl[\int_X g^p\,d\mu\biggr]^{1/p}
$$
{\rm (\textbf{Minkowski inequality})}.
\end{theorem}

\begin{remark}
Equality in the H\"older inequality holds if and only if there exist
$\alpha,\beta\geq 0$, not both equal to 0, such that $\alpha f^p =
\beta g^q$ a.e.
\end{remark}

\begin{exercise}
Prove that equality in the Minkowski inequality holds if and only if
there exist $\alpha,\beta\geq 0$, not both equal to 0, such that
$\alpha f = \beta g$ a.e.
\end{exercise}

\begin{exercise}
Suppose $\mu(\Omega)=1$ and suppose $f$ and $g$ are two positive
measurable functions on $\Omega$ such that $fg\geq 1$. Prove that
$$
 \int_\Omega f \, d\mu \cdot \int_\Omega g \, d\mu\geq 1.
$$
\end{exercise}


\begin{exercise}
Suppose $\mu(\Omega)=1$ and $h\colon \Omega\to [0, \infty]$ is
measurable. Denote $A = \int_\Omega h \, d\mu$. Prove that
$$
  \sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}\, d\mu \leq 1+A.
$$
\end{exercise}

\begin{exercise}[Bonus]
If $\bm$ is Lebesgue measure on $[0,1]$ and if $h$ is a continuous
function on $[0,1]$ such that $h=f'$, then the inequalities in the
previous exercise have a simple geometric interpretation. From this,
conjecture (for general $\Omega$) under what conditions on $h$
equality can hold in either of the above inequalities, and prove
your conjecture.
\end{exercise}

Recall that $L^1_\mu(X) =\{ f\colon \int_X |f|\, d\mu <\infty\}$.

\begin{definition}
For a measurable function $f \colon X \to \IC$ and $p >0$ we define
its \textbf{``p-norm''} by
$$
   \|f\|_p = \Bigl[ \int_X|f|^p \, d\mu\Bigr]^{1/p}.
$$
We also define the space of p-integrable functions by
$$
  L^p_{\mu}(X) =\bigl\{ f\colon  \|f\|_p <\infty\bigr\}
$$
\end{definition}

If $\mu$ is the Lebesgue measure on $\IR^k$, we will simply write
$L^p(\IR^k)$.

\begin{definition}
A special case: $\mu$ is a counting measure on a countable set $A$.
Then we identify $A$ with $\IN$ and functions $f\colon A \to \IC$
with complex sequences $\xi = \{\xi_n\}$. Then
$$
   \|\xi\|_p = \biggl[\sum_{n=1}^{\infty}|\xi_n|^p\biggr]^{1/p}
$$
and
$$
  \ell^p = \bigl\{ \xi\colon  \|\xi\|_p <\infty\bigr\}
$$
\end{definition}

\begin{proposition}
If $1\leq p <\infty$, then $L^p_\mu$ is a complex vector space.
\end{proposition}

Next we investigate the case $p=\infty$. Let $g \colon X \to
[0,\infty]$ be measurable. The set
$$
  S = \bigl\{\alpha\in \IR\colon \mu\bigl(g^{-1}(
  \alpha,\infty]\bigr)=0\bigr\}
$$
is either empty or a closed interval $[\beta,\infty)$. If
$S=\emptyset$ we set $\beta = \infty$.

\begin{definition}
The value of $\beta$ is called the \textbf{essential supremum} and
denoted by
$$
   \text{ess-sup} \, g  = \beta.
$$
The \textbf{infinity-norm} of a function $f\colon X\to \IC$ is
defined by
$$
 \|f\|_\infty = \, \text{ess-sup} \, |f|.
$$
We also denote by
$$
  L^\infty_\mu(X)=\bigl\{ f\colon  \|f\|_\infty <\infty\bigr\}
$$
the space of \textbf{essentially bounded functions}.
\end{definition}

Note: $|f(x)|\leq \|f\|_\infty$ a.e. A special case: $\ell^\infty$
is the space of bounded sequences.

\begin{proposition}
$L^\infty_\mu$ is a complex vector space.
\end{proposition}

\begin{theorem}
Let $p$ and $q$ be complex conjugate exponents, including the
limiting cases $p=1$, $q=\infty$ and $p=\infty$, $q=1$. Then for
every $f \in L^p$ and $g\in L^q$ we have $fg\in L^1$ and
$$
   \|fq\|_1 \leq \|f\|_p \, \|g\|_q.
$$
\end{theorem}

\begin{exercise}
When does one get equality in $\|fg\|_1 \leq \|f\|_\infty \,
\|g\|_1$?
\end{exercise}

\begin{theorem}
Let $1\leq p \leq \infty$. Then for every $f,g \in L^p$ we have
$f+g\in L^q$ and
$$
   \|f+q\|_p \leq \|f\|_p + \|g\|_p.
$$
We also have
$$
   \|cf\|_p=|c| \, \|f\|_p.
$$
\end{theorem}

Nonetheless, $\|\cdot\|_p$ is not a norm as there are nonzero
functions $f$ with $\|f\|_p=0$. But we will `reduce' the $L^p$ space
so that $\|\cdot\|_p$ will be a true norm.

Let $V$ be a complex vector space and $W \subset V$ a subspace. We
say that $v_1 \sim v_2$ iff $v_1-v_2 \in W$. This is an equivalence
relation. Denote by
$$
  [v] = \{v_1\in V\colon v-v_1\in W\}
$$
the equivalence class containing $v$ and
$$
   V/W = \{[v]\colon v\in V\}
$$
the set of equivalence classes. Then $V/W$ is a vector space with
$$
   [v]+[w]=[v+w],\qquad c[v]=[cv].
$$
We call $V/W$ the quotient space.

\begin{definition}
Let $(X,\fM,\mu)$ be a measure space and
$$
  \cN_\mu=\{f\colon X\to\IC\ \ \text{measurable},
  \ \ f=0\ \ \text{a.e.}\}
$$
For every $1\leq p\leq\infty$ define
$$
    \cL^p_\mu = L^p_\mu/\cN_\mu.
$$
\end{definition}

Note: $\cL^p_\mu$ is obtained from $L^p_\mu$ by identifying
functions that coincide a.e.

\begin{theorem}
Let $1\leq p \leq \infty$. Then $\cL^p_\mu$ is a vector space with
norm $\|\cdot\|_p$.
\end{theorem}

\begin{exercise}
Suppose $f\colon X\to \IC$ is measurable and $\|f\|_\infty>0$.
Define
$$
  \varphi(p)=\int_X |f|^p \, d\mu = \|f\|_p^p \qquad (0< p < \infty)
$$
and consider the set $E=\{p\colon \varphi(p) < \infty\}$. Each of
the following questions is graded as a separate exercise. Question
(c) and (e) are \textbf{bonus} problems.
\begin{itemize} \item[(a)] Let $r < p < s$ and $r,s\in E$. Prove that
$p \in E $. \item[(b)]Prove that $\log\varphi$ is convex in the
interior of $E$ and that $\varphi$ is continuous on $E$.
\item[(c)] By (a), $E$ is connected. Is $E$ necessarily open?
Closed? Can $E$ consist of a single point? Can $E$ be any connected
subset of $(0, \infty)$? \item[(d)] If $r < p < s$, prove that
$\|f\|_p \leq \max\bigl(\|f\|_r, \|f\|_s\bigr)$. Show that this
implies the inclusion $L^r_\mu \cap L^s_\mu \subset L^p_\mu$.
\item[(e)] Assume that $\|f\|_r < \infty$ for some $r < \infty$ and
prove that $\|f\|_p \to \|f\|_\infty$ as $p \to \infty$.
\end{itemize}
\end{exercise}

Recall that if $V$ is a vector space with a norm $\|\cdot\|$, we can
define distance (metric) on $V$ by
$$
   d(u,v)=\|u-v\|.
$$
Thus the space $\cL^p_\mu$ with norm $\|\cdot\|_p$ becomes a metric
space.

\begin{definition}
A metric is \textbf{complete} if every Cauchy sequence converges to
a limit. A vector space $V$ with norm $\|\cdot\|$ that induces a
complete metric on it is called \textbf{Banach space}.
\end{definition}

\begin{theorem}
Let $1\leq p \leq \infty$. Then $\cL^p_\mu$ is a Banach space.
\end{theorem}

\begin{corollary}
Let $1\leq p \leq \infty$. If $f_n\to f$ in the $L^p_\mu$ metric,
then there is a subsequence $\{f_{n_k}\}$ such that $f_{n_k} \to f$
a.e.
\end{corollary}

\begin{exercise}
Let $\mu(X) = 1$. Each of the following questions is graded as a
separate exercise.\begin{itemize} \item[(a)] Prove that $\|f\|_r
\leq \|f\|_s$ if $0 < r < s \leq \infty$. \item[(b)] Under what
condition does it happen that $0 < r < s \leq \infty$ and $\|f\|_r =
\|f\|_s < \infty$? \item[(c)] Prove that $L^r_\mu \supset L^s_\mu$
if $0<r<s$. If $X=[0,1]$ and $\bm$ is the Lebesgue measure, show
that $L^r_\bm \neq L^s_\bm$.
\end{itemize}
\end{exercise}

\begin{exercise}[Bonus]
For some measures, the relation $r < s$ implies $L^r(\mu) \subset
L^s(\mu)$; for others, the inclusion is reversed; and there are some
for which $L^r(\mu)$ does not contain $L^s(\mu)$ if $r \ne s$. Give
examples of these situations, and find conditions on $\mu$ under
which these situations will occur.
\end{exercise}

\begin{exercise} (a) Show that
$\int_0^{\frac{\pi}{2}}\sqrt{x\sin x}\, dx < \frac{\pi}{2\sqrt{2}}$;\\
${\ \ \ \ \ \ \ }$\ \ \ \ \ \ \ \ \ \ \ \ \ (b) Show that
$\bigl[\int_0^1 x^{\frac{1}{2}}(1-x)^{-\frac{1}{3}}\, dx\bigr]^3
\leq \frac{8}{5}$.
\end{exercise}

