1. Measurable Space and Integration

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real analysis   Real and Complex Analysis  

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

The [Real Analysis] series of posts is my memo on the lecture Real Analysis (Spring, 2021) by Prof. Insuk Seo. The lecture follows the table of contents of Real and Complex Analysis (3rd ed.) by Rudin, with minor changes in order.

In the first chapter, we define measurablility, measure, Borel space and integration with respect to a measure. The first part will be a mere list of definitions and properties and I will skip proofs that are easy to be shown.


Measurable space

We want to define a “measure” that measures the “volumes” of subsets of $\mathbb{R}.$ By volumes, i meant a function $\mu$ with the following properties.

\[\begin{aligned} &\text{(i) } \mu([0,1]) = 1.\\ &\text{(ii) } A,B \text{ are disjoint } \implies \mu(A\cup B) = \mu(A) + \mu(B). \\ &\text{(iii) } A \equiv B \implies \mu(A) = \mu(B). \end{aligned} \tag{1}\]

We start by defining measurable spaces: candidates for a domain of “measure” functions.


For a set $X$, a collection of subsets of $X$ $\mathfrak{M}$ is a $\sigma$-algebra of $X$ if
(i) $X \in \mathfrak{M}.$
(ii) $A \in \mathfrak{M} \implies A^c \in \mathfrak{M}.$
(iii) $A_1,A_2,\cdots \in \mathfrak{M} \implies \cup_{i=1}^\infty A_i \in \mathfrak{M}.$

We say a set $A$ is measurable if $A \in \mathfrak{M}.$ For any set $X$, we can easily derive two $\sigma$-algebras.

  1. $\mathfrak{M}_1 := \{\phi, X\}.$
  2. $\mathfrak{M}_2 := 2^X.$

The first one is called the trivial $\sigma$-algebra, and the second is called the discrete $\sigma$-algebra. Measures in trivial $\sigma$-algebra of $\mathbb{R}$ is unimportant since their properties are not so complicated. Measure with property (1) in discrete $\sigma$-algebra of $\mathbb{R}$ cannot be defined and it is proved as the Banach-Tarski paradox. Our focus of interest will be non-trivial $\sigma$-algebras that are inbetween these two. As an example of such, we can easily prove that $\mathfrak{M}_3 := \{A \subset X: A \text{ or } A^c \text{ is countable}\}$ is a $\sigma$-algebra. ($\mathfrak{M}_3$ is sometimes called a countable-cocountable $\sigma$-algebra.)

The following properties are direct from the definition and basic properties of set operations.

(i) $\phi \in \mathfrak{M}.$
(ii) $A_1,A_2,\cdots \in \mathfrak{M} \implies \cap_{i=1}^\infty A_i \in \mathfrak{M}.$
(iii) $A, B \in \mathfrak{M} \implies A\setminus B \in \mathfrak{M}.$


Measurable functions


For a set $X$, a collection of subsets $\tau$ is a topology on $X$ if
(i) $\phi, X \in \tau.$
(ii) $A_i \in \tau,~ i \in I \implies \cup_{i \in I} A_i \in \tau.$
(iii) $A_1,\cdots,A_n \in \tau \implies \cap_{i=1}^n A_i \in \tau.$

We call $(X, \tau)$ a topological space and say a subset $V$ is open in $X$ if $V \in \tau.$ Similar to that from the above, we can define trivial and discrete topology for all sets. In addition, we can always define topology from a metric space $(X, d)$ by defining open sets as sets that are countable unions of open balls.

Properties of functions usually follows, and named after, that of image/preimage of them. Measurable functions are functions that have inverse images of open sets as measurable sets.


$(X, \mathfrak{M})$ is a measurable space, $(Y, \tau)$ is a topological space. $f: X \to Y$ is a measurable function if $f^{-1}(V) \in \mathfrak{M}$ for all $V \in \tau.$

Our first theorem states that composition with continuous function preserves measurability of functions.

$X$ is a measurable space and $Y$ is a topological space. $g: Y \to Z$ is continuous. Then
(i) $f: X \to Y$ is continous $\implies$ $(g\circ f)$ is continuous.
(ii) $f: X \to Y$ is measurable $\implies$ $(g \circ f)$ is measurable.

Another useful theorem is about composition with multivariate function.

$f, g : X \to \mathbb{R}$ are measurable functions. $\Phi: \mathbb{R}^2 \to Y$ is a continuous mapping to topological space $Y.$ Then $h(x) := \Phi\left(f(x), g(x)\right)$ is measurable.

