0x022 Real Analysis

1. Measures
2. Integration
3. Differentiation
4. Product Measures
- 4.1. Products of Measure Spaces
  - 4.1.1. Monotone Class Theorem
5. Real and Complex Measures
- 5.1. Total Variation
6. Reference

1. Measures

We need to know how to measure the "size" or "volume" of subsets of a space $X$ before we can integrate function $f$.

There is an analogy between:

topological space, open set, continuous function
measurable space, measurable set, measurable function.

The main conclusions here are the Littlewood three principles:

Every measurable set is nearly a finite sum of intervals
(Luzin) Every function (of class L^p) is nearly continuous
(Egorov) Every convergent sequence of functions is nearly uniformly convergent.

1.1. Measurable Space

1.1.1. Sigma Algerbra

Definition ($\sigma$-algebra) $X$ is a set and $S$ is a set of subsets of $S$, $S$ is called $\sigma$-algebra when the following conditions are satisfied:

$\emptyset \in S$
$E \in S \implies X \setminus E \in S$
$E_1, E_2, ... \in S \implies \bigcup_{k=1}^{\infty} E_i \in S$

From the previous definition, it is easy to show that $\sigma$-algebra is closed under countable union

trivial $\sigma$-algebra

Suppose $X$ is a set, then $\{ \emptyset, X \}$ is a $\sigma$-algebra, the power set of $X$, $2^X$ is also a $\sigma$-algebra.

$\sigma$-algebra vs topology

As mentioned, $\sigma$-algebra is closed under countable union and countable intersection. It should be contrast with a topology $\tau \subset \mathcal{P}(X)$, where

$\emptyset \in \tau, X \in \tau$
closed under finite intersection: $U_i \in \tau \implies \cap_{i=1}^{n} U_i \in \tau$
closed under union of arbitrary collection: $U_{\alpha} \in \tau \implies \cup_{\alpha} U_{\alpha} \in \tau$

Definition (measurable space; measurable set) A measurable space is an ordered pair $(X, S)$, where $X$ is a set and $S$ is a $\sigma$-algebra on $X$. An element of $S$ is called an $S$-measurable set.

1.1.2. Construction

Definition (generation of sigma algebra) If $\mathcal{F}$ is any collection of subsets of a set $X$, then the $\sigma$-algebra generated by $\mathcal{F}$ is

\[\sigma(\mathcal{F}) = \cap \{ \mathcal{A} \subset \mathcal{P}(X) | \mathcal{F} \subset \mathcal{A} \land \mathcal{A} \text{ is sigma algebra} \}\]

The $\sigma$-algebra generated by open subsets is the Borel subsets

Definition (Borel $\sigma$-algebra on topological space) Let $(X, \mathcal{T})$ be a topological space. The Borel $\sigma$-algebra is

\[\mathcal{B}(X) = \sigma(\mathcal{T})\]

In the case of metric space, Borel subsets might be described generatively. We apply the transfinite induction to the collection of open sets by repeating countable union and countable intersection.

An important example of Borel $\sigma$-algebra is on the set of real. It is, for example, used in the probability theory.

Definition (Borel subsets on R) The smallest $\sigma$-algebra on $R$ containing all open subsets of $R$ is called the collection of Borel subsets of $R$. An element of this $\sigma$-algebra is called a Borel set.

example of Borel sets

every open, half, closed subset is a Borel set.

$F_\sigma, G_\delta$ are Borel set.

every countable subset $\{ x_1, x_2, ... \}$ of $R$ is a Borel set because it can be written as

\[B = \cup^{\infty}_i \{ x_i \}\]

cardinality of Borel subsets on R

The cardinality of the collection of Borel sets is equal to the continuum

\[\aleph_1 2^{\aleph_0} = 2^{\aleph_0}\]

1.1.3. Completion

Borel subsets are not complete, completing it by enlarging it by including all subsets of sets of measure 0 introduces the Lebesgue measurable subsets. This is a special case of completing measure.