Next we approximate $L^p$ functions by simple functions and
continuous functions. Let
$$
  S=\bigl\{s\colon X\to\IC\ \ \text{simple, measurable},\ \
  \mu(s\neq 0)<\infty\bigr\}.
$$

\begin{theorem}
$S$ is dense in $L^p$ for every $1\leq p< \infty$.
\end{theorem}

Note: this is not true for $p=\infty$ (for example, $X=\IR$ and
$f\equiv 1$).

Recall that $C_c(\IR)$ denotes the space of continuous functions
$f\colon \IR\to \IC$ with compact support. And again, in $\IR$ we
use the Lebesgue measure $\bm$.

\begin{theorem}
$C_c$ is dense in $L^p_{\mu}$ for every $1\leq p< \infty$.
\end{theorem}

Note: this is not true for $p=\infty$ (for example, $X=\IR$ and
$f\equiv 1$).

\begin{remark}
Step functions $f\colon \IR \to \IC$ are dense in $L^p$.
\end{remark}

We can say that the $L^p$ space, for $1<p<\infty$, is the completion
(in the $p$-metric) of the space of continuous functions. This is
not true for $p=\infty$.

\begin{definition}
A function $f\colon \IR\to\IC$ is said to \textbf{vanish at
infinity} if $f(x) \to 0$ as $|x|\to \infty$. The space of
continuous functions vanishing at infinity is denoted by $C_0(\IR)$.
\end{definition}

Note that $C_c(\IR) \subset C_0(\IR)$.

\begin{theorem}
The completion of $C_c(\IR)$ in the $\|\cdot\|_\infty$-metric is
$C_0(\IR)$ (modulo the identification of equivalent functions).
\end{theorem}

\begin{corollary}
$C_0(\IR)$ with norm $\|\cdot\|_\infty$ is a Banach space.
\end{corollary}

\begin{exercise}[Bonus]
Suppose $1 < p < \infty$ and $f\in L^p((0, \infty))$ relative to the
Lebesgue measure. Define
$$
     F(x)=\frac{1}{x}\int_0^x
     f(t)\, dt \ \ \ \ \ \ (0 < x < \infty).
$$
Prove Hardy's inequality
$$
    \|F\|_p \leq \frac{p}{p-1}\|f\|_p
$$
which shows that the mapping $f\to F$ carries $L^p$ into $L^p$.
[Hint: assume first that $f\geq 0$ and $f\in C_c((0,\infty))$, i.e.\
the support of $f$ is a finite closed interval $[a,b] \subset
(0,\infty)$. Then use integration by parts:
$$
   \int_{\varepsilon}^A F^p(x)\, dx = -p\int_{\varepsilon}^A
   F^{p-1}xF'(x)\, dx
$$
where $\varepsilon<a$ and $A>b$. Note that $xF'=f-F$, and apply
H\"older inequality to $\int F^{p-1} f\, dx$.]
\end{exercise}

\section{Complex measures}


Given two measures $\mu_1$ and $\mu_2$ on $(X,\fM)$, we can define a
new measure $\mu = \mu_1+\mu_2$ by $\mu(A) = \mu_1(A) + \mu_2(A)$
for all $A \in \fM$ (it is easy to check that $\mu$ is a measure).
Also for any measure $\mu_1$ on $(X,\fM)$ and a scalar $c\geq 0$ we
can define a measure $\mu = c\mu_1$ by $\mu(A) = c\mu_1(A)$ for all
$A \in \fM$. Thus the collection of measures on $(X,\fM)$ is
`almost' a vector space, except we cannot subtract measures or
multiply them by negative scalars.

Also recall that given a measure $\mu$ on $(X,\fM)$ and a measurable
function $f\geq 0$ on $X$ we can define a measure $\lambda$ by
$\lambda(A) = \int_A f\, d\mu$ for all $A \in \fM$. (Recall that we
write $d\lambda = f\, d\mu$ and call $f$ the density of $\lambda$.)
But what if $f$ takes negative values? Or what if $f$ takes complex
values?

\begin{definition}
A function $\mu \colon \fM \to \IC$ is called a \textbf{complex
measure} if it is $\sigma$-additive, i.e.\ if for any countable
collection of pairwise disjoint measurable sets
$\{E_i\}_{i=1}^\infty$ we have
$$
   \mu \bigl( \cup_{i=1}^{\infty} E_i\bigr) =
   \sum_{i=1}^{\infty} \mu(E_i).
$$
Note: we say that $\{E_i\}_{i=1}^\infty$ make a \textbf{partition}
of $E =\cup_{i=1}^{\infty} E_i$ if $E_i$'s are pairwise disjoint.
\end{definition}

As opposed to complex measures, we will call earlier defined
measures \textbf{positive measures}. Observe that not every positive
measure is a complex measure (because positive measures can take
infinite values).

\begin{lemma}
The series $\sum_{i=1}^{\infty} \mu(E_i)$ in the above definition
must converge absolutely.
\end{lemma}

\begin{definition}
Let $\mu$ be a complex measure on $(X,\fM)$. Define the function
$|\mu|$ on $\fM$ by
$$
   |\mu| (E) =
   \sup \sum_{i=1}^{\infty} |\mu(E_i)|.
$$
where the supremum is taken over all measurable partitions of $E$.
Then $|\mu|$ is called the \textbf{total variation} of $\mu$.
\end{definition}

\begin{theorem}
Let $\mu$ be a complex measure on $(X,\fM)$.\begin{itemize}
\item[\rm (a)] The total variation $|\mu|$ is a positive measure on
$(X,\fM)$. It satisfies $|\mu(E)| \leq |\mu|(E)$ for every $E\in
\fM$;
\item[\rm (b)] If $\lambda$ is a positive measure on $\fM$ with
$|\mu(E)| \leq \lambda(E)$ for all $E\in \fM$, then $|\mu|(E) \leq
\lambda(E)$ for all $E\in \fM$.
\end{itemize}
\end{theorem}

Remark: the property (b) means that $|\mu|$ is the smallest positive
measure that `dominates' $\mu$ (in the sense of (b)).

\begin{lemma}
If $z_1,\ldots,z_N \in \IC$, then there is a subset $S \subset
\{1,\ldots,N\}$ such that
$$
  \biggl| \sum_{k\in S} z_k \biggr|
  \geq \frac{1}{\pi} \sum_{k=1}^N |z_k|.
$$
\end{lemma}

\begin{theorem}
If $\mu$ be a complex measure on $(X,\fM)$, then $|\mu|$ is a finite
measure, i.e.\ $|\mu|(X) <\infty$.
\end{theorem}

Observe that the range of $\mu$ is bounded, since $|\mu(E)| \leq
|\mu|(E) \leq |\mu|(X)$; i.e.\ all the values of $\mu$ lie in a
closed disk $D$ of radius $R=|\mu|(X)$, i.e.\ $\mu(E)\in D$ for all
$E\in \fM$.

\begin{exercise}
Let $(X, \mathcal{M}, \mu)$  be a measure space. For $f \in L^1_\mu$
define
$$
     \mu_f(E)= \int_{E}f\, d \mu \text{.}
$$
Show:\\
${\ \ \ \ \ \ \ }$(a) $\mu_f$ is a complex measure;\\
${\ \ \ \ \ \ \ }$(b)
$|\mu_f|=\mu_{|f|}$, assuming that $f$ is real-valued;\\
${\ \ \ \ \ \ \ }$(c) $|\mu_f|=\mu_{|f|}$, now for a general $f \in
L^1(\mu)$.
\end{exercise}

We denote by $\IM(X,\fM)$ the set of all complex measures on
$(X,\fM)$. For any two measures $\mu_1,\mu_2 \in \IM(X,\fM)$ we can
define a new measure $\mu = \mu_1+\mu_2$ by $\mu(A) = \mu_1(A) +
\mu_2(A)$ for all $A \in \fM$ (it is easy to check that $\mu$ is a
measure). Also for any measure $\mu_1 \in \IM(X,\fM)$ on $(X,\fM)$
and a scalar $c\in\IC$ we can define a measure $\mu = c\mu_1$ by
$\mu(A) = c\mu_1(A)$ for all $A \in \fM$.

