11.2 Measure spaces

中文

Given a measurable space $(X,\mathcal{M})$ , a measure $\mu$ on $X$ is a function $\mu :\mathcal{M}\to \overline{\mathbb{R}}$ such that

(Meas4) $\mu (A)\ge 0$ for all measurable sets $A,$ , and $\mu (\varnothing )=0$

(Meas5) If $A_1,A_2,\dots \in \mathcal{M}$ is any mutually disjoint sequence (finite or countably infinite) of measurable sets such that $A_i\cap A_j=\varnothing \mathrm{if}i\ne j$ , then

$$\mu (A_1\cup A_2\cup A_3\cup \dots )=\sum _i\mu (A_i).$$

A function $\mu$ satisfying property (Meas5) is often referred to as being countably additive. If the series on the right-hand side is not convergent, it is given the value $\infty$. A measure space is a triple $(X,\mathcal{M},\mu )$ consisting of a measurable space $(X,\mathcal{M})$ ) together with a measure $\mu .$

Exercise

Show that if $B\subset A$ are measurable sets, then $\mu B\le \mu A$

Example 11.4

An occasionally useful measure on a sigma-algebra defined on a set $X$ is the Dirac measure. Let $a$ be any fixed point of $X,$ and set

$${\delta }_a(A)=\{\begin{array}{ll}1& \text{ if }a\in A,\\ 0& \text{ if }a\notin A.\end{array}$$

(Meas4) holds trivially for $\mu ={\delta }_a$ , and (Meas5) follows from the obvious fact that the union of a disjoint family of sets $\left\{A_i\right\}$ can contain $a$ if and only if $a\in A_j$ for precisely one member $A_j$ . This measure has applications in distribution theory, Chapter 12.

Example 11.5

The branch ofmathematics known as probability theory is best expressed in terms of measure theory. A probability space is a measure space $(\Omega ,\mathcal{M},P)$ , where $P(\Omega )=1$. Sets $A\in \Omega$ are known as events and $\Omega$ is sometimes referred to as the universe. This ‘universe’ is usually thought of as the set of all possible outcomes of a specific experiment. Note that events are not outcomes of the experiment, but sets of possible outcomes. The measure function $P$ is known as the probability measure on $\Omega ,$ and $P(A)$ is the probability of the event $A$. All events have probability in the range $0\le P(A)\le 1$ . The element $\varnothing$ has probability 0, $P(\varnothing )=0$ , and is called the impossible event. The entire space $\Omega$ has probability 1; it can be thought of as the certainty event.

The event $A\cup B$ is referred to as either $A$ or $B$, while $A\cap B$ is $A$ and $B$. Since $P$ is additive on disjoint sets and $A\cup B=(A-(A\cap B))\cup (B-(A\cap B))\cup (A\cap B)$ , the probabilities are related by

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

The two events are said to be independent if $P(A\cap B)=P(A)P(B)$ . This is by no means always the case.

We think of the probability of event $B$ after knowing that $A$ has occurred as the conditional probability $P(B|A)$ , defined by

$$P(B|A)=\frac{P(A\cap B)}{P(A)}.$$

Events $A$ and $B$ are independent if and only if $P(B|A)=P(B)-$ in other words, the probability of $B$ in no way depends on the occurrence of $A$.

For a finite or countably infinite set of disjoint events $H_1,H_2,\dots$ . partitioning $\Omega$ (some times called a hypothesis), $\Omega ={\bigcup }_i{H}_i$ , we have for any event $B$

$$B=\bigcup _{i=1}^{\infty }(H_i\cap B).$$

Since the sets in this countable union are mutually disjoint, the probability of $B$ is

$$P(B)=\sum _{i=1}^{\infty }P(H_i\cap B)=\sum _{i=1}^{\infty }P(B|H_i)P(H_i).$$

This leads to Bayes’ formula for the conditional probability of the hypothesis $H_i$ given the outcome event $B$ ,

$$\begin{array}{c}P(H_i|B)=\frac{P(B\cap H_i)}{P(B)}\\ =\frac{P(H_i)P(B|H_i)}{P(B)}\\ =\frac{P(H_i)P(B|H_i)}{{\sum }_{k=1}^{\infty }P(B|H_k)P(H_k)}.\end{array}$$