The proof uses the general fact that any open set in $\mathbb{R}^2$ can be represented as a countable union of rectangles. Theorem (1.8) is important since it results in common operations such as addition to retain measurability of functions. In addition, since any complex function $f$ can be represented with two real functions $u, v$ as $f = u + iv$, it is (mostly) enough to study real functions.

If $f: X \to \mathbb{C}$ is measurable, there exists a complex measurable function $\alpha: X \to \mathbb{C}$ such that
(i) $|\alpha|=1.$
(ii) $f = \alpha |f|.$

Let $\alpha = \varphi(f + \chi_E)$, where $E = \{f=0\}$ and $\varphi(z) = z/|z|.$ Then $\alpha$ is measurable and satisfies the conditions.

Limits of measurable functions are also measurable, if exists. For the discussion, we think of functions in extended real number system $\overline{\mathbb{R}}$. To prove so, we need this lemma.

For $f: X \to \overline{\mathbb{R}},$ $f$ is measurable if and only if $f^{-1}(a, \infty] \in \mathfrak{M}$ for all $a \in \mathbb{R}.$

With the help of the lemma, we get

Let $(f_n: X \to \overline{\mathbb{R}})_{n=1}^\infty$ be a sequence of measurable functions. Then the followings are also measurable.
(i) $\sup_n f_n.$
(ii) $\inf_n f_n.$
(iii) $\limsup_n f_n.$
(iv) $\liminf_n f_n.$

Since $\inf_n f_n = - \sup_n (-f_n)$ and $\liminf,$ $\limsup$ are defined as $\sup$ and $\inf,$ it is enough to prove that $\sup_n f_n$ is measurable. It comes directly from $$ \{\sup_n f_n > a \} = \{ f_n > a \text{ for some } n\in\mathbb{N}\}. $$

And finally,

If $f_n$'s are measurable and $\lim_n f_n = f,$ then $f$ is measurable.

Using the facts, we can generalize positive functions to functions. Details are on the remark.


(i) If $f,g$ are measurable, then $\max\{f,g\}$ and $\min\{f,g\}$ are measurable.
(ii) $f$ is measurable if and only if $f^+, f^-$ are measurable, where $$ f^+ := \max\{f, 0\},~ f^- := -\min\{f, 0\}. $$

Rationale behind decomposition of $f$ into $f^+$ and $f^-$ not by other positive functions are by the following proposition.

Let $g, h$ be positive functions so that $f = g - h.$ Then $g \ge f^+$ and $h \ge f^-.$

Thus $f^+$ and $f^-$ is a positive functions that decomposes $f$ without superflous information.


Borel $\sigma$-algebra

In real number system, in many cases we want open sets to be measurable sets. Borel $\sigma$-algebra is the minimal $\sigma$-algebra that contains all open sets. The existence of it is trivial by the theorem.

For a set $X$ and a collection of subsets $\mathcal{F},$ there exists the smallest $\sigma$-algebra $\mathfrak{M}^*$ that contains $\mathcal{F}.$

$$ \mathfrak{M}^* := \bigcap_{\mathcal{F} \subset \mathfrak{M},\\\mathfrak{M} \text{ is a } \sigma\text{-alg}} \mathfrak{M}. $$

For a topological space $(X, \tau),$ $\mathcal{B}:= \cap_{\tau \subset \mathfrak{M}} \mathfrak{M}$ is the Borel $\sigma$-algebra. $B \in \mathcal{B}$ is a borel set. For a topological space $Y,$ $f:X\to Y$ is a borel function if $f^{-1}(V) \in \mathcal{B}$ for all open $V \subset Y.$

Borel $\sigma$-algebras are important non-trivial $\sigma$-algebras that are inbetween $\mathfrak{M}_1$ and $\mathfrak{M}_2.$ Notable examples of Borel sets include, but not limited to $F_\sigma$ and $G_\delta$ sets. By definition, all continuous functions are Borel functions. On $\mathbb{R},$ Borel sets include sets of the form $(a, b],$ [a, b) and $\{\frac{1}{n}: n\in\mathbb{N}\}.$ In fact, almost all sets that we can imagine are Borel sets.

The next theorem generalizes the results until now.