\begin{theorem}
$\IM(X,\fM)$ is a vector space with a norm $\|\mu\| = |\mu|(X)$.
\end{theorem}

\begin{exercise}[Bonus]
Prove that the space $\IM(X,\fM)$ with the norm $\|\mu\|$ is a
Banach space, i.e.\ it is a complete metric space (every Cauchy
sequence converges to a limit). Hint: given a Cauchy sequence of
complex measures $\{\mu_n\}$ you need to construct the limit measure
$\mu$ and prove that $\|\mu_n - \mu\| \to 0$ as $n \to \infty$.
\end{exercise}


\begin{definition}
Let $\mu$ be a real measure on $(X,\fM)$ (i.e.\ a real-valued
complex measure). Define
$$
   \mu^+ = \tfrac 12 (|\mu|+\mu)
   \qquad\text{and}\qquad
   \mu^- = \tfrac 12 (|\mu|-\mu)
$$
These are called \textbf{positive and negative variations} of $\mu$.
\end{definition}

Note that $\mu^+$ and $\mu^-$ are positive measures that satisfy
$$
   \mu = \mu^+ - \mu^-
   \qquad\text{and}\qquad
   |\mu| = \mu^+ + \mu^-.
$$
The first formula is called \textbf{Jordan decomposition} of $\mu$.
We will prove that $\mu^+$ and $\mu^-$ are \emph{minimal} measures
that satisfy Jordan decomposition.

\begin{definition}
Let $\mu$ be a positive measure and $\lambda$ a complex measure on
$(X,\fM)$.\begin{itemize}
\item[\rm (a)] we say that $\lambda$ is \textbf{absolutely
continuous} with respect to $\mu$ and write $\lambda \ll \mu$ if $$\
\forall E\in \fM \colon\ \mu(E)=0\ \Rightarrow \lambda(E)=0;$$
\item[\rm (b)] we say that $\lambda$ is \textbf{concentrated on}
$A \in \fM$ if $$\ \forall E\in \fM \colon\ \lambda(E)=\lambda(E\cap
A);$$
\item[\rm (c)] we say that two complex measures $\lambda_1$ and
$\lambda_2$ are \textbf{mutually singular} and write $\lambda_1
\perp \lambda_2$ if $\lambda_1$ is concentrated on $A_1$ and
$\lambda_2$ is concentrated on $A_2$ such that $A_1 \cap A_2 =
\emptyset$.
\end{itemize}
\end{definition}

\noindent Note: $\lambda$ is concentrated on $A$ if and only if $$\
\forall E\in \fM \colon\ A\cap E = \emptyset \ \Rightarrow
\lambda(E)=0.$$ Note also that if $\lambda \ll \mu$ and $\mu$ is
concentrated on $A$, then $\lambda$ is concentrated on $A$ as well.
Suppose that $0\neq c\in \IC$; then $\lambda \ll \mu
\Longleftrightarrow c\lambda \ll \mu$; also $\lambda$ is
concentrated on $A$ iff $c\lambda$ is concentrated on $A$.

\begin{proposition}
Let $\mu$ be a positive measure and $\lambda$'s complex measures on
$(X,\fM)$.\begin{itemize}
\item[\rm (a)] if $\lambda$ is concentrated on $A$, then $|\lambda|$
is concentrated on $A$;
\item[\rm (b)] if $\lambda_1 \perp \lambda_2$, then $|\lambda_1| \perp
|\lambda_2|$;
\item[\rm (c)] if $\lambda_1 \perp \mu$ and $\lambda_2 \perp \mu$,
then $\lambda_1+\lambda_2 \perp \mu$;
\item[\rm (d)] if $\lambda_1 \ll \mu$ and $\lambda_2 \ll \mu$,
then $\lambda_1+\lambda_2 \ll \mu$;
\item[\rm (e)] if $\lambda \ll \mu$,
then $|\lambda| \ll \mu$;
\item[\rm (f)] if $\lambda_1 \ll \mu$ and $\lambda_2 \perp \mu$,
then $\lambda_1 \perp \lambda_2$;
\item[\rm (g)] if $\lambda \ll \mu$ and $\lambda \perp \mu$,
then $\lambda = 0$;
\end{itemize}
\end{proposition}

Example: let $\mu$ be a positive measure on $(X,\fM)$ and $f \in
L^1_\mu$. Then the measure $\mu_f$ defined by
$$
   \mu_f (E) = \int_E f\, d\mu
$$
is absolutely continuous with respect to $\mu$, i.e.\ $\mu_f \ll
\mu$.

\begin{exercise}
Let  $\mu$ be a positive measure on $(X,\fM)$ and $f,g \in L^1_\mu$.
Define $\mu_f$ and $\mu_g$ as above. Prove that \begin{itemize}
\item[\rm (a)] $\mu_f$ is concentrated on $A$ if and only if
$\mu\{x \in A^c \colon f(x) \neq 0\} = 0$;
\item[\rm (b)] $\mu_f \perp \mu_g$ if and only if
$\mu\{x\in X \colon f(x)g(x) \neq 0\}=0$;
\item[\rm (c)] if $f \geq 0$, then $$\ \mu\ll\mu_f
\ \Longleftrightarrow\ f(x)>0\ \text{for } \mu-\text{a.e.}\ x\in
X.$$
\end{itemize}
\end{exercise}

\begin{exercise}
Let $\lambda$ be a positive measure on $(X,\fM)$. Prove that
$\lambda$ is concentrated on $A$ if and only if $\lambda(A^c)=0$.
Give a counterexample to this statement in the case of a complex
measure $\lambda$.
\end{exercise}

\begin{definition}
A positive measure $\mu$ on $(X,\fM)$ is said to be
$\sigma$-\textbf{finite} if there exist a countable sequence of sets
$E_n \in \fM$ such that $X = \cup_n E_n$ and $\mu(E_n)<\infty$ for
all $n$.
\end{definition}

\begin{theorem}[Lebesgue-Radon-Nikodym]
Let $\mu$ be a positive $\sigma$-finite measure and $\lambda$ a
complex measure on $(X,\fM)$.\begin{itemize}
\item[\rm (a)] There exist a unique pair of complex measures
$\lambda_{\rm a}$ and $\lambda_{\rm s}$ such that $$\lambda =
\lambda_{\rm a} + \lambda_{\rm s}$$ and $\lambda_{\rm a} \ll \mu$
and $\lambda_{\rm s} \perp \mu$;
\item[\rm (b)] There is a unique $h \in L^1_\mu$ (up to
$\mu$-equivalence) such that $$\lambda_{\rm a} (E) = \int_E h\,
d\mu$$ i.e.\ $\lambda_{\rm a} = \mu_h$.
\end{itemize}
\end{theorem}

Note that $\lambda_{\rm a} \perp \lambda_{\rm s}$. The part (a) is
called the \textbf{Lebesgue decomposition} of $\lambda$. The
function $h$ is called the \textbf{Radon-Nikodym derivative} of
$\lambda_{\rm a}$ with respect to $\mu$, we sometimes write
$d\lambda_{\rm a} = h\, d\mu$.

In particular, if $\lambda \ll \mu$, then $\lambda = h\, d\mu$ for
some $h \in L^1_{\mu}$.

The Lebesgue-Radon-Nikodym theorem can be extended to the case where
both $\mu$ and $\lambda$ are positive $\sigma$-finite measures. But
we cannot go beyond $\sigma$-finiteness, as the following exampke
shows:

Example: let $X=[0,1]$, $\mu$ the Lebesgue measure, and $\lambda$
the counting measure on $X$. Then $\lambda$ cannot be decomposed as
$\lambda_{\rm a} + \lambda_{\rm s}$. Moreover, we have $\mu \ll
\lambda$ but there is no $h\in L^1_{\lambda}$ such that $d\mu = h\,
d\lambda$.

\begin{theorem}
Let $\mu$ be a positive measure and $\lambda$ a complex measure on
$(X,\fM)$. Then the following are equivalent: \begin{itemize}
\item[\rm (a)] $\lambda \ll \mu$; \item[\rm (b)] $\forall
\varepsilon>0$ $\exists \delta>0\colon$ $\mu(E)<\delta
\Longrightarrow |\lambda(E)| < \varepsilon$. \end{itemize}
\end{theorem}

The property (b) justifies the name \emph{absolute continuity}.