Theorem 11.3 · Continuity of measure from below

Let $E_1,E_2,...$ . be a sequence ofmeasurable sets, which is increasing in the sense that $E_n\subset E_{n+1}$ for all $n=1,2,\dots$ . Then

$$E=\bigcup _{n=1}^{\infty }{E}_n\in \mathcal{M}and\mu E=\underset{n\to \infty }{\lim }\mu E_n.$$

Proof

$E$ is measurable by condition (Meas3). Set

$$F_1=E_1,F_2=E_2-E_1,\dots ,F_n=E_n-E_{n-1},\dots .$$

The sets $F_n$ are all measurable and disjoint, $F_n\cap F_m=\varnothing$ if $n\ne m$ . Since $E_n=F_1\cup$ $F_2\cup \cdots \cup F_n$ we have by (Meas5)

$$\underset{n\to \infty }{\lim }\mu E_n=\sum _{i=1}^{\infty }\mu (F_n)=\mu \left(\bigcup _{k=1}^{\infty }{F}_k\right)=\mu (E).$$

Lebesgue measure

Every open set $U$ on the real line is a countable union ofdisjoint open intervals. This follows from the fact that every rational point $r\in U$ lies in a maximal open interval $(a,b)\subseteq U$ where $a<r<b$ . These intervals must be disjoint, else they would not be maximal, and there are countably many of them since the rational numbers are countably infinite. On the real line $\mathbb{R}$ set $\mu (I)=b-a$ for all open intervals $I=(a,b)$ . This extends uniquely by countable additivity (Meas5) to all open sets of $\mathbb{R}$. By finite additivity $\mu$ must take the value $b-a$ on all intervals, open, closed or half-open. For example, for any $ϵ>0$

$$(a-ϵ,b)=(a-ϵ,a)\cup [a,b),$$

and since the right-hand side is the union of two disjoint Borel sets we have

$$b-a+ϵ=a-a+ϵ+\mu ([a,b)).$$

Hence the measure of the left-closed interval $[a,b)$ is

$$\mu ([a,b))=b-a.$$

Using

$$[a,b)=\{a\}\cup (a,b)$$

we see that every singleton has zero measure, $\mu (\{a\})=0$ , and

$$(a,b]=(a,b)\cup \{b\},[a,b]=(a,b)\cup \{a\}\cup \{b\}⟹\mu ((a,b])=\mu ([a,b])=b-a.$$

Exercise

Show that if $A\subset \mathbb{R}$ is a countable set, then $\mu (A)=0$

Exercise

Show that the set of finite unions of left-closed intervals $[a,b)$ is closed with respect to the operation of taking differences of sets.

For any set $A\subset \mathbb{R}$ we define its outer measure

$${\mu }^{\ast }(A)=\inf \{\mu (U)\mid U\text{ is an open set in }\mathbb{R}\text{ and }A\subseteq U\}.\text{\hspace{1em}(11.1)}$$

While outer measure can be defined for arbitrary sets $A$ of real numbers it is not really a measure at all, for it does not satisfy the countably additive property (Meas5). The best we can do is a property known as countable subadditivity: if $A_1,A_2,\dots \in 2^X$ is any mutually disjoint sequence of sets then

$${\mu }^{\ast }\left(\bigcup _i{A}_i\right)\le \sum _{i=1}^{\infty }{\mu }^{\ast }(A_i).\text{\hspace{1em}(11.2)}$$

Proof

Let $ϵ>0$ . For each $n=1,2,\dots$ . there exists an open set $U_n\supseteq A_n$ such that

$$\mu (U_n)\le {\mu }^{\ast }(A_n)+\frac{ϵ}{2^n}.$$

Since $U=U_1\cup U_2\cup ..$ . covers $A=A_1\cup A_2\cup \dots$

$${\mu }^{\ast }(A)\le \mu (U)\le \mu (U_1)+\mu (U_2)+\cdots \le \sum _{i=1}^{\infty }{\mu }^{\ast }(A_n)+ϵ.$$

As this is true for arbitrary $ϵ>0$ , the inequality Eq. (11.2) follows immediately.