It can be interpreted as the smallest $\sigma$-algebra on $R$ containing the Borel sets and the sets with outer measure 0

Definition (Lebesgue Measurable set) A set $A \subset R$ is called Lebesgue Measurable if there exists a Borel set $B \subset A$ such that $| A \setminus B|=0$

There are several equivalent definitions:

For each $\epsilon > 0$, there exists a closed set $F \subset A$ with $|A \setminus F | < \epsilon$
There exist closed sets $F_1, F_2, ..., F_2$ contained in $A$ such taht $| A \setminus \cup_{k=1}^\infty F_k| = 0$
For each $\epsilon > 0$, there exists an open set $A \subset G$ such that $G \setminus A < \epsilon$
There exist open sets $G_1, G_2, ...$ containing $A$ such taht $| \cap_{k=1}^\infty G_k \setminus A| = 0$

1.2. Measure Space

1.2.1. Measure

Definition (measure) Suppose $X$ is a set and $S$ is a $\sigma$-algebra on $X$. A measure $\mu$ on $(X,S)$ is a function such that

(non-negativity): $\mu(E) \geq 0$
(empty set has measure 0) $\mu(\emptyset) = 0$
(countable additivity) $\mu(\bigcup_{k=1}^{\infty} E_k) = \sum_{k=1}^{\infty} \mu(E_k)$

Definition (measure space) A measure space is an ordered triple $(X, S, \mu)$, where $X$ is a set, $S$ is a $\sigma$-algebra, and $\mu$ is a measure on $(X,S)$

counting measure

If $X$ is a set, then counting measure is the measure $\mu$ defined on the $\sigma$-algebra of all subsets by setting

\[\mu(A) = \begin{cases} |A| & \text{if A is finite} \\ \infty & \text{if A is infinite} \end{cases}\]

where $|A|$ is the cardinality of the set $|A|$

outer measure

outer measure is NOT a measure on $(\mathbb{R}, \mathcal{F})$ where $\mathcal{F}$ is all subsets of $\mathbb{R}$ as it does not satisfy finitely additivity.

but it is a measure on $(\mathbb{R}, \mathcal{B})$ where $\mathcal{B}$ is all Borel subsets

Dirac measure

Suppose $(X, S)$ is a measurable space and $c \in X$, the Dirac measure $\delta_c$ on $(X,S)$ is defined by

\[\delta_c(E) = \begin{cases} 1 & \text{ if } c \in E \\ 0 & \text{ otherwise} \end{cases}\]

This is one way (another way is through the generalized function) to define Dirac function rigorously, where Lebesgue integral with respect to this measure is

\[\int f(x) \delta \{ dx \} = f(0)\]

where $f$ is a continuously compactly supported function

Lemma (properties of measure) Measure has the following properties

(measure preserves order) Suppose $(X, S, \mu)$ is a measure space, $D, E \in S$ and $D \subset E$, then $\mu(D) \leq \mu(E)$
(countable subadditivity) Suppose $(X, S, \mu)$ is a measure space and $E_1, E_2, ... \in S$, then $\mu(\cup_{k=1}^{\infty} E_i) \leq \sum_{k=1}^{\infty} \mu(E_i)$
(measure of increasing sequence) Suppose $(X, S, \mu)$ is a measure space and $E_1 \subset E_2 \subset$ is an increasing sequence of sets in $S$. Then $\mu(\cup_{k=1}^{\infty} E_k) = \lim_{k \to \infty} \mu(E_k)$
(measure of a union) Suppose $(X, S, \mu)$ is a measure space and $D,E \in S, \mu(D \cap E) < \infty$, then $\mu (D \cup E) = \mu(D) + \mu(E) - \mu(D \cap E)$

Note the countable additivity property applies to disjoint countable unions, the following countable subadditivity deals with countable unions that may not be disjoint unions.