$(X, \mathfrak{M})$ is a measurable space, $Y, Z$ are topological spaces, and $Y$ is equipped with Borel $\sigma$-algebra $\mathcal{B}.$ Then
(i) For $f: X \to Y,$ $\Omega:=\{E \subset Y: f^{-1}(E) \in \mathfrak{M}\}$ is a $\sigma$-algebra.
(ii) For a measurable $f: X \to Y,$ $f^{-1}(E) \in \mathfrak{M}$ for $E \in \mathcal{B}.$
(iii) For a measurable $f: X \to Y$ and a Borel $g: Y \to Z,$ $h:=(g \circ f)$ is measurable.


Measures

For simplicity, denote $\overline{\mathbb{R}^+}=[0, \infty].$

For a measurable space $(X, \mathfrak{M}),$ a function $\mu: \mathfrak{M} \to \overline{\mathbb{R}^+}$ is a measure, if
(i) ($\sigma$-additivity) $A_1,A_2,\cdots \in \mathfrak{M}$ are disjoint $\implies \mu(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i).$
(ii) $\mu(E) < \infty$ for some $E \in \mathfrak{M}.$

We say a measurable space equipped with a measure $(X, \mathfrak{M}, \mu)$ a measure space. The second condition is for non-trivial result (to avoid $\mu(E)=\infty$ for all measurable $E$). Throughout this lecture, we merely denote “measures” for positive measures unless otherwise noted. If codomain of such function is $\mathbb{R},$ we call it a real measure. Similarly, for such function of codomain $\mathbb{C},$ we call it a complex measure.

(i) $\mu(\phi) = 0.$
(ii) $A_1,\cdots,A_n$ are disjoint $\implies$ $\mu(\cup_{i=1}^n A_i) = \sum_{i=1}^n \mu(A_i).$
(iii) $A \subset B$ $\implies$ $\mu(A) \le \mu(B).$
(iv) (continuity from below) $A_n \uparrow A$ $\implies$ $\mu(A_n) \uparrow \mu(A).$
(v) (continuity from above) $A_n \downarrow A,$ $\mu(A_1) < \infty$ $\implies$ $\mu(A_n) \downarrow \mu(A).$

(i) through (iii) are trivial. For (iv), consider $B_i := A_i \setminus A_{i-1}$ for $i \ge 2.$ For (v), consider $C_i := A_1 \setminus A_i.$

Examples of measures include counting measure and Lebesgue measure. Counting measure $c$ counts the number of elements of sets. To be specific, For a measurable space $(X, \mathfrak{M})$ where $\mathfrak{M}=2^X,$

\[c(E) = \begin{cases} |A|,~ A \text{ is finite.} \\ \infty,~ A \text{ is infinite.} \end{cases}\]

Lebesgue measure $m$ (on $\mathbb{R}$) is the one that satisfy (1) and will be our focus in the later chapter.

\[m\left((a,b)\right) = b-a,~~ a<b\in\mathbb{R}.\]


Lebesgue integration

Now that we have a measure space, we define integration with respect to measures, or the Lebesgue integration. When it comes to defining the integration, since any measurable $f$ can be decomposed into positive functions $f^+$ and $f^-,$ it suffices to define integrals for positive functions.



$s: X \to \mathbb{R},$ $s(x) = \sum_{i=1}^n \alpha_i \chi_{A_i}(x)$ is a simple function, if $A_1,\cdots,A_n \in \mathfrak{M}$ is a partition of $X$ and $\alpha_i \ne \alpha_j$ for all $i \ne j.$


A simple function $s$ is measurable, since measurability of $A_i$’s implies measurability of $\chi_{A_i}$’s. We use a construction scheme called the standard machine to define Lebesgue integral. Detailed flow and definition is described in the earlier blog post, so I will not cover it again.

As I discussed at the end of the section of the linked post, (seamingly straightforward) additive property

\[\int f + g ~d\mu = \int f d\mu + \int g d\mu\]

requires a convergence theorem to be proved. We will prove it right at the next section. For now, additive property for integrals of simple functions can be easily shown.


Let $s,t: X \to \overline{\mathbb{R}^+}$ be simple functions and $E \in \mathfrak{M}.$ Then $$ \int_E s+t ~d\mu = \int_E s d\mu + \int_E t d\mu. $$


Convergence theorems

Monotone convergence theorem

Convergence theorems will be the most used utility throughout all chapters. Monotone convergence theorem (MCT) will be our starting point. Proof of MCT requires two lemmas.