Exercise

Show that outer measure satisfies $(\mathrm{Meas}4),{\mu }^{\ast }(\varnothing )=0$

Exercise

For an open interval $I=(a,b)$ show that ${\mu }^{\ast }(I)=\mu (I)=b-a$

Exercise

Show that

$$A\subseteq B⟹{\mu }^{\ast }(A)\le {\mu }^{\ast }(B).\text{\hspace{1em}(11.3)}$$

Following Carathéodory, a set $E$ is said to be Lebesgue measurable if for any open interval $I=(a,b)$

$$\mu (I)={\mu }^{\ast }(I\cap E)+{\mu }^{\ast }(I\cap E^c).\text{\hspace{1em}(11.4)}$$

At first sight this may not seem a very intuitive notion.

What it is saying is that when we try to cover the mutually disjoint sets $I\cap E$ and $I-E=I\cap E^c$ with open intervals, the overlap of the two sets of intervals can be made ‘arbitrarily small’ (see Fig. 11.2). From now on we will often refer to Lebesgue measurable sets simply as measurable.

Figure 11.2 Lebesgue measurable set

Theorem 11.4 · Splitting outer measure by a measurable set

If $E$ is measurable then for any set $A$, measurable or not,

$${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c).$$

Proof

Given $ϵ>0.$ , let $U$ be an open set such that $A\subseteq U$ and

$$\mu (U)<{\mu }^{\ast }(A)+ϵ.\text{\hspace{1em}(11.5)}$$

Since $A=(A\cap E)\cup (A\cap E^c)$ is a union of two disjoint sets, Eq. (11.2) gives

$${\mu }^{\ast }(A)\le {\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c).$$

Setting $U={\bigcup }_n{I}_n$ where $I_n$ are a finite or countable collection of disjoint open intervals, we have by Eq. (11.3)

$$\begin{array}{l}{\mu }^{\ast }(A)\le {\mu }^{\ast }(U\cap E)+{\mu }^{\ast }(U\cap E^c)\\ \le {\sum }_n{\mu }^{\ast }(I_n\cap E)+{\mu }^{\ast }(I_n\cap E^c)\\ ={\sum }_n{\mu }^{\ast }(I_n)\text{ since }E\text{ is measurable. }\end{array}$$

Using the inequality Eq. (11.5)

$${\mu }^{\ast }(A)\le {\mu }^{\ast }(U\cap E)+{\mu }^{\ast }(U\cap E^c)<{\mu }^{\ast }(A)+ϵ$$

for arbitrary $ϵ>0$ . This proves the desired result

Corollary 11.5 · Splitting outer measure over disjoint measurable sets

If $E_1,E_2,\dots ,E_n$ is any family ofdisjoint measurable sets then

$${\mu }^{\ast }(E_1\cup E_2\cup \cdots \cup E_n)=\sum _{i=1}^n{\mu }^{\ast }(E_i).$$

Proof

If $E$ and $F$ are disjoint measurable sets, setting $A=E\cup F$ in Eq. (11.4) gives

$${\mu }^{\ast }(E\cup F)={\mu }^{\ast }((E\cup F)\cap E)+{\mu }^{\ast }((E\cup F)\cap E^c)={\mu }^{\ast }(E)+{\mu }^{\ast }(F).$$

The result follows by induction on $n$.

Theorem 11.6 · Lebesgue measurable sets and Lebesgue measure

The set of all Lebesgue measurable sets $\mathcal{L}$ is a sigma-algebra, and ${\mu }^{\ast }$ is a measure on $\mathcal{L}.$

Proof

The empty set is Lebesgue measurable, since for any open interval $I$,

$${\mu }^{\ast }(I\cap \varnothing )+{\mu }^{\ast }(I\cap \mathbb{R})={\mu }^{\ast }(\varnothing )+{\mu }^{\ast }(I)={\mu }^{\ast }(I)=\mu (I).$$

If $E$ is a measurable set then so is $E^c$ , on substituting in Eq. (11.4) and using $(E^c{)}^c=$ $E$. Hence conditions (Meas1) and (Meas2) are satisfied. We prove (Meas3) in severa stages.