If $\lambda$ is a positive but not finite measure, then (a) and (b)
are not equivalent. For example, let $X=(0,1)$, $\mu$ the Lebesgue
measure, and $\lambda(E) = \int_E t^{-1}\, dt$. Then $\lambda \ll
\mu$, but the property (b) does not hold.

\begin{theorem}[Polar representation]
Let $\mu$ be a complex measure on $(X,\fM)$. Then there is a
measurable function $h \colon X \to \IC$ such that $|h|=1$ and $d\mu
= h\, d|\mu|$.
\end{theorem}

\begin{theorem}
Let $\mu$ be a positive measure on $(X,\fM)$. Suppose $g \in
L^1_{\mu}$ and $d\lambda = g\, d\mu$. Then $d|\lambda| = |g|\,
d\mu$.
\end{theorem}

\begin{theorem}[Hahn decomposition]
Let $\mu$ be a real-valued measure on $(X,\fM)$. Then there is a
decomposition $X=A\cup B$, $A\cap B=\emptyset$, and
$$
   \mu^+(E)=\mu(A\cap E), \qquad \mu^-(E)=\mu(B\cap E)
$$
for any $E \in \fM$.
\end{theorem}

\begin{exercise}
Let $f$ be a real-valued and integrable function on $[0,1]$. Prove
that there exists $c \in [0,1]$ such that
$$\int_{[0,c]}f\,d\bm = \int_{[c,1]}f\,d\bm \text{.}$$
\end{exercise}

\begin{exercise}
Let $f, f_k \in L^1_m{[0,1]}$, for $k=1,2,\ldots$. Suppose $f_k(x)
\rightarrow f(x)$ a.e.\ on $[0,1]$, and $\int_{[0,1]} |f_k|\, dm
\rightarrow \int_{[0,1]} |f|\, dm$. Show that $$\int_{[0,1]} |f_k
-f| \, dm \rightarrow 0.$$ Hint: look at $|f| + |f_k| - |f-f_k|$.
\end{exercise}

\begin{exercise}
Do the previous exercise with $L^1$ replaced by $L^2$.
\end{exercise}

Let $X$ and $Y$ be vector spaces with norms and $\Lambda \colon X
\to Y$ a linear mapping.

\begin{definition}
The \textbf{norm} of $\Lambda$ is
\begin{align*}
  \|\Lambda\| &= \sup\{\|\Lambda x\|\colon
  x\in X,\ \ \|x\|=1\} \\
  &= \sup\{\|\Lambda x\|/\|x\|\colon
  x\in X,\ \ x\neq 0\}
\end{align*}
We say that $\Lambda$ is \textbf{bounded} if $\|\Lambda\|<\infty$.
\end{definition}

\begin{lemma}
We have $\|\Lambda\| = \inf\{c>0\colon \|\Lambda x\|\leq c\|x\| \ \
\forall x\in X\}$; furthermore, $\inf$ can be replaced with $\min$.
\end{lemma}

\begin{theorem}
The following conditions are equivalent: \begin{itemize} \item[\rm
(a)] $\Lambda$ is bounded; \item[\rm (b)] $\Lambda$ is continuous on
$X$; \item[\rm (c)] $\Lambda$ is continuous at some point $x\in X$.
\end{itemize}
\end{theorem}

\begin{definition}
In a special case, where $Y=\IC$, we deal with linear
\textbf{functionals} $L\colon X\to \IC$ and call
$$
  X^\ast = \{\text{all bounded linear functionals}\ \ L\colon X\to \IC\}
$$
the \textbf{dual space} (to $X$). It is a vector space with norm
$\|L\|$.
\end{definition}

Consider $L^p_{\mu}(X)$ on a measure space $(X,\fM,\mu)$ with $1\leq
p\leq \infty$. Let $q$ be a conjugate exponent. Given a function
$g\in L^q_{\mu}(X)$, we construct a linear functional $\Phi_g \colon
L^p_{\mu}(X) \to \IC$ by
$$
   \Phi_g(f) = \int_X fg\, d\mu
$$
It is bounded (by H\"older inequality) and $\|\Phi_g\| \leq \|g\|_q$
(here $\|g\|_q$ denotes the norm in the $L^q$ space).

\begin{theorem}
Let $\mu$ be a finite or $\sigma$-finite measure and $1\leq p <
\infty$. Then for every bounded linear functional $\Phi \colon
L^p_{\mu}(X) \to \IC$ there exists a unique $g\in L^q_{\mu}(X)$ (up
to equivalence) such that
$$
   \Phi(f) = \int_X fg\, d\mu \qquad \forall f\in L^p_{\mu}(X),
$$
i.e.\ $\Phi = \Phi_g$; furthermore, $\|\Phi\| = \|g\|_q$.
\end{theorem}

In other words, $L^q_{\mu}$ is isometrically isomorphic to the dual
space $L^p_{\mu}(X)^\ast$.

For $1<p<\infty$, the theorem holds without $\sigma$-finiteness.

This theorem does not hold for $p=\infty$, see the next exercise.

\begin{exercise}
Let $X=[0,1]$ and $\mu$ the Lebesgue measure. Show that
$L^{\infty}(X)^\ast \supset L^1(X)$, but $L^{\infty}(X)^\ast \neq
L^1(X)$ (in the sense $g \rightarrow \Phi_g$). (Hint: Use the
following consequence of the Hahn-Banach theorem: If $X$ is a Banach
space and $A \subset X$ is a closed subspace of $X$, with $A\neq X$,
then there exists $f\in X^\ast$ with $f \neq 0$, and $f(x)=0$ for
all $x \in A$.)
\end{exercise}

\begin{exercise}
Suppose $\mu$ is a positive measure on $X$ and $\mu(X)< \infty$. Let
$f \in L^\infty_{\mu}$ and $\|f\|_{\infty}>0$. Define $$\alpha_n =
\int_X |f|^n\, d\mu \ \ \ \ \ \ \ (n=1,2,3, \dots) \text{.}$$ Prove
that
$$\lim_{n \rightarrow \infty} \frac{\alpha_{n+1}}{\alpha_{n}}
= \|f\|_{\infty} \text{.}$$ (Hint: Start with $\|f\|_{\infty}=1$,
use H\"older inequality and the result of Exercise~51 (e).)
\end{exercise}

\section{Differentiation}

\begin{definition}
Let $\mu$ be a complex measure on $\IR$ (with Borel
$\sigma$-algebra). Then $F(x) = \mu\bigl ((-\infty,x) \bigr)$ is
called the \textbf{distribution function} of the measure $\mu$.
\end{definition}

We denote by $\bm$ the Lebesgue measure on $\IR$ (and on $\IR^k$).

\begin{theorem}
The following two conditions are equivalent: \begin{itemize}
\item[\rm (a)] $F$ is differentiable at $x\in\IR$ and $F'(x)=A$;
\item[\rm (b)] $\forall \varepsilon>0$ $\exists \delta>0 \colon$
$\left|\frac{\mu(I)}{\bm(I)} - A\right| <\varepsilon$ for every open
interval $I \subset \IR$ such that $x \in I$ and $\bm(I)<\delta$.
\end{itemize}
\end{theorem}

Motivated by this we will define derivatives of measures in $\IR^k$.

\begin{definition}
For $x\in \IR^k$ and $r>0$ we denote by
$$
   B(x,r) = \{y\in \IR^k\colon |y-x|<r\}
$$
the \textbf{open ball} of radius $r$ centered on $x$. If $\mu$ is a
complex measure on $\IR^k$ (with Borel $\sigma$-algebra), then we
denote
$$
   (Q_r\mu)(x) = \frac{\mu\bigl(B(x,r)\bigr)}
   {\bm\bigl(B(x,r)\bigr)}
$$
and define the \textbf{symmetric derivative} of $\mu$ at $x$ by
$$
   (D\mu)(x) = \lim_{r\to 0} (Q_r\mu)(x)
$$
(if the limit exists).
\end{definition}


\begin{exercise}
Let $\mu$ be a complex Borel measure on $\IR$ and assume that its
symmetric derivative $(D\mu)(x)$ exists at some $x_0 \in \IR$. Does
it follow that its distribution function $F(x)= \mu((-\infty, x))$
is differentiable at $x_0$?
\end{exercise}


To study derivatives of measures, we introduce more definitions.

\begin{definition}
If $\mu$ is a complex measure on $\IR^k$ (with Borel
$\sigma$-algebra), then the \textbf{maximal function} $M\mu \colon
\IR^k \to [0,\infty]$ defined by
$$
  (M\mu)(x) = \sup_{0<r<\infty} (Q_r|\mu|)(x).
$$
\end{definition}

\begin{lemma}
$M\mu$ is lower semicontinuous, i.e.\ the set $\{x\colon
M\mu(x)>a\}$ is open for every $a\in\IR$.
\end{lemma}

\begin{lemma}[Covering]
Let $W = \cup_{i=1}^N B(x_i,r_i)$ be a finite union of open balls.
Then there is $S \subset \{1,2, \ldots, N\}$ such that
\begin{itemize}
\item[\rm (a)] the balls $(x_i,r_i)$, $i\in S$ are disjoint;
\item[\rm (b)] $W \subset \cup_{i\in S} B(x_i,3r_i)$; \item[\rm (c)]
$\bm(W) \leq 3^k \sum_{i\in S} \bm\bigl(B(x_i,r_i)\bigr)$.
\end{itemize}
\end{lemma}

\begin{theorem} \label{10.7}
Let $\mu$ be a complex measure on $\IR^k$. Then $\forall \lambda>0$
$$
  \bm(M\mu>\lambda) \leq \frac{3^k\|\mu\|}{\lambda}
$$
where $\|\mu\| = |\mu|(\IR^k)$.
\end{theorem}

In other words, the maximal function cannot be large on a large set.

