11.1 Measurable spaces and functions

中文

Measurable spaces

Given a set $X$, a sigma-algebra $\mathcal{M}$ on $X$ consists of a collection ofsubsets, known as measurable sets, satisfying

(Meas1) The empty set $\varnothing$ is a measurable set, $\varnothing \in \mathcal{M}$

(Meas2) If $A$ is measurable then so is its complement:

$$A\in \mathcal{M}⟹A^c=X-A\in \mathcal{M}.$$

(Meas3) $\mathcal{M}$ is closed under countable unions:

$$A_1,A_2,A_3,\cdots \in \mathcal{M}⟹\bigcup _i{A}_i\in \mathcal{M}.$$

The pair $(X,\mathcal{M})$ is known as a measurable space. Although there are similarities between these axioms and (Top1)–(Top3) for a topological space, (Meas2) is distinctly different in that the complement ofan open set is a closed set and is rarely open. It follows from (Meas1) and (Meas2) that the whole space $X={\varnothing }^c$ is measurable. The intersection of any pair of measurable sets is measurable, for

$$A\cap B=(A^c\cup B^c{)}^c.$$

Also, $\mathcal{M}$ is closed with respect to taking differences,

$$A-B=A\cap B^c=(A^c\cup B{)}^c.$$

Exercise

Show that any countable intersection of measurable sets is measurable

Example 11.1

Given any set $X$, the collection $\mathcal{M}=\{\varnothing ,X\}$ is obviously a sigma-algebra. This is the smallest sigma-algebra possible. By contrast, the largest sigma-algebra is the set of all subsets $2^X$ . All interesting examples fall somewhere between these two extremes.

It is trivial to see that the intersection of any two sigma-algebras $\mathcal{M}\cap {\mathcal{M}}^{\prime }$ is another sigma-algebra – check that properties (Meas1)–(Meas3) are satisfied by the sets common to the two sigma-algebras. This statement extends to the intersection of an arbitrary family of sigma-algebras, ${\cap }_{i\in I}{\mathcal{M}}_i$ . Hence, given any collection of subsets $\mathcal{A}\subset 2^X$ , there is a unique ‘smallest’ sigma-algebra $\mathcal{S}\supseteq \mathcal{A}$ . This is the intersection of all sigma-algebras that contain $\mathcal{A}.$ It is called the sigma-algebra generated by $\mathcal{A}.$ . For a topological space $X,$ , the sigma-algebra generated by the open sets are called Borel sets on $X$. They include all open and all closed sets and, in general, many more that are neither open nor closed.

Example 11.2

In the standard topology on the real line $\mathbb{R}$ every open set is a countable union of open intervals. Hence the Borel sets are generated by the set of all open intervals $\{(a,b)|a<b\}$ . Infinite left-open intervals such as $(a,\infty )=(a,a+1)\cup (a,a+2)\cup$ $(a,a+3)\cup ..$ . are Borel sets by (Meas3), and similarly all intervals $(-\infty ,a)$ are Borel. The complements of these sets are the infinite right or left-closed intervals $(-\infty ,a]$ and $[a,\infty )$ . Hence all closed intervals $[a,b]=(-\infty ,b]\cap [a,\infty )$ are Borel sets.

Exercise

Prove that the sigma-algebra of Borel sets on $\mathbb{R}$ is generated by (a) the infinite left-open intervals $(a,\infty )$, (b) the closed intervals $[a,b]$.

Exercise

Prove that all singletons $\{a\}$ are Borel sets on $\mathbb{R}$

If$(X,\mathcal{M})$ and $(Y,\mathcal{N})$ are two measurable spaces then we define the product measurable space $(\mathbf{X}\times \mathbf{Y},\mathcal{M}\otimes \mathcal{N})$ , by setting the sigma-algebra $\mathcal{M}\otimes \mathcal{N}$ on $X\times Y$ to be the sigma-algebra generated by all sets of the form $A\times B$ where $A\in \mathcal{M}$ and $B\in \mathcal{N}$

Measurable functions

Given two measurable spaces $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ , a map $f:X\to Y$ is said to be a measurable function if the inverse image of every measurable set is measurable:

$$A\in \mathcal{N}⟹f^{-1}(A)\in \mathcal{M}.$$

This definition mirrors that for a continuous function in topological spaces.

Theorem 11.1 · Borel measurability of continuous functions

If $X$ and $Y$ are topological spaces and $\mathcal{M}$ and $\mathcal{N}$ are the sigma-algebras of Borel sets, then every continuous function $f:X\to Y$ is Borel measurable.