Firstly, if $E$ and $F$ are measurable sets then $E\cup F$ is measurable. For, $I=(I\cap E\cup$ $F))\cup \left(I\cap (E\cup F{)}^c\right)$ and using Eq. (11.2) gives

$${\mu }^{\ast }(I)\le {\mu }^{\ast }(I\cap (E\cup F))+{\mu }^{\ast }(I\cap (E\cup F{)}^c).\text{\hspace{1em}(11.6)}$$

The sets in the two arguments on the right-hand side can be decomposed as

$$I\cap (E\cup F)=((I\cap F)\cap E)\cup ((I\cap F)\cap E^c)\cup ((I\cap F^c)\cap E)$$

and

$$I\cap (E\cup F{)}^c=I\cap (E^c\cap F^c)=(I\cap F^c)\cap E^c.$$

Again using Eq. (11.2) gives

$$\begin{array}{l}{\mu }^{\ast }(I\cap (E\cup F))+{\mu }^{\ast }(I\cap (E\cup F{)}^c)\\ \le {\mu }^{\ast }((I\cap F)\cap E)+{\mu }^{\ast }((I\cap F)\cap E^c)+{\mu }^{\ast }((I\cap F^c)\cap E)+{\mu }^{\ast }((I\cap F^c)\cap E^c)\end{array}$$

On setting $A=I\cap F$ and $A=I\cap F^c$ respectively in Theorem 11.4 we have

$${\mu }^{\ast }(I\cap (E\cup F))+{\mu }^{\ast }(I\cap (E\cup F{)}^c)\le {\mu }^{\ast }(I\cap F)+{\mu }^{\ast }(I\cap F^c)={\mu }^{\ast }(I),$$

since $F$ is measurable. Combining this with the inequality Eq. (11.6) we conclude that for al intervals $I$

$${\mu }^{\ast }(I\cap (E\cup F))+{\mu }^{\ast }(I\cap (E\cup F{)}^c)={\mu }^{\ast }(I)=\mu (I),$$

which shows that $E\cup F$ is measurable. Incidentally it also follows that the intersection $E\cap F=(E^c\cup F^c{)}^c$ and the difference $F-E=F\cap E^c$ is measurable. Simple induction shows that any finite union of measurable sets $E_1\cup E_2\cup \dots \cup E_n$ is measurable.

Let $E_1,E_2,\dots$ . be any sequence of disjoint measurable sets. Set

$$S_n=\bigcup _{i=1}^n{E}_i\text{ and }S=\bigcup _{i=1}^{\infty }{E}_i.$$

By subadditivity Eq. (11.2),

$${\mu }^{\ast }(S)\le \sum _{i=1}^{\infty }{\mu }^{\ast }(E_i)$$

and since $S\supset S_n$ we have, using Corollary 11.5,

$${\mu }^{\ast }(S)\ge {\mu }^{\ast }(S_n)=\sum _{i=1}^n{\mu }^{\ast }(E_i).$$

Since this holds for all integers $n$ and the right-hand side is a monotone increasing series,

$${\mu }^{\ast }(S)=\sum _{i=1}^{\infty }{\mu }^{\ast }(E_i).\text{\hspace{1em}(11.7)}$$

If the series does not converge, the right-hand side is assigned the value $\infty$.

Since the $E_i$ are disjoint sets, so are the sets $I\cap E_i$ . Furthermore by Corollary 11.5

$$\sum _{i=1}^n{\mu }^{\ast }(I\cap E_i)={\mu }^{\ast }\left(\bigcup _{i=1}^n(I\cap E_i)\right)={\mu }^{\ast }\left(I\cap \bigcup _{i=1}^n{E}_i\right)\le {\mu }^{\ast }(I).$$