1.2.2. Outer Measure

Definition (length of open interval) The length $l(I)$ of an open interval $I$ is defined by

\[ l(I) = \begin{cases} b-a & \text{for } I = (a,b) \\ 0 & \text{if } I = \emptyset \\ \infty & \text{if } I = (-\infty, a) \text{ or } I = (a, \infty) \\ \infty & \text{if } I = (-\infty, \infty) \end{cases}\]

Definition (outer measure, $|A|$) The outer measure $|A|$ of a set $A \subset R$ is defined by

\[ |A| = \inf \{ \sum_{k=1}^{\infty} l(I_k): I_1, I_2, ... \text{ are open intervals and covers } A \}\]

countable set has measure 0

Every countable subset of $R$ has outer measure 0. Each item can be covered by a open interval

\[I_k = (a_k - \epsilon/2^k, a_k+ \epsilon/2^k)\]

where

\[\sum l(I_k) = 2 \epsilon\]

Cantor set, uncountable set whose measure is 0

Note uncountable set can also have measure 0. The example is the Cantor set.

Lemma (properties of outer measure)

(outer measure preserves order) $A \subset B \implies |A| \leq |B|$
(outer measure is translation invariant) Suppose $t \in R, A \subset R$, then $|t+A| = |A|$
(countable subadditivity) Suppose $A_1, A_2, ...$ is a sequence of subsets of $R$. Then $|\bigcup_{k=1}^{\infty} A_k | \leq \sum_{k=1}^{\infty} |A_k|$

Theorem (outer measure of a closed interval) Suppose $a, b \in R, a < b$, then

\[|[ a, b ] | = b-a\]

note that this can be proved with Heine-Borel

nonadditivity of outer measure

There exists disjoint subsets $A, B \subset R$ such that

\[|A \cup B| \neq |A| + |B|\]

note that this can be proved using AoC

Outer measure, despite its name, is not a measure on the $\sigma$-algebra of all subsets of $R$. It is, however, a measure when restricted to the Borel subsets of $R$ (and also a somewhat larger class of sets called the Lebesgue measurable sets)

Theorem (outer measure is a measure on Borel sets) Outer measure is a measure on $(R,B)$ where $B$ is the $\sigma$-algebra of Borel subsets of $R$

1.2.3. Lebesgue Measure

Definition (Lebesgue Measure wrt Lebesgue Measurable set) Lebesgue Measure is the measure on $(R, L)$ where $L$ is the $\sigma$-algebra of Lebesgue measurable subsets of $R$, that assigns to each Lebesgue measurable set its outer measure.

1.3. Measurable Function

Measurable function is a function between the underlying sets of two measurable spaces that preserves the structure of the spaces: the preimage of any measurable set is measurable. It is similar to the continuous function which preserves the topological structure.

Note that measurability of a function depends on the $\sigma$-algebra.

Definition (measurable function) Suppose $(X,S)$ is a measurable space. A function $f: X \to R$ is called S-measurable function if for every Borel set $B \subset R$

\[f^{-1}(B) \in S\]

In a more general definition, let $(X, \Sigma), (Y, T)$ be measurable spaces. A function $f: X \to Y$ is said to be measurable iff for every $E \in T$, the preimage of $E$ under $f$ is in $\Sigma$

\[f^{-1}(E) \in \Sigma\]

measurable functions

If $S = \{ \emptyset, X \}$, then the only $S$-measurable function from $X \to R$ are the constant functions.

If $S$ is the power set of $X$, then every function from $X \to R$ is S-measurable.

If $S$ is the set of Borel subsets, then we call it the Borel measurable function

Definition (Borel measurable function) Suppose $X \subset R$. A function $X \to R$ is called Borel measurable if $f^{-1}(B)$ is a Borel set for every Borel set $B \subset R$

1.3.1. Construction

Checking the preimage of every Borel subset is not easy, the following one is a condition to check a much smaller collection of subsets. Recall this is analogous to the topological continuity condition to check the inverse of basis element instead of checking every open set

Criterion (measurable function) Suppose $(X,S)$ is a measurable space and $f: X \to R$ is a measurable function iff

\[f^{-1}((a, \infty)) \in S\]

Intuitively, the inverse set preserves $\sigma$-algebra, i.e. $\{ A \subset R | f^{-1}(A) \in S \}$ is a $\sigma$-algebra if $S$ is a $\sigma$-algebra, under the given condition, it can construct all Borel sets in it.

continuous functions are Borel measurable

Every continuous real-valued function defined on a Borel subset is a Borel measurable function.