Proof

Let ${\mathcal{O}}_X$ and ${\mathcal{O}}_Y$ be the families of open sets in $X$ and $Y$ respectively. We adopt the notation $f^{-1}(A)\equiv \{f^{-1}(A)|A\in \mathcal{A}\}$ for any family of sets $\mathcal{A}\subseteq 2^Y$ . Since $f$ is continuous, $f^{-1}({\mathcal{O}}_Y)\subseteq {\mathcal{O}}_X$ . The sigma-algebras of Borel sets on the two spaces are $\mathcal{M}=\mathcal{S}({\mathcal{O}}_X)$ and $\mathcal{N}=\mathcal{S}({\mathcal{O}}_Y)$ . To prove $f$ is Borel measurable we must show that $f^{-1}(\mathcal{N})\subseteq \mathcal{M}$ Let

$${\mathcal{N}}^{\prime }=\{B\subseteq Y\mid f^{-1}(B)\in \mathcal{M}\}\subset 2^Y.$$

This is a sigma-algebra on $Y$, for $f^{-1}(\varnothing )=\varnothing$ and

$$f^{-1}\left(B^c\right)=(f^{-1}(B){)}^c,f^{-1}(\bigcup _i{B}_i)=\bigcup _i{f}^{-1}(B_i).$$

Hence ${\mathcal{N}}^{\prime }\supseteq {\mathcal{O}}_Y$ for $f^{-1}({\mathcal{O}}_Y)\subseteq {\mathcal{O}}_X\subset \mathcal{M}$ . Since $\mathcal{N}$ is the sigma-algebra generated by ${\mathcal{O}}_Y$ we must have that ${\mathcal{N}}^{\prime }\supseteq \mathcal{N}$ . Hence $f^{-1}(\mathcal{N})\subseteq f^{-1}({\mathcal{N}}^{\prime })\subseteq \mathcal{M}$ as required. -

Exercise

If $f:X\to Y$ and $g:Y\to Z$ are measurable functions between measure spaces, show tha the composition $g\circ f:X\to Z$ is a measurable function

If $f:X\to \mathbb{R}$ is a measurable real-valued function on a measurable space $(X,\mathcal{M})$ , where $\mathbb{R}$ is assumed given the Borel structure of Example 11.2, it follows that the set

$$\{x\mid f(x)>a\}=f^{-1}((a,\infty ))$$

is measurable in $X$. Since the family of Borel sets on the real line is generated by the intervals $(a,\infty )$ (see the exercise following Example 11.2), this can actually be used as a criterion for measurability: $f:X\to \mathbb{R}$ is a measurable function iff for any $a\in \mathbb{R}$ the set $\{x\mid f(x)>a\}$ is measurable.

Exercise

Prove the sufficiency of this condition [refer to the proof of Theorem 11.1].

Example 11.3

If$(X,\mathcal{M})$ is a measurable space then the characteristic function ${\chi }_A:X\to \mathbb{R}$ of a set $A\subset X$ is measurable if and only if $A\in \mathcal{M}$ , since for any $a\in \mathbb{R}$

$$\{x\mid {\chi }_A(x)>a\}=\{\begin{array}{ll}X& \text{ if }a<0,\\ A& \text{ if }0\le a<1,\\ \varnothing & \text{ if }a\ge 1.\end{array}$$

Exercise

Show that for any $a\in R$ , the set $\{x\in X\mid f(x)=a\}$ is a measurable set of $X.$

If $f:X\to \mathbb{R}$ is a measurable function then so is its modulus $\mid f\mid$ , since the continuous function $x\mapsto \left|x\right|$ on $\mathbb{R}$ is necessarily Borel measurable, and $|f|$ is the composition function $|\cdot |\circ f$ . Similarly the function $f^a$ for $a>0$ is measurable, and $1/f$ is measurable if $f(x)\ne 0$ for all $x\in X$ . If $g:X\to \mathbb{R}$ is another measurable function then the function $f+g$ is measurable since it can be written as the composition $f=\rho \circ F$ where $F:X\to {\mathbb{R}}^2=$ $\mathbb{R}\times \mathbb{R}$ and $\rho :{\mathbb{R}}^2\to \mathbb{R}$ are the maps

$$F:x\mapsto (f(x),g(x))\text{ and }\rho (a,b)=a+b.$$

The function $F$ is measurable since the inverse image of any product of intervals $I_1,I_2$ is

$$F^{-1}(I_1\times I_2)=f^{-1}(I_1)\cap g^{-1}(I_2),$$

which is a measurable set in $X$ since $f$ and $g$ are assumed measurable functions, while the map $\rho$ is evidently continuous on ${\mathbb{R}}^2$

Exercise

Show that for measurable functions $f,g:X\to \mathbb{R}$ , the function $fg$ is measurable.

An important class of functions are the simple functions: measurable functions $h:X\to \mathbb{R}$ that take on only a finite set ofextended real values $a_1,a_2,\dots ,a_n$ . Since $A_i=h^{-1}(\{a_i\})$ is a measurable subset of $X$ for each $a_i^{}$ , we can write a simple function as a linear combination of measurable characteristic functions

$$h=a_1{\chi }_{A_1}+a_2{\chi }_{A_2}+\cdots +a_n{\chi }_{A_n}\text{ where all }{a}_i\ne 0.$$

Some authors use the term step function instead of simple function, but the common convention is to preserve this term for simple functions $h:\mathbb{R}\to \mathbb{R}$ in which each set $A_i$ is a union of disjoint intervals (Fig. 11.1).