\begin{definition}
\textbf{Weak} $L^1$ is the space of measurable functions
$$
  L^1_W (\IR^k) = \{f\colon \IR^k \to \IC \text{ such that }
  \lambda\, \bm\{|f|>\lambda\} \text{ is bounded on } \lambda
  \in (0,\infty) \}
$$
\end{definition}

\begin{lemma}
We have $L^1_{\bm}(\IR^k) \subset L^1_W(\IR^k)$.
\end{lemma}

Note: there are functions in $L^1_W(\IR^k)$ that are not in
$L^1_{\bm}(\IR^k)$. For example, $f=1/x$ for $k=1$.

\begin{definition}
For every $f\in L^1_W (\IR^k)$ define the \textbf{maximal function}
$Mf \colon \IR^k \to [0,\infty]$ by
$$
  (Mf)(x) = \sup_{0<r<\infty} \frac{1}{\bm(B(x,r))}\,
  \int_{B(x,r)}|f|\, d\bm.
$$
Note that $\bm(B(x,r)) = \bm(B(0,r))$ does not depend on $x$.
\end{definition}

We have $Mf = M\mu$, where $d\mu = f\, d\bm$. Thus
theorem~\ref{10.7} states that the `maximal function' $M$ induces an
operator $L^1_{\bm} (\IR^k) \to L^1_W (\IR^k)$, which is bounded
(its bound is $\leq 3^k$): for every $f \in L^1_{\bm}(\IR^k)$ and
every $\lambda>0$
$$
  \bm\{Mf>\lambda\} \leq 3^k\lambda^{-1}\|f\|_1.
$$

\begin{definition}
If $f\in L^1_{\bm} (\IR^k)$, then any point $x\in \IR^k$ for which
$$
  \lim_{r\to 0} \frac{1}{\bm(B(x,r))}
  \int_{B(x,r)} |f(y)-f(x)|\, d\bm(y) =0
$$
is called a \textbf{Lebesgue point} of $f$.
\end{definition}

Roughly speaking, $x$ is a Lebesgue point if $f$ does not oscillate
too much near $x$, in average sense.

\begin{lemma}
If $f$ is continuous at $x$, then $x$ is a Lebesgue point of $f$.
\end{lemma}

\begin{lemma}
If $x$ is a Lebesgue point of $f$, then
$$
   f(x) =  \lim_{r\to 0} \frac{1}{\bm(B(x,r))}
   \int_{B(x,r)} f\, d\bm
$$
(but the converse is not true).
\end{lemma}

\begin{exercise}
Let $f \in L^1_{\bm} (\IR^k)$. Show that $|f(x)| \leq (Mf)(x)$ at
every Lebesgue point $x$ of $f$.
\end{exercise}

\begin{exercise}
Let
\begin{displaymath}
f(x) = \left\{
\begin{array}{ll}
0 & \textrm{if \ \ $x<0\ $ or $\ x>1$}\\
1 & \textrm{if \ \ $0<x<1$}\\
\end{array} \right. \text{.}
\end{displaymath}
Is it possible to define $f(0)$ and $f(1)$ such that $0$ and $1$
become Lebesgue points of $f$?
\end{exercise}

\begin{exercise}[Bonus]
Construct a function $f \colon \IR \to \IR$ such that $f(0)=0$ and
$0$ is a Lebesgue point of $f$, but for every $\varepsilon>0$
$$
   \bm\bigl\{x\in\IR \colon |x|<\varepsilon\
   \text{ and }\ |f(x)|\geq 1\bigr\}>0,
$$
i.e.\ $f$ is essentially discontinuous at $0$.
\end{exercise}

\begin{theorem}
If $f \in L^1(\IR^k)$, then almost every point $x\in \IR^k$ is a
Lebesgue point of $f$.
\end{theorem}

\begin{theorem}
Let $\mu$ be a complex Borel measure on $\IR^k$ such that $\mu \ll
\bm$. Then the Radon-Nokodym derivative $f = d\lambda/d\bm$ is equal
to the symmetric derivative of $\mu$, i.e.\ $f = D\mu$ $\bm$-a.e.
\end{theorem}

Next we explore some other ways of computing symmetric derivatives.

\begin{definition}
Let $x \in \IR^k$. We say that Borel sets $E_i \subset \IR^k$
\textbf{shrink nicely to} $x$ if there exists $\alpha>0$ such that
for some sequence of balls $B(x,r_i)$ with $r_i \to 0$ we have $E_i
\subset B(x,r_i)$ and $m(E_i) \geq \alpha \bm(B(x,r_i))$ for all
$i$.
\end{definition}

Examples: any sequence of intervals $I_i \supset x$ in $\IR^k$ such
that $\bm(I_i)\to 0$ shrinks to $x$ nicely. More generally: any
sequence of balls or cubes in $\IR^k$ must shrink nicely. But the
sequence of rectangles $[-\tfrac 1i, \tfrac 1i] \times [-\tfrac
1{i^2}, \tfrac 1{i^2}]$ does not shrink nicely to $0$.

\begin{theorem} \label{LP}
If $f \in L^1(\IR^k)$ and $x\in \IR^k$ is a Lebesgue point of $f$,
then for any sequence $\{E_i\}$ that shrinks nicely to $x$ we have
$$
   f(x) = \lim_{i\to\infty} \frac{1}{\bm(E_i)}
   \int_{E_i} f\, d\bm.
$$
\end{theorem}

Recall the Fundamental Theorem of Calculus:\\

\noindent (a) if $f\colon [a,b] \to \IR$ is continuous and $F(x) =
\int_a^x f(t)\, dt$, then $F'(x)=f(x)$ for all $x\in (a,b)$.\\

\noindent (b) if $f\colon [a,b] \to \IR$ is continuously
differentiable, then
$$
   f(x)-f(a) = \int_a^x f'(t)\, dt
$$
for all $x\in [a,b]$.\\

Note that in both parts the Riemannian integral is used. Our goal is
to extend this theorem to Lebesgue integrable functions.

\begin{theorem}[FTC in $L^1$, easy part]
If $f \in L^1(\IR)$ and $F(x) = \int_{(-\infty,x]} f\, d\bm$ for
$x\in \IR$, then $F'(x) = f(x)$ at every Lebesgue point $x$ of $f$.
\end{theorem}

\begin{definition}
Let $E \subset \IR^k$ be a measurable set and $x\in \IR^k$. Then
$$
  \lim_{r\to 0} \frac{\bm(E\cap B(x,r))}{\bm(B(x,r))}
$$
is the \textbf{metric density} of $E$ at $x$ (if the limit exists).
\end{definition}

\begin{corollary}
For every measurable set $E \subset \IR^k$ the metric density is 1
at a.e.\ point $x\in E$ and 0 at a.e.\ point $x\in E^c$.
\end{corollary}

\begin{corollary}
If $\varepsilon>0$, then there is no measurable set $E \subset \IR$
such that $$\epsilon < \frac{\bm(E\cap I)}{\bm(I)}<1-\varepsilon$$
for every finite nontrivial interval $I \subset \IR$.
\end{corollary}

In fact, if $\epsilon < \frac{\bm(E\cap I)}{\bm(I)}$ for every
finite nontrivial interval $I \subset \IR$, then $\bm(E^c)=0$. And
if $\frac{\bm(E\cap I)}{\bm(I)}<1-\varepsilon$ for every finite
nontrivial interval $I \subset \IR$, then $\bm(E)=0$.

Our next goal is to extend to $L^1$ the second part of the FTC. We
want to find a class of functions $f \colon [a,b] \to \IR$ such that
\beq
     f(x)-f(a) = \int_{[a,x]} f'\, d\bm
     \qquad \forall x\in [a,b] \tag{FTC-2}
\eeq
Obviously, $f$ must be differentiable (at least a.e.) and $f'$ must
be integrable (i.e.\ $f' \in L^1 ([a,b])$). But is this enough?

Answer: No! Counterexample: Cantor function (devil's staircase)
constructed via the middle-third Cantor set. This example shows that
(FTC-2) may fail if $f$ experiences `rapid growth on tiny sets'.

\begin{definition}
A function $f\colon [a,b] \to \IC$ is said to be \textbf{absolutely
continuous (a.c.)} on an interval $[a,b]$ if $\forall \varepsilon>0$
$\exists \delta>0$ such that if $n\in \IN$ and $$(\alpha_1,\beta_1),
\ldots, (\alpha_n,\beta_n)$$ are disjoint subintervals of $[a,b]$,
then
$$
   \sum_{i=1}^n (\beta_i-\alpha_i) < \delta\quad
   \Longrightarrow\quad
   \sum_{i=1}^n |f(\beta_i)-f(\alpha_i)| < \varepsilon.
$$
\end{definition}

Note: if $f$ is absolutely continuous, then $f$ is continuous, even
uniformly continuous (but not vice versa).

\begin{lemma}
If $f\colon [a,b] \to \IC$ is differentiable at almost all $x \in
[a,b]$, $f' \in L^1([a,b])$, and (FTC-2) holds, then $f$ is
absolutely continuous.
\end{lemma}

We will show that the converse is also true.