Let $s: X \to \overline{\mathbb{R}^+}$ be a simple function. Define $\varphi(E) = \int_E s~d\mu$ for all measurable $E.$ Then $\varphi$ is a measure on $(X, \mathfrak{M}).$
We write $E_n \uparrow E$ if $E = \cup_{n=1}^\infty E_n.$ If $E_n \uparrow E$ and $s$ is a simple function, then $\int_{E_n} s d\mu \uparrow \int_E s d\mu.$


Let $(f_n)_{n=1}^\infty$ be a sequence of positive measurable funnctions such that $f_n \uparrow f$ to a positive function $f.$ Then
(i) $f$ is measurable.
(ii) $\int_X f_n d\mu \uparrow \int_X f d\mu.$

(i) Trivial since the pointwise limit of measurable functions is measurable.
(ii) $$ \int_X f_n d\mu \le \int_X f d\mu,~ \forall n. $$ Thus $$ \lim_n \int_X f_n d\mu \le \int_X f d\mu \tag{1}. $$ Now given a function $s$ such that $0 \le s \le f$ and $0<c<1,$ observe that $E_n := \\{ f_n \ge c\cdot s \\}$ is a measurable set. Since $f_n \uparrow f,$ $E_n \uparrow X$ as $n \to \infty.$ $$ \begin{aligned} &\int_{E_n} c \cdot s ~d\mu \le \int_{E_n} f_nd\mu \le \int_X f_nd\mu \\ \overset{n\to\infty}\implies& \int_X c\cdot s ~d\mu \le \lim_n \int_X f_n d\mu \\ \overset{c\to1, \\ \sup_{0\le s\le f}}\implies& \int_X f d\mu \le \lim_n \int_X f_n d\mu. \end{aligned} \tag{2} $$ By (1) and (2), we get the desired result.


Results from MCT

Now the additive property of Lebesgue integral is direct from MCT.


Let $f,g: X \to \overline{\mathbb{R}^+}$ be measurable functions. Then $$ \int_X f+g ~d\mu = \int_X f d\mu + \int_X gd\mu. $$


Interchangeability between integral and infinite sum is also clear for positive measurable functions.


Let $(f_n)_{n=1}^\infty$ be a sequence of positive measurable functions. Then $$ \sum_{n=1}^\infty \int_X f_nd\mu = \int_X \sum_{n=1}^\infty f_n d\mu. $$


Similar to lemma 13, a measure can be constructed using any positive measurable function. (It will be covered in detail later as a Radon-Nikodym derivative.)


Let $f: X \to \overline{\mathbb{R}^+}$ be a measurable function and $E \in \mathfrak{M}.$ Define $$ \varphi(E) := \int_E f~d\mu, $$ then $\varphi$ is a measure and $$ \int_X g ~d\varphi = \int_X g \cdot f~d\mu. $$


Fatou’s lemma

In order to use MCT, a sequence $f_n$ must converges to some function $f.$ Fatou’s lemma relieves the condition and gives useful tool.


Let $(f_n)_{n=1}^\infty$ be a sequence of measurable functions. Then $$ \int_X \liminf_n f_n d\mu \le \liminf_n \int_X f_n d\mu. $$

$$ g_n := \inf_{k \ge n} f_k $$ Use MCT with the fact $f_n \ge g_n.$


Dominated convergence theorem

Another useful convergence theorem is dominated convergence theorem (DCT).


Let $f,g \in L^1(\mu),$ then
(i) $cf \in L^1(\mu)$ for $c\in\mathbb{R}$ and $\int cf~d\mu = c\int f ~d\mu.$
(ii) $f+g \in L^1(\mu)$ and $\int f+g~d\mu = \int f ~d\mu + \int g ~d\mu.$
(iii) $|\int f ~d\mu| \le \int |f|d\mu.$


Let $(f_n)_{n=1}^\infty$ be a sequence of measurable functions such that $f_n \to f.$ Suppose there exists a measurable $g$ such that $|f_n| \le g$ for all $n.$ Then
(i) $f \in L^1(\mu).$
(ii) $\int_X |f_n-f| ~d\mu \to 0$ as $n \to \infty.$
(iii) $\int_X f_nd\mu \to \int_X f~d\mu$ as $n\to\infty.$

(i) Clearly, $|f| \le g.$
(ii) $$ |f_n - f| \le |f_n| + |f| \le 2g, \\ 2g - |f_n - f| \ge 0. $$ By Fatou's lemma, $$ \begin{aligned} \int_X \liminf_n \left( 2g - |f_n - f|\right) d\mu &= \int_X 2g~d\mu \\ &\le \liminf_n \int_X 2g-|f_n-f| ~d\mu \\ &= \int_X 2g~d\mu - \limsup_n \int_X|f_n - f| ~d\mu \end{aligned} $$ hence $$ \limsup_n \int_X |f_n - f| ~d\mu \le 0. $$ (iii) Trivial by traingle inequality.