Proof: $(a, \infty)$ is open, by the property of continuity, we know the preimage is also open and must be included in Borel set by definition.

monotone functions are Borel measurable

Every increasing function defined on a Borel subset of $R$ is a Borel measurable function.

Criterion (operations preserving measurable functions) Measurable functions are closed under composition, algebraic operations

Suppose $(X, S)$ is a measurable space and $f,g: X \to R$ are S-measurable. Then

$f+g$, $f-g$, $fg$ are S-measurable functions
if $\forall{x \in X} g(x) \neq 0$, then $f/g$ is an S-measurable function

functions of random variable

This criterion basically characterizes why functions of random variables is still a random variable.

For example, Suppose $X_i$ are (measurable) random variables, then all the followings are valid random variables

$X_1 + X_2 + ... + X_n$
$X_1^2$
$cX_1$
$X_1X_2$
$\inf{X_i}$

Criterion (measurable functions and limit) Suppose $f_1, f_2, ...$ is a sequence of S-measurable functions from $X$ to $R$. Suppose $\lim_{k \to \infty} f_k(x)$ exists for each $x \in X$. Let $f: X \to $R$ be

\[f(x) = \lim_{k \to \infty} f_k(x)\]

Then $f$ is also an S-measurable function

measurable function vs continuous function

Recall continuous function is not closed under pointwise limit:

Let $f_n(x)=x^n$ on the set $x \in [0,1]$ where $f_n(x)$ is continous function. when $n \to \infty$, it converges to 0 for $0 < x < 1$, 1 for $x=1$, which is not a continuous function

1.3.2. Convergence

Egorov Theorem roughly states that pointwise convergence of a sequence of measurable functions is close to uniform convergence.

Theorem (Egorov) Suppose $(X,S,\mu)$ is a measurable space and $\mu(X) < \infty$. Suppose $f_1, f_2, ...$ is a sequence of $S$-measurable function from $X$ to $R$ that converges pointwise on $X$ to a function $f$. Then for every $\epsilon > 0$, there exists a set $E \in S$ such that $\mu(X \setminus E ) < \epsilon$ and $f_1, f_2...$ converges uniformly to $f$ on $E$

Theorem (Luzin) Suppose $g: R \to R$ is a Borel measurable function. Then for every $\epsilon > 0$ there exists a closed set $F \subset R$ such that $| R \setminus F| < \epsilon$ and $g|_{F}$ is continuous function on $F$

2. Integration

Riemann Integral is not good enough. It has several deficiencies as follows:

Riemann integration does not handle functions with many discontinuities
Riemann integration does not handle unbounded functions
Riemann integration does not work well with limits

space of Riemann integrable function is not complete

Let $r_1, r_2, ...$ be a sequnece of rational number in $[0,1]$, define $f_k: [0, 1] \to R$ by $f_k(x) = 1$ if $x \in \{ r_1, ..., r_k \}$ otherwise 0.

$f_k$ is Riemann integrable, but its limit $f$ is not Riemann integrable.

For a measure space $(X, S, \mu)$, we aims at defining $\int_{A} f(x) d\mu$ for suitable measurable function $f: X \to R$ and for any $A \in S$ in a systematic manner.

2.1. Integration with respect to a Measure

Definition (S-partition) Suppose $S$ is a $\sigma$-algebra on a set $X$. An $S$-partition of $X$ is a finite collection $A_1, ..., A_m$ of disjoint sets in $S$ such that $A_i \cup .... \cup A_m = X$

Definition (lower Lebesgue sum) Suppose $(X, S, \mu)$ is a measure space, $f: X \to [0, \infty]$ is a S-measurable function and $P$ is a partition $A_1, ..., A_m$ of $X$. The lower Lebesgue sum $\mathcal{L}(f,P)$ is defined by

\[\mathcal{L}(f, P) = \sum_{j=1}^{m} \mu(A_j) \inf_{A_j} f\]

We first think about reducing integration to the simple case of integrating non-negative function $f^{+}$, later it can be extended to a general function $f$ by expressing it using two non-negative function $f = f^{+} - f^{-}$

Definition (Lebesgue Integral, integral of a nonnegative function) Suppose $(X, S, \mu)$ is a measure space and $f: X \to [0, \infty]$ is an $S$-measurable function. The integral of $f$ with respect to $\mu$ is defined by

\[ \int f d\mu = \sup \{ \mathcal{L}(f, P) | P \text{ is an S-partition of }X \}\]

Next, we think about reducing it to the case of integrating simple functions.