\begin{theorem}
Let $I = [a,b]$ and $f \colon I \to \IR$ be a continuous and
nondecreasing function. Then the following properties are
equivalent:
\begin{itemize}
\item[\rm (a)] $f$ is absolutely continuous; \item[\rm (b)] $f$
maps sets of measure zero to sets of measure zero; \item[\rm (c)]
$f$ is differentiable a.e.\ on $I$, $f' \in L^1(I)$ and (FTC-2)
holds.
\end{itemize}
\end{theorem}

\begin{definition}
The \textbf{total variation function} $V_a^x$ of a function $f
\colon [a,b] \to \IC$ is defined by
$$
  V_a^x = \sup \sum_{i=1}^n |f(t_i) - f(t_{i-1})|
$$
where the supremum is taken over all $n$ and all ordered sequences
$$
   a=t_0<t_1<t_2<\cdots<t_{n-1}<t_n=x.
$$
If $V_a^b<\infty$, then $f$ is said to be of \textbf{bounded
variation} on $[a,b]$, denoted by $f\in BV[a,b]$, and the value of
$V_a^b$ is called the \textbf{total variation} of $f$.
\end{definition}

\begin{lemma}
Suppose $f(x)$ is of bounded variation on $[a,b]$. Then
\begin{itemize}
\item[\rm (a)] $V_a^x$ is non-decreasing in $x$;
\item[\rm (b)] if $f$ is real-valued, then $V+f$ and $V-f$ are
non-decreasing;
\end{itemize}
\end{lemma}

\begin{corollary}
Let $f \colon [a,b] \to \IR$ be of bounded variation on $[a,b]$.
Then there are (strictly) monotonically increasing functions $u,v
\colon [a,b] \to \IR$ such that $f = u-v$.
\end{corollary}

\begin{lemma}
Suppose $f(x)$ is absolutely continuous on $[a,b]$. Then
\begin{itemize}
\item[\rm (a)] $f$ is of bounded variation on $[a,b]$;
\item[\rm (b)] $V$, $V+f$, and $V-f$ are absolutely continuous
on $[a,b]$.
\end{itemize}
\end{lemma}

\begin{theorem}
A function $f \colon [a,b] \to \IC$ is absolutely continuous if and
only if $f$ is differentiable for a.e.\ $x\in [a,b]$, $f'\in
L^1([a,b])$, and (FTC-2) holds.
\end{theorem}

This theorem answers the question posed around (FTC-2).

\medskip Example: Consider a function
$$
   f(x) = \left\{ \begin{array}{cl}
   x^2 \cos\tfrac{\pi}{x^2} & \ \ \text{for}\ \ 0<x\leq 1\\
   0 & \ \ \text{for}\ \ x=0
   \end{array}\right.
$$
It is differentiable at every point $x\in [0,1]$ (including $x=0$),
but $f'$ does not belong to $L^1 ([0,1])$. So this function is not
absolutely continuous.

The following theorem can be stated without proof:

\begin{theorem}
Let $f \colon [a,b] \to \IC$ be differentiable at every point $x\in
[a,b]$ and $f'\in L^1([a,b])$. Then (FTC-2) holds.
\end{theorem}

Note: the differentiability at \emph{every} point $x\in [a,b]$ is
assumed.

\medskip Example: Consider a function
$$
   f(x) = \left\{ \begin{array}{cl}
   x^2 \cos\tfrac{\pi}{x} & \ \ \text{for}\ \ 0<x\leq 1\\
   0 & \ \ \text{for}\ \ x=0
   \end{array}\right.
$$
It is differentiable at every point $x\in [0,1]$ (including $x=0$)
and $f'$ is bounded, hence $f' \in L^1 ([0,1])$. So this function,
unlike the one in the previous example, is absolutely continuous.

\begin{exercise}
Let $\alpha>0$, $\beta>0$ and
$$
   f(x) = \left\{ \begin{array}{cl}
   x^{\alpha} \cos\tfrac{\pi}{x^\beta} & \ \ \text{for}\ \ 0<x\leq 1\\
   0 & \ \ \text{for}\ \ x=0
   \end{array}\right.
$$
Show that if $\alpha > 1$ and $\alpha > \beta$, then $f$ is
absolutely continuous on $[0, 1]$. For an extra credit, show that
the assumption $\alpha >1$ is redundant.
\end{exercise}

\begin{exercise}
Let $f,g$ be absolutely continuous on $[a,b]$. Show that\\
${\ \ \ \ \ \ \ }$(a) $fg$ is absolutely continuous on $[a,b]$.\\
${\ \ \ \ \ \ \ }$(b) if $f(x) \neq 0$  for all $x \in [a,b]$, then
$1/f$ is absolutely continuous on $[a,b]$.\\
${\ \ \ \ \ \ \ }$(c) Does (a) remain true if $[a,b]$ is replaced by
$\IR?$
\end{exercise}

\begin{exercise}
A function $f \colon [a,b] \to \IC$ is said to be Lipschitz
continuous if $\exists L>0$ such that $|f(x) - f(y)| \leq L |x-y|$
for all $x,y\in [a,b]$. Prove that if $f$ is Lipschitz continuous,
then $f$ is absolutely continuous and $|f'| \leq L$ a.e. Conversely,
if $f$ is absolutely continuous and $|f'| \leq L$ a.e., then $f$ is
Lipschitz continuous (with that constant $L$).
\end{exercise}

\begin{proposition}
Let $f \colon [a,b] \to \IR$ be absolutely continuous. Then it maps
sets of measure zero to sets of measure zero.
\end{proposition}

Note: the converse is not true, even if $f$ is continuous or even
differentiable (see the first example above). Also, this proposition
does not extend to functions of bounded variations (a
counterexample: devil's staircase maps the Cantor set onto the
entire unit interval).

\begin{corollary}
Let $f \colon [a,b] \to \IR$ be absolutely continuous. Then it maps
Lebesgue measurable sets to Lebesgue measurable sets.
\end{corollary}

The following theorem can be stated without proof:

\begin{theorem}
Let $f \colon [a,b] \to \IR$ be monotonically increasing. Then it is
differentiable a.e.\ and $f(b) - f(a) \geq \int_{[a,b]} f' \, d\bm$.
\end{theorem}

Note: a strict inequality is possible (example: devil's staircase).

\begin{exercise}
Let $f \colon [a,b] \to \IR$ be absolutely continuous. Prove that
$V_a^x \leq \int_{[a,x]} |f'|\, d\bm$. For an extra credit: is the
equality always true?
\end{exercise}

\begin{exercise}
Let $f \colon [0,1] \to \IR$ be absolutely continuous on $[\delta,
1]$ for each $\delta>0$ and continuous and of bounded variation on
$[0,1]$. Prove that $f$ is absolutely continuous on $[0, 1]$. [Hint:
use the result of the previous exercise to verify that $f' \in
L^1([0,1])$.]
\end{exercise}

\begin{exercise}[Bonus]
Let $f \colon [a,b] \to \IC$ be absolutely continuous. Prove that
$|f|$ is absolutely continuous and
$$
  \Bigl|\frac{d}{dx}|f(x)|\Bigr| \leq |f'(x)|
$$
for a.e.\ $x \in [a,b]$. [Hint: first, use triangle inequality in
the form $\bigl| |p| - |q| \bigr| \leq |p-q|$, for complex numbers
$p,q$, to show that $|f|$ is AC. Then prove the above inequality for
$x \in [a,b]$ that are Lebesgue points for both $f'$ and $|f|'$; use
Theorem~\ref{LP}.]
\end{exercise}

Let $A\colon \IR^k \to \IR^k$ be a linear transformation defined by
a $k \times k$ matrix, which we also call $A$. Then for any
measurable set $E \subset \IR^k$ we have
$$
   \bm (A(E)) = c\, \bm(E)
$$
where the scaling factor is $c = |\det A|$. In particular, every
isometry (translation, rotation, reflection) preserves the Lebesgue
measure.

For a (nonlinear) transformation $T \colon V \to \IR^k$ defined on
an open set $V \subset \IR^k$ we define its derivative $T'(x)=A$ at
a point $x\in V$ by
$$
   \lim_{h\to 0} \frac{\|T(x+h)-T(x)-Ah\|}{\|h\|}=0
$$
(where of course $h \in \IR^k$), if the limit exists. Note that the
derivative $T'(x)$ is a $k\times k$ matrix. (If you define the
transformation $T$ coordinate-wise, by $k$ functions of $k$
variables, then $T'(x)$ will be the matrix of their partial
derivatives.)

\begin{lemma}
Let $T \colon V \to \IR^k$ be differentiable at $x\in V$. Then
$$
  \lim_{r\to 0} \frac{\bm(T(B(x,r)))}{\bm(B(x,r))}=|\det T'(x)|
$$
\end{lemma}

The following is a (global) change of variables rule:

\begin{theorem}
Let $T \colon V \to \IR^k$ be differentiable at every point $x\in V$
and one-to-one. Then for any measurable function $f\colon \IR^k \to
[0,+\infty]$
$$
   \int_{T(V)} f\, d\bm = \int_V (f\circ T) \, |\det T'|\, d\bm
$$
\end{theorem}

The factor $\det T'$ is called the Jacobian (of the transformation
$T$).

Note: a continuous (not differentiable) transformation may map
measurable sets into nonmeasurable sets. Example: devil's staircase.