Hence the series ${\sum }_{i=1}^{\infty }{\mu }^{\ast }(I\cap E_i)$ is convergent and for any $ϵ>0$ there exists an integer $n$ such that

$$\sum _{i=n}^{\infty }{\mu }^{\ast }(I\cap E_i)<ϵ.$$

Now since $S_n\subset S$

$$I\cap S=I\cap \left(S_n\cup (S-S_n)\right)=(I\cap S_n)\cup \left(I\cap (S-S_n)\right)$$

and by subadditivity Eq. (11.2),

$$\begin{array}{l}\mu (I)={\mu }^{\ast }((I\cap S)\cup (I\cap S^c))\\ \le {\mu }^{\ast }(I\cap S)+{\mu }^{\ast }(I\cap S^c)\\ \le {\mu }^{\ast }((I\cap S_n))+{\mu }^{\ast }(I\cap (S-S_n))+{\mu }^{\ast }(I\cap (S_n{)}^c)\\ =\mu (I)+{\mu }^{\ast }(I\cap {\bigcup }_{i=n+1}^{\infty }{E}_i)\\ =\mu (I)+{\sum }_{i=n+1}^{\infty }{\mu }^{\ast }(I\cap E_i)\\ <\mu (I)+ϵ.\end{array}$$

Since $ϵ>0$ is arbitrary

$$\mu (I)={\mu }^{\ast }(I\cap S)+{\mu }^{\ast }(I\cap S^c),$$

which proves that $S$ is measurable.

If $E_1,E_2,\dots$ . are a sequence of measurable sets, not necessarily disjoint, then se

$$F_1=E_1,F_2=E_2-E_1,\dots ,F_n=E_n-(E_1\cup E_2\cup \cdots \cup E_{n-1}),\dots$$

These sets are all measurable and disjoint, and

$$\bigcup _{i=1}^{\infty }{E}_i=\bigcup _{i=1}^{\infty }{F}_i.$$

The union of any countable collection $\{E_n\}$ of measurable sets is therefore measurable, proving that $\mathcal{L}$ is a sigma-algebra. The outer measure ${\mu }^{\ast }$ is a measure on $\mathcal{L}$ since it satisfies ${\mu }^{\ast }(\varnothing )=0$ and is countably additive by Eq. (11.7). -

Theorem 11.6 shows that $(\mathbb{R},\mathcal{L},\mu ={\mu }^{\ast }{|}_{\mathcal{L}})$ is a measure space. The notation $\mu$ for Lebesgue measure agrees with the earlier convention $\mu ((a,b))=b-a$ on open intervals. All open sets are measurable as they are disjoint unions of open intervals. Since the Borel sets form the sigma-algebra generated by all open sets, they are included in $\mathcal{L}.$ . Hence every Borel set is Lebesgue measurable. It is not true however that every Lebesgue measurable set is a Borel set.

A property that holds everywhere except on a set of measure zero is said to hold almost everywhere, often abbreviated to a.e. . For example, two functions $f$ and $g$ are said to be equal a.e. if the set of points where $f(x)\ne g(x)$ is a set of measure zero. It is sufficient for the set to have outer measure zero, ${\mu }^{\ast }(A)=0$ , in order for it to have measure zero (see Problem 11.8).

Lebesgue measure is defined on cartesian product spaces ${\mathbb{R}}^n$ in a similar manner. We give the construction for $n=2$ . We have already seen that the sigma-algebra ofmeasurable sets on the product space ${\mathbb{R}}^2=\mathbb{R}\times \mathbb{R}$ is defined as that generated by products of measurable sets $E\times E^{\prime }$ where $E$ and $E^{\prime }$ are Lebesgue measurable on $\mathbb{R}$. The outer measure of any set $A\subset {\mathbb{R}}^2$ is defined as

$${\mu }^{\ast }(A)=\sup \sum _i\mu (I_i)\mu (I_i^{\prime })$$

where $I_i$ and $I_i^{\prime }$ are any finite or countable family of open intervals of $\mathbb{R}$ such that the union ${\bigcup }_i{I}_i\times I_i^{\prime }$ covers $A.$ . The outer measure of any product of open intervals is clearly the product of their measures, ${\mu }^{\ast }(I\times I^{\prime })=\mu (I)\mu (I^{\prime })$ ). We say a set $E\subset {\mathbb{R}}^2$ is Lebesgue measurable if, for any pair of open intervals $I,I^{\prime }$

$$\mu (I)\mu (I^{\prime })={\mu }^{\ast }((I\times I^{\prime })\cap E)+{\mu }^{\ast }((I\times I^{\prime })\cap E^c).$$