Integral of a simple function

Suppose $(X, S, \mu)$ is a measure space and $E \in S$, Then

\[\int \chi_{E} d\mu = \mu(E)\]

Additionally, suppose $E_1, ..., E_n$ are disjoint sets in $S$

\[\int (\sum_{k=1}^n c_k \chi_{E_k}) d\mu = \sum_{k=1}^n c_k \mu(E_k)\]

Integral wrt to counting measure

Suppose $\mu$ is counting measure on $Z^+$ and $b_1, b_2, ...$ is a sequence of nonnegative numbers. Think of $b$ as the function from $Z^+$ to $[0, \infty)$ defined by $b(k) = b_k$,

\[\int b d\mu = \sum_{k=1}^\infty b_k\]

Theorem (monotone convergence) Suppose $(X, S, \mu)$ is a measure space and $0 \leq f_1 \leq f_2 \leq ...$ is an increasing sequence of $S$-measurable functions. Define $f: X \to [0, \infty]$ by

\[f(x) = \lim_{k \to \infty} f_k(x)\]

Then

\[\lim_{k \to \infty} \int f_k(x) d\mu = \int f(x) d\mu\]

Theorem (addivity of integration) Suppose $(X, S, \mu)$ is a measure space and $f, g: X \to R$ are S-measurable functions such that $\int |f| < \infty, \int |g| < \infty$. Then

\[\int (f+g) d\mu = \int f d\mu + \int g d\mu\]

Theorem (integration is homogeneous) Suppose $(X, S, \mu)$ is a measure space and $f: X \to [-\infty, \infty]$ is a function such that $\int f d\mu$ is defined. If $c \in R$, then

\[\int c f d\mu = c \int f d\mu\]

2.2. Limits of Integrals

Definition (integration on a subset) Suppose $(X, S, \mu)$ is a measure space and $E \in S$. If $f: X \to [-\infty, \infty]$ is an $S$-measurable function, then $\int_{E} f d\mu$ is defined by

\[\int_{E} f d\mu = \int \chi_{E} f d\mu\]

Definition (almost every) Suppose $(X, S, \mu)$ is a measure space. A set $E \in S$ is said to contain $\mu$-almost every element of $X$ if $\mu( X \setminus E)=0$.

Theorem (Dominated Convergence Theorem) Suppose $(X, S, \mu)$ is a measure space, $f: X \to [-\infty, \infty]$ is $S$-measurable, and $f_1, f_2, ...$ are $S$-measurable functions from $X$ to $[-\infty, \infty]$ such that

\[\lim_{k \to \infty} f_k (x) = f(x)\]

for almost every $x \in X$. If there exists an $S$-measurable function $g: X \to [0, \infty]$ such that

\[\int g d\mu < \infty \land | f_k (x) | \leq g(x) \]

for every $k \in Z^+$ and almost every $x \in X$, then

\[\lim_{k \to \infty} \int f_k(x) d\mu = \int f d\mu\]

Dominated Convergence Theorem can easily prove the Bounded Convergence theorem:

Bounded convergence Theorem

Suppose $(X, S, \mu)$ is a measure space with $\mu(X) < \infty$, $f_1, f_2, ...$ is a sequence of $S$-measurable functions converging pointwisely to $f$. If there exists a $c \in (0, \infty)$ such that

\[(\forall{k,x}) |f_k(x)| \leq c\]

then we can exchange limit and integral

\[\lim_{k \to \infty} \int f_k d\mu = \int f d\mu\]

2.3. Space L1 of integrable functions

The integrable functions form a vector space, and this vector space is complete in the appropriate norm.