\section{Integration on product spaces}


\begin{definition}
Let $X$ and $Y$ be two sets. Its \textbf{Cartesian product} $X
\times Y$ is the set of all (ordered) pairs $(x,y)$, with $x\in X$
and $y\in Y$. If $A\subset X$ and $B\subset Y$, then the set
$A\times B \subset X\times Y$ is called a \textbf{rectangle}.
\end{definition}

\begin{definition}
Let $(X,\fM)$ and $(Y,\fN)$ be two measurable spaces. For every $A
\in \fM$ and $B \in \fN$ the set $A \times B$ is called a
\textbf{measurable rectangle} in $X \times Y$. Any finite union $E =
R_1 \cup R_2 \cup \cdots \cup R_n$ of disjoint measurable rectangles
is called an \textbf{elementary set}. The collection of elementary
sets is denoted by $\cE$.
\end{definition}

\begin{lemma}
$\cE$ is an algebra, i.e.\ finite unions, intersections and
differences of elementary sets are elementary sets.
\end{lemma}

We denote by $\fM \times \fN$ the (minimal) $\sigma$-algebra in $X
\times Y$ generated by elementary sets (equivalently, by measurable
rectangles).


\begin{definition}
Let $E \subset X \times Y$ and $x\in X$, $y\in Y$. Then
$$
  E_x = \{y' \colon (x,y')\in E\} \subset Y
$$
is called the $x$-\textbf{section} of $E$ and
$$
  E^y = \{x' \colon (x',y)\in E\} \subset X
$$
is called the $y$-\textbf{section} of $E$.
\end{definition}

\begin{theorem}
If $E \in \fM \times \fN$, then $E_x \in \fN$ and $E^y \in \fM$ for
every $x\in X$ and $y\in Y$.
\end{theorem}

\begin{definition}
A \textbf{monotone class} $\cM$ is a collection of sets with the
following properties: \begin{itemize} \item[(i)] $A_i \in \cM,\ A_i
\subset A_{i+1}\ (i=1,2,\ldots)\ \Longrightarrow\ \cup_{i=1}^\infty
A_i \in \cM$; \item[(ii)] $B_i \in \cM,\ B_i \supset B_{i+1}\
(i=1,2,\ldots)\ \Longrightarrow\ \cup_{i=1}^\infty B_i \in \cM$.
\end{itemize}
\end{definition}

Note: $\sigma$-algebras are monotone classes, but not vice versa.
The intersection of monotone classes is a monotone class.

\begin{lemma}
If $X$ is a set and $\cF$ a collection of subsets of $X$, then there
exists a smallest monotone class $\cM$ that contains $\cF$.
\end{lemma}

\begin{theorem}
The $\sigma$-algebra $\fM \times \fN$ is the smallest monotone class
containing $\cE$.
\end{theorem}

For any function $f \colon X\times Y \to Z$ and $x\in X$ we denote
by $f_x$ a function on $Y$ defined by $f_x(y) = f(x,y)$. Similarly,
for any $y\in Y$ we denote by $f^y$ a function on $X$ defined by
$f^y(x) = f(x,y)$.

\begin{theorem}
Let $Z$ be a topological space and $f \colon X\times Y \to Z$ a
measurable function (with respect to the $\sigma$-algebra $\fM\times
\fN$). Then
\begin{itemize}
\item[\rm (i)] $f_x$ is $\fN$-measurable on $Y$ for each $x\in X$; \item[\rm (ii)]
$f^y$ is $\fM$-measurable on $X$ for each $y\in Y$;
\end{itemize}
\end{theorem}

Next we construct the product of measures.

\begin{theorem}
Let $(X,\fM,\mu)$ and $(Y,\fN,\lambda)$ be two measure spaces and
let $Q \in \fM \times \fN$. Define
\begin{align*}
   \varphi(x)&=\lambda(Q_x)\quad\text{for every  }x\in X\\
   \psi(x)&=\mu(Q^y)\quad\text{for every  }y\in Y
\end{align*}
Then $\varphi$ is $\fM$-measurable, $\psi$ is $\fN$-measurable, and
$$
  \int_X \varphi \, d\mu = \int_Y \psi\, d\lambda.
$$
\end{theorem}

\begin{definition}
Let $(X,\fM,\mu)$ and $(Y,\fN,\lambda)$ be two measure spaces with
$\sigma$-finite measures. Then we can define a measure, $\mu \times
\lambda$, on $\fM \times \fN$ by
$$
   (\mu \times \lambda) (Q) =
   \int_X \lambda(Q_x) \, d\mu(x) = \int_Y \mu(Q^y)\, d\lambda(y).
$$
The measure $\mu \times \lambda$ is called the \textbf{product} of
the measures $\mu$ and $\lambda$.
\end{definition}

Observe that $\mu \times \lambda$ is also a $\sigma$-finite measure.

We can rewrite the above formula as
$$
   (\mu \times \lambda) (Q) =
   \int_X \biggl(\int_Y \chi_Q(x,y)\, d\lambda(y)\biggr) \, d\mu(x)=
   \int_Y \biggl(\int_X \chi_Q(x,y)\, d\mu(x)\biggr) \, d\lambda(y).
$$

\begin{exercise}
Given measure spaces $(X,\fM,\mu)$ and $(Y,\fN,\lambda)$ with
$\sigma$-finite measures, show that $\mu \times \lambda$ is the
unique measure on $\fM \times \fN$ such that
$$
   (\mu \times \lambda)(A \times B)=
   \mu(A)\lambda(B)
$$
for all measurable rectangles $A \times B$ in $X \times Y$.
\end{exercise}

\begin{theorem}[Fubini]
Let $(X,\fM,\mu)$ and $(Y,\fN,\lambda)$ be two measure spaces with
$\sigma$-finite measures and let $f$ be an $(\fM \times
\fN)$-measurable function on $X\times Y$. Then
\begin{itemize}
\item[\rm (a)] If $0\leq f \leq \infty$ and if
$$
   \varphi(x) = \int_Y f_x\, d\lambda,\qquad
   \psi(y) = \int_X f^y\, d\mu,
$$
then $\varphi$ is $\fM$-measurable, $\psi$ is $\fN$-measurable, and
\beq
  \int_X \varphi \, d\mu =
  \int_{X\times Y} f\, d(\mu \times \lambda) =
  \int_Y \psi\, d\lambda.
  \tag{Fubini}
\eeq
\item[\rm (b)] If $f$ is complex-valued and $\int_X \varphi^\ast \,
d\mu < \infty$, where $\varphi^\ast (x) = \int_Y |f|_x\, d\lambda$,
then $f \in L^1_{\mu\times\lambda}$.
\item[\rm (c)] If $f \in L^1_{\mu\times\lambda}$, then
$$
   f_x \in L^1_{\lambda}\quad{\rm for}\ {\rm a.e.}\ \ x\in X,
$$
$$
   f^y \in L^1_{\mu}\quad{\rm for}\ {\rm a.e.}\ \ y\in Y,
$$
$$
    \varphi(x)= \int_Y f_x\, d\lambda \in L^1_{\mu},
$$
$$
    \psi(y)= \int_X f^y\, d\mu \in L^1_{\lambda},
$$
and the above equation {\rm (Fubini)} holds.
\end{itemize}
\end{theorem}

Note that equation (Fubini) can be written as
\begin{align}
  \int_{X\times Y} f\, d(\mu \times \lambda) &=
   \int_X \biggl(\int_Y f(x,y)\, d\lambda(y)\biggr) \, d\mu(x)\nonumber\\
   &=\int_Y \biggl(\int_X f(x,y)\, d\mu(x)\biggr) \,
   d\lambda(y)\tag{DoubleIter}
\end{align}
(a double integral is represented by iterated integrals).

\begin{corollary}
If one iterated integral in {\rm (DoubleIter)} exists, then all the
three exist and are equal.
\end{corollary}

\begin{exercise}
Let $a_n \geq 0$ for $n=1,2,\ldots$, and for $t\geq 0$ let
$$
    N(t)=\#\{n: a_n > t\}
$$
Prove that
$$
   \sum_{n=1}^{\infty} a_n = \int_0^{\infty} N(t)\, dt
$$
For an extra credit, find a formula for $\sum_{n=1}^{\infty}
\phi(a_n)$ in terms of $N(t)$ as defined above, if $\phi:[0,\infty)
\rightarrow [0,\infty)$ is locally absolutely continuous (i.e., AC
on every finite interval), non-decreasing, and $\phi(0)=0$ (Hint:
use the above fact with $\phi(x)=x$).
\end{exercise}

\begin{exercise}
Let $f\in L^1([0,1])$ and $f \geq 0$. Show that
$$
   \int_0^1 \frac{f(y)}{|x-y|^{1/2}}\, dy
$$
is finite for a.e.\ $x \in [0,1]$ and integrable.
\end{exercise}

\begin{exercise}
Use Fubini's theorem and the relation
$$
   \frac{1}{x}=\int_0^{\infty}{e}^{-xt}\, dt\quad\text{for}\ \ x>0
$$
to prove that
$$
   \lim_{A\rightarrow \infty} \int_0^A
   \frac{\sin x}{x} \, dx=\frac{\pi}{2}
$$
\end{exercise}

It is instructive to see a few counterexamples that show that the
assumptions of Fubini's theorem cannot be dispensed with.