Complete measure space


We say a measure space $(X, \mathfrak{M}, \mu)$ is complete, if $$ E \in \mathfrak{M},~ \mu(E)=0 \implies N \in \mathfrak{M} \text{ and } \mu(N) = 0,~ \forall N \subset E. $$


One can build a complete measure space from a given measure space. We call such complete measure space a completion of it.


$$ \mathfrak{M}^* := \left\{E: A\subset E\subset B \text{ for some } A,B\in\mathfrak{M} \text{ s.t. } \mu(B\setminus A)=0\right\} \\ \mu^*(E) := \mu(A),~ \forall E \in \mathfrak{M}^* $$ Then (i) $\mu^*$ is a well-defined measure and (ii) $(X, \mathfrak{M}^*, \mu^*)$ is a complete measure space.

(i) It is enough to show that for given sets $A, B, A', B'$ such that $$ A\subset E\subset B,~ \mu(B-A) = 0 \\ \text{ and } A'\subset E\subset B', \mu(B'-A')=0, $$ we get $\mu(A) = \mu(A').$ $$ A-A' \subset E-A' \subset B'-A' \implies \mu(A-A') \le \mu(B'-A')=0, \\ A'-A \subset E-A \subset B-A \implies \mu(A-A') \le \mu(B-A)=0. $$ Thus the result follows.
(ii) First we show that $\mathfrak{M}^*$ is a $\sigma$-algebra. It is not difficult to show that $X \in \mathfrak{M}^*$ and $E^c \in \mathfrak{M}^*$ for $E \in \mathfrak{M}.$ Suppose $E_1,E_2,\cdots \in \mathfrak{M}^*.$ Then for given $i,$ there exist $$ A_i \sub E_i \sub B_i,~ \mu(B_i-A_i) = 0. $$ Then $$ \cup_{i=1}^\infty A_i \sub \cup_{i=1}^\infty E_i \sub \cup_{i=1}^\infty B_i,~ \cup_{i=1}^\infty A_i, \cup_{i=1}^\infty B_i \in\mathfrak{M}^* $$ and $$ \begin{aligned} \mu(\cup_{i=1}^\infty B_i - \cup_{i=1}^\infty A_i) &\le \sum_{i=1}^\infty \mu(B_i-A_i) = 0. \end{aligned} $$ (iii) $$ \begin{aligned} \mu^*(\cup_{i=1}^\infty E_i) = \mu(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i) = \sum_{i=1}^\infty \mu^*(E_i). \end{aligned} $$


Sets of Measure zero

Almost everywhere


we call that the statement $S$ holds almost everywhere (a.e.), if there exists $E \in \mathfrak{M}$ such that $S$ hold on $E$ and $\mu(E^c) = 0.$



Let $E \in \mathfrak{M}$ such that $\mu(E^c)=0.$ Then $f: E \to Y$ is a measurable function on $X,$ if $$ f^{-1}(V\cap E) \in \mathfrak{M},~ \forall \text{ open } V \sub Y. $$

If we define $\tilde f$ as an extension of $f$ such that $\tilde f(x) = 0$ for all $x \in E^c,$ then $\tilde f$ is measurable in terms of previous definition.

If the measure space is complete, any arbitrary extension of $f$ is measurable and \(\int_X \tilde f ~d\mu = \int_E f ~d\mu.\) One can show that properties and theorems we discussed earlier in this chapter also holds for conditions relieved to be almost everywhere.


Almost everywhere on $E$


The statement $S$ holds a.e. on $E \in \mathfrak{M}$, if $$ \mu \left( \{x \in E: S \text{ does not hold at } x\} \right). $$ That is, $$ \mu (S^c \cap E) = 0. $$


(i) $f \ge 0$ on $E \in \mathfrak{M},$ $\int_E f d\mu=0$ $\implies$ $f=0$ a.e. on $E.$
(ii) $f \in L^1(\mu),$ $\int_E f d\mu = 0,$ $\forall E \in \mathfrak{M}$ $\implies$ $f=0$ a.e.


References

  • Rudin. 1986. Real and Complex Analysis. 3rd edition. McGraw-Hill.
  • Real Analysis (Spring, 2021) @ Seoul National University, Republic of Korea (instructor: Prof. Insuk Seo).