Definition ($||f||_1, \mathcal{L}^1(\mu)$) Suppose $(X, S, \mu)$ is a measure space. If $f: X \to [-\infty, \infty]$ is $S$-measurable. then the $\mathcal{L}^1$-norm of $f$ is defined by

\[||f||_1 = \int |f| d\mu\]

The Lebesgue space $\mathcal{L}^1(\mu)$ is defined by

\[\mathcal{L}^1(\mu) = \{ f: f \text{ is an S-measurable function from X to R such that } ||f||_1 < \infty\]

3. Differentiation

4. Product Measures

4.1. Products of Measure Spaces

Definition (product of $\sigma$-algebra, $S \otimes T$) Suppose $(X, S), (Y, T)$ are measurable spaces. Then the product $S \otimes T$ is defined to be the smallest $\sigma$-algebra on $X \times Y$ contains

\[\{ A \times B | A \in S, B \in T \}\]

Definition (cross sections of sets) Suppose $X, Y$ are sets and $E \subset X \times Y$. Then for $a \in X, b \in Y$, the cross sections $[E]_a, [E]^b$ are defined by

\[[E]_a = \{ y \in Y | (a, y) \in E \}\]

\[[E]^b = \{ x \in X | (x, b) \in E \}\]

Lemma (cross sections of measurable sets are measurable) Suppose $S, T$ is a $\sigma$-algebra on $X, Y$. If $E \in S \otimes T$,

\[(\forall a \in X) [E]_a \in T\]

\[(\forall b \in Y) [E]^b \in S\]

Definition (cross section of functions) Suppose $f: X \times Y \to R$, then for $a \in X, b in Y$, the cross section function $[f]_a: Y \to R, [f]^b: X \to R$ is defined as

\[[f]_a(y) = f(a, y)\]

\[[f]^b(x) = f(x, b)\]

Lemma (cross sections of measurable functions are measurable) Suppose $f: X \times Y \to R$ is an $S \otimes T$-measurable function, then, $[f]_a$ is a $T$-measurable function on $Y$ for every $a \in X$. Similar conclusion applies to $[f]^b$

4.1.1. Monotone Class Theorem

Definition (semi-algebra) Suppose $W$ is a set and $A$ is a set of subsets of $W$. A collection $A$ is said to be a semi-algebra if

if $E, F \in A$, then $E \cap F \in A$
if $E \in A$, then $W \setminus E$ is a finite disjoint union of sets of $A$

Definition (algebra) Suppose $W$ is a set and $A$ is a set of subsets of $W$. Then $A$ is called an algebra on $W$ if the following three conditions are satisfied

$\emptyset \in A$
if $E \in A$, then $W \setminus E \in A$
if $E,F \in A$, then $E \cup F \in A$

It means it is closed under complementation and under finite unions

intervals

Collection of finite unions of intervals is an algebra, however collection of countable unions of intervals is not an algebra

Definition (monotone class) Suppse $W$ is a set and $\mathcal{M}$ is a set of subsets of $W$. Then $\mathcal{M}$ is called a monotone class on $W$ when

If $E_1 \subset E_2 \subset ...$ is an increasing sequence of sets in $\mathcal{M}$, then $\cup_{k=1}^\infty E_k \in \mathcal{M}$
If $E_k$ is a decreasing sequence of sets in $\mathcal{M}$, then $\cap_{k=1}^{\infty} E_k \in \mathcal{M}$

5. Real and Complex Measures

5.1. Total Variation

Definition (total variation measure) Suppose $\mu$ is a complex measure on a measurable space $(X,S)$. The total variation measure $|\mu|$ is the function $|\mu|: S \to [0, \infty]$ defined by

\[|\mu|(E) = \sup \{ \sum_{i=1}^n |\mu(E_i)| : E_i \text{ are disjoint sets in $S$ such that } \cup_i^n E_i \subset E \}\]

6. Reference

[1] Axler, Sheldon. "Measure, Integration & Real Analysis." (2020): 411.