First, let $X=Y=\IN$ and $\mu=\lambda$ counting measure. Then
$\mu\times\lambda$ is the counting measure on $X\times Y$. Define a
function $f(x,y)$ on $X\times Y$ as follows: $f(i,j)=a_{ij}$ for
$i,j\geq 1$, where
$$
   a_{ij}=\left\{
   \begin{array}{cl} +1 & \text{if}\ \ i=j\\
   -1 & \text{if}\ \ i=j+1\\
   0 & \text{otherwise} \end{array}\right.
$$
Then
$$
   \int_X \biggl(\int_Y f(x,y)\, d\lambda(y)\biggr) \, d\mu(x)=
   \sum_i \sum_j a_{ij} = 1
$$
but
$$
   \int_Y \biggl(\int_X f(x,y)\, d\mu(x)\biggr) \, d\lambda(y)
   =\sum_j \sum_i a_{ij} =0.
$$
Incidentally, we found a sequence $\{a_{ij}\}$ such that $\sum_i
\sum_j a_{ij} \neq \sum_j \sum_i a_{ij}$. The reason why Fubini's
theorem does not apply is that $f \notin L^1_{\mu\times\lambda}$.

See a Lebesgue-measure version of this example in Rudin (page 166).

The product measure $\mu \times \lambda$ is not necessarily
complete. In fact, if there is a non-empty null set $A \subset X$
and a non-measurable set $B \subset Y$, then $A\times B$ is not
measurable but it is a subset of a null set $A \times Y$. In
particular, if $X = Y = \IR$ and $\mu = \lambda = \bm$ the Lebesgue
measure, then $\bm \times \bm$ is not a complete measure, thus it is
not a Lebesgue measure on $\IR^2$.

\begin{theorem}
Let $\bm_k$ denote the Lebesgue measure on $\IR^k$. Then the
completion of $\bm_s \times \bm_t$ is the Lebesgue measure
$\bm_{s+t}$.
\end{theorem}

There is an alternative version of Fubini's theorem, adapted to
complete measures (in particular, to Lebesgue measures on $\IR^k$):

\begin{theorem}
Let $(X,\fM,\mu)$ and $(Y,\fN,\lambda)$ be two measure spaces with
complete $\sigma$-finite measures. Let $\bigl(X\times Y, (\fM\times
\fN)^\ast, (\mu\times\lambda)^\ast\bigr)$ be the completion of the
measure $\mu \times \lambda$. Let $f$ be an $(\fM \times
\fN)^\ast$-measurable function on $X\times Y$. Then $f_x$ is
$\fN$-measurable for almost every $x\in X$, $f^y$ is
$\fM$-measurable for almost every $y\in Y$, and all the other
conclusions of Fubini's theorem hold.
\end{theorem}

The following interesting lemma is used in the proof:

\begin{lemma}
Let $\nu$ be a positive measure on a $\sigma$-algebra $\fM$ in $X$,
and $(X,\fM^\ast,\mu^\ast)$ be its completion. Then for every
$\fM^\ast$-measurable function $f$ there exists a $\fM$-measurable
function $g$ such that $f=g$ $\nu$-almost everywhere.
\end{lemma}


\begin{definition}
Let $f\colon \IR \to \IC$ and $g\colon \IR \to \IC$ be two Lebesgue
measurable functions. Then their \textbf{convolution} $h = f *g$ is
defined by
$$
   h(x) = \int_{-\infty}^{\infty} f(x-t) \, g(t) \, dt
$$
(whenever the integral exists).
\end{definition}

\begin{theorem}
Let $f,g \in L^1(\IR)$. Then
$$
   \int_{-\infty}^{\infty} |f(x-t) \, g(t)| \, dt < \infty
$$
for almost every $x\in \IR$ and $h=f*g\in L^1(\IR)$ with
\beq
   \|h\|_1 \leq \|f\|_1 \|g\|_1
   \tag{Young's inequality}
\eeq
\end{theorem}

We note that it is not always true that $fg \in L^1(\IR)$. For
example, consider $f = g = x^{-1/2} \chi_{[0,1]}$, then $f,g \in
L^1(\IR)$ but $fg \notin L^1(\IR)$.

If we assume that $f,g \geq 0$, then $\|h\|_1 = \|f\|_1 \|g\|_1$.

\begin{exercise}
Suppose $f \in L^1(\IR)$ and $g \in L^p(\IR)$ for some $1\leq p \leq
\infty$. Show that $f*g$ exists at a.e. $x\in \IR$ and $f*g \in
L^p(\IR)$, and prove that
$$
   \|f*g\|_p \leq \|f\|_1 \|g\|_p.
$$
Hints: the case $p=\infty$ is simple and can be treated separately.
If $p<\infty$, then use H\"older inequality and argue as in the
proof of the previous theorem.
\end{exercise}

\begin{exercise}
Let $f \in L^1(\IR)$ and
$$
   g(x)= \int_{\IR} f(y)\,e^{-(x-y)^2}\, dy.
$$
Show that $g \in L^p(\IR)$, for all $1\leq p \leq \infty$, and
estimate $\|g\|_p$ in terms of $\|f\|_1$. You can use the following
standard fact: $\int_{-\infty}^\infty e^{-x^2}\, dx=\sqrt{\pi}$.
\end{exercise}

\begin{exercise}[Bonus]
Let $E=[1, \infty)$ and $f \in L^2_{\bm} (E)$. Also assume that $f
\geq 0$ a.e.\ and define
$$
          g(x) = \int_E f(y)\, e^{-xy} \, dy.
$$
Show that $g\in L^1(E)$ and
$$
  \|g\|_1 \leq c\,\|f\|_2
$$
for some $c<1$. Estimate the minimal value of $c$ the best you can.
\end{exercise}

\begin{definition}
Let $(X,\fM,\mu)$ be a measure space with a $\sigma$-finite measure
and $f\colon X \to [0,\infty]$ a measurable function. Then
$$
     g(t)=\mu(\{x\in X\colon f(x)>t\}) = \mu\{f>t\}
$$
is called the \textbf{distribution function} of $f$.
\end{definition}

(In probability theory, where $\mu(X)=1$, the distribution function
is defined by $F(t) = \mu\{f\leq t\}=1-g(t)$.)

Note that the distribution function is monotonic non-increasing,
thus it is a Borel measurable function.


\begin{theorem}
Let $f$ and $\mu$ be as above. Suppose $\varphi \colon [0,\infty]
\to [0,\infty]$ is a monotonic non-decreasing function that is
absolutely continuous on every finite interval $[0,T]$ and satisfies
$\varphi(0)=0$ and $\lim_{t\to\infty} \varphi(t) = \varphi(\infty)$.
Then
$$
   \int_X (\varphi \circ f)\, d\mu = \int_0^{\infty}
   \mu\{f>t\}\varphi'(t)\, dt.
$$
\end{theorem}

In particular, if $\varphi(t)=t$ we obtain a useful formula
$$
   \int_X f\, d\mu = \int_0^{\infty} \mu\{f>t\}\, dt,
$$
which is sometimes given as a definition of the Lebesgue integral.

\begin{exercise}
Let $f \colon \IR \to \IR$ be Lebesgue measurable. Prove that\\
${\ \ \ }$(a) $A=\{(x,y) \in \IR^2 \ |\ y < f(x)\}$ is
Lebesgue measurable (in the two-dimensional sense)\\
${\ \ \ }$(b) Let $f\geq 0$. Is it always true that
$\int_{\IR} f\, d\bm$ equals the Lebesgue measure of $A$?\\
${\ \ \ }$(c) $\{(x,y) \in \IR^2 \ |\ y = f(x)\}$
is a null set.
\end{exercise}

\begin{definition}
For every Lebesgue measurable function $f \colon \IR^k \to \IC$
define the \textbf{maximal function} $Mf \colon \IR^k \to
[0,\infty]$ by
$$
  (Mf)(x) = \sup_{0<r<\infty} \frac{1}{\bm(B(x,r))}\,
  \int_{B(x,r)}|f|\, d\bm
$$
whenever the integral exists.
\end{definition}

Note: earlier we defined this function only for $f\in L^1 (\IR^k)$,
and in that case we proved $Mf \in L^1_W (\IR^k)$.

\begin{theorem}
If $f\in L^1 (\IR^k)$ and $Mf \in L^1 (\IR^k)$, then $f =0$ a.e.
\end{theorem}

\begin{exercise}
Let $f \colon \IR^2 \to \IR$ be such that $f_x$ is Borel-measurable
for every $x\in \IR$ and $f^y$ is continuous for every $y \in \IR$.
Prove that $f$ is Borel-measurable. (See hint on p.\ 176 in Rudin.)
\end{exercise}

\begin{theorem}[Hardy--Littlewood]
Let $1<p \leq \infty$. If $f\in L^p (\IR^k)$, then $Mf \in L^p
(\IR^k)$.
\end{theorem}

\end{document}
\end