Figure 11.1 Simple (step) function

Let $f$ and $g$ be any pair of measurable functions from $X$ into the extended reals $\overline{R}.$ . The function $h=\sup (f,g)$ defined by

$$h(x)=\{\begin{array}{ll}f(x)& \text{if}f(x)\ge g(x)\\ g(x)& \text{if}g(x)>f(x)\end{array}$$

is measurable, since

$$\{x\mid h(x)>a\}=\{x\mid f(x)>a\text{ or }g(x)>a\}=\{x\mid f(x)>a\}\cup \{x\mid g(x)>a\}$$

is measurable. Similarly, $\inf (f,g)$ is a measurable function. In particular, if $f$ is a measurable function, its positive and negative parts

$$f^{+}=\sup (f,0)\text{ and }{f}^{-}=-\inf (f,0)$$

are measurable functions.

Exercise

Show that $f=f^{+}-f^{-}\mathrm{and}|f|=f^{+}+f^{-}.$

A simple extension of the above argument shows that sup $f_n$ is a measurable function for any countable set of measurable functions $f_1,f_2,f_3,\dots$ . with values in the extended real numbers $\overline{\mathbb{R}}.$ . We define the lim sup as

$$\mathrm{limsup}{f}_n(x)=\underset{n\ge 1}{\inf }{F}_n(x)\text{ where }{F}_n(x)=\underset{k\ge n}{\sup }{f}_k(x).$$

The lim sup always exists since the functions $F_n$ are everywhere monotone decreasing,

$$F_1(x)\ge F_2(x)\ge F_3(x)\ge \dots$$

and therefore have a limit if they are bounded below or approach $-\infty$ if unbounded below. Similarly we can define

$$\mathrm{liminf}{f}_n(x)=\underset{n\ge 1}{\sup }{G}_n(x)\text{ where }{G}_n(x)=\underset{k\ge n}{\inf }{f}_k(x).$$

It follows that if $f_n$ is a sequence of measurable functions then $\mathrm{limsup}{f}_n$ and $\mathrm{liminf}{f}_n$ are also measurable. By standard arguments in analysis $f_n(x)$ is a convergent sequence if and only if $\mathrm{limsup}{f}_n(x)=\mathrm{liminf}{f}_n(x)=\lim f_n(x)$ . Hence the limit of any convergent sequence of measurable functions $f_n(x)\to f(x)$ ) is measurable. Note that the convergence need only be ‘pointwise convergence’, not uniform convergence as is required in many theorems in Riemann integration.

Theorem 11.2 · Approximation of measurable functions by simple functions

Any measurable function $f:X\to \mathbb{R}$ is the limit of a sequence of simple functions. The sequence can be chosen to be monotone increasing at all positive values of $f,$ and monotone decreasing at negative values.

Proof

Suppose $f$ is positive and bounded above, $0\le f(x)\le M$ . For each integer $n=$ $1,2,\dots$ . let $h_n$ be the simple function

$$h_n=\sum _{k=0}^n\frac{kM}{2^n}{\chi }_{A_k}$$

where $A_k$ is the measurable set

$$A_k=\left\{x\mid \frac{kM}{2^n}\le f(x)<\frac{k+1}{2^n}\right\}.$$

These simple functions are increasing, $0\le h_1(x)\le h_2(x)\le ..$ . and $|f(x)-h_n(x)|<$ $M/2^n$ . Hence $h_n(x)\to f(x)$ for all $x\in X$ as $n\to \infty$

If $f$ is any positive function, possibly unbounded, the functions $g_n=\inf (h_n,n)$ are positive and bounded above. Hence for each $n$ there exists a simple function $h_n^{\prime }$ such that $|g_n-h_n^{\prime }|<1/n$ . The sequence of simple functions $h_n^{\prime }$ clearly converges everywhere to $f(x)$ . To obtain a monotone increasing sequence of simple functions that converge to $f,$ , set

$$f_n=\sup (h_1^{\prime },h_2^{\prime },\dots ,h_n^{\prime }).$$

If $f$ is not positive, construct simple function sequences approaching the positive and negative parts and use $f=f^{+}-f^{-}$ -

Problems

Problem 11.1

If $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ are measurable spaces, show that the projection maps ${\mathrm{pr}}_1:X\times$ $Y\to X$ and ${\mathbf{pr}}_2:X\times Y\to Y$ defined by ${\mathsf{pr}}_1(x,y)=x$ and ${\mathsf{pr}}_2(x,y)=y$ are measurable functions.

Problem 11.2

Find a step function $s$($x$) that approximates $f(x)=x^2$ uniformly to within $\epsilon >0$ on [0, 1], in the sense that $|f(x)-s(x)|<\epsilon$ everywhere in [0, 1].

Problem 11.3

Let $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ be measurable functions and $E\subset X$ a measurable set. Show that

$$h(x)=\{\begin{array}{ll}f(x)& \text{ if }x\in E\\ g(x)& \text{ if }x\notin E\end{array}$$

is a measurable function on $X.$

Problem 11.4

If $f,g:\mathbb{R}\to \mathbb{R}$ are Borel measurable real functions show that $h(x,y)=f(x)g(y)$ is a measurable function $h:{\mathbb{R}}^2\to \mathbb{R}$ with respect to the product measure on ${\mathbb{R}}^2$