As for the real line, outer measure ${\mu }^{\ast }$ is then a measure on ${\mathbb{R}}^2$ . Lebesgue measure on highe dimensional products ${\mathbb{R}}^n$ is completely analogous. We sometimes denote this measure by ${\mu }^n$

Example 11.6

The Cantor set, Example 1.11, is a closed set since it is formed by taking the complement of a sequence of open intervals. It is therefore a Borel set and is Lebesgue measurable. The Cantor set is an uncountable set of measure 0 since the length remaining after the $n$th step in its construction is

$$1-\frac{1}{3}-2{\left(\frac{1}{3}\right)}^2-2^2{\left(\frac{1}{3}\right)}^3-\cdots -2^{n-1}{\left(\frac{1}{3}\right)}^n={\left(\frac{2}{3}\right)}^n\to 0.$$

Its complement is an open subset of[0, 1] with measure 1 – that is, having ‘no gaps’ between its component open intervals.

Example 11.7

Not every set is Lebesgue measurable, but the sets that fail are non constructive in character and invariably make use of the axiom ofchoice. A classic example is the following. For any pair of real numbers $x$, $y\in I=(0,1)$, set $xQy$ if and only if $x-y$ is a rational number. This is an equivalence relation on $I$ and it partitions this set into disjoint equivalence classes $Q_x=\{y\mid y-x=r\in \mathbb{Q}\}$ where $\mathbb{Q}$ is the set ofrational numbers. Assuming the axiom of choice, there exists a set $T$ consisting of exactly one representative from each equivalence class $Q_x$ . Suppose it has Lebesgue measure $\mu (T)$ . For each rationa number $r\in (-1,1)$ let $T_r=\{x+r\mid x\in T\}$ . Every real number $y\in I$ belongs to some $T_r$ since it differs by a rational number $r$ from some member of $T$ . Hence, since $|r|<1$ for each such $T_r$ , we must have

$$(-1,2)\supset \bigcup _r{T}_r\supset (0,1).$$

The sets $T_r$ are mutually disjoint and all have measure equal to $\mu (T)$ ). Ifthe rational numbers are displayed as a sequence $r_1,r_2,...$ . then

$$3\ge \mu (T_1)+\mu (T_2)+\cdots =\sum _{i=1}^{\infty }\mu (T)\ge 1.$$

This yields a contradiction either for $\mu (T)=0\mathrm{or}\mu (T)>0;$ ; in the first case the sum is $0,$ in the second it is .

Problems

Problem 11.5

Show that every countable subset of $\mathbb{R}$ is measurable and has Lebesgue measure zero.

Problem 11.6

Show that the union of a sequence of sets of measure zero is a set ofLebesgue measure zero.

Problem 11.7

If ${\mu }^{\ast }(N)=0$ show that for any set $E$, ${\mu }^{\ast }(E\cup N)={\mu }^{\ast }(E-N)={\mu }^{\ast }(E)$ ). Hence show that $E\cup N$ and $E-N$ are Lebesgue measurable if and only if $E$ is measurable.

Problem 11.8

A measure is said to be complete ifevery subset of a set of measure zero is measurable. Show that if $A\subset \mathbb{R}$ is a set of outer measure zero, ${\mu }^{\ast }(A)=0$ , then $A$ is Lebesgue measurable and has measure zero. Hence show that Lebesgue measure is complete.

Problem 11.9

Show that a subset $E$ of $\mathbb{R}$ is measurable if for all $ϵ>0$ there exists an open se $U\supset E$ such that ${\mu }^{\ast }(U-E)<ϵ$

Problem 11.10

If $E$ is bounded and there exists an interval $I\supset E$ such that

$${\mu }^{\ast }(I)={\mu }^{\ast }(I\cap E)+{\mu }^{\ast }(I-E)$$

then this holds for all intervals, possibly even those overlapping $E$.

Problem 11.11

The inner measure ${\mu }_{\ast }(E)$ of a set $E$ is defined as the least upper bound of the measures of all measurable subsets of $E$. Show that ${\mu }_{\ast }(E)\le {\mu }^{\ast }(E)$

For any open set $U\supset E$ , show that

$$\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$$

and that $E$ is measurable with finite measure if and only if ${\mu }_{\ast }(E)={\mu }^{\ast }(E)<\infty$