2.2 Radon measures and integrationDAFA3E4936D2

Proof. Assume first that $\mu (\Omega )<\infty $, and let $F\subset \Omega $ be relatively closed; we may assume that $F\neq \varnothing $. The open sets $U_ j=\{ x\in \Omega :\operatorname {dist}(x,F)<j^{-1}\} $ decrease to $F$, so $\mu (U_ j)\to \mu (F)$ because $\mu $ is finite. Hence $F$ can be approximated in measure from outside by open sets. Passing to complements, every open set can be approximated from inside by relatively closed sets.

Let $\mathcal A$ be the class of Borel sets $E$ such that for every $\varepsilon >0$ there are an open set $U$ and a relatively closed set $F$ with $F\subset E\subset U$ and $\mu (U\setminus F)<\varepsilon $. We have just seen that $\mathcal A$ contains the relatively closed sets, and $\mathcal A$ is closed under complements because the condition is symmetric. To see that $\mathcal A$ is closed under countable unions, let $E=\bigcup _ jE_ j$ with $E_ j\in \mathcal A$, and choose $F_ j\subset E_ j\subset U_ j$ with $\mu (U_ j\setminus F_ j)<\varepsilon 2^{-j}$. The open set $U=\bigcup _ jU_ j$ satisfies $\mu (U\setminus E)<\varepsilon $. Since $\mu (\bigcup _{j\leq N}E_ j)\to \mu (E)$, there is $N$ such that the relatively closed set $F=\bigcup _{j\leq N}F_ j$ satisfies $\mu (E\setminus F)\leq \mu (E\setminus \bigcup _{j\leq N}E_ j)+\sum _{j\leq N}\mu (E_ j\setminus F_ j)<2\varepsilon $, and hence $\mu (U\setminus F)<3\varepsilon $. Thus $\mathcal A$ is a $\sigma $-algebra containing the open sets, so it contains every Borel set $E$. If $F\subset E$ is relatively closed and $(K_ j)$ is a compact exhaustion of $\Omega $, then the sets $F\cap K_ j$ are compact and $\mu (F\setminus K_ j)\to 0$. Hence every Borel set can be approximated in measure from outside by open sets and from inside by compact sets.

In general, let $(K_ j)$ be a compact exhaustion of $\Omega $, put $K_0=\varnothing $, and let $V_ j=\operatorname{Int}K_{j+1}$. The measures $\mu _ j(E)=\mu (E\cap V_ j)$ are finite, hence Radon by the first part. For inner regularity, let $E$ be a Borel set and $\varepsilon >0$. A compact set $K\subset E$ with $\mu _ j(E\setminus K)<\varepsilon $ satisfies $\mu (K)\geq \mu _ j(K)>\mu (E\cap V_ j)-\varepsilon $, and $\mu (E\cap V_ j)\to \mu (E)$ as $j\to \infty $. For outer regularity, note that the Borel sets $B_ j=K_ j\setminus K_{j-1}$ partition $\Omega $ and that $E\cap B_ j\subset V_ j$. By outer regularity of $\mu _ j$ there is an open set $O_ j\subset V_ j$ containing $E\cap B_ j$ with $\mu (O_ j\setminus (E\cap B_ j))<\varepsilon 2^{-j}$; it suffices to intersect an approximating open set with $V_ j$. Then $O=\bigcup _ jO_ j$ is an open set containing $E$ with $\mu (O\setminus E)<\varepsilon $. Finally, $\mu $ is $\sigma $-finite because $\Omega =\bigcup _ jK_ j$ and each $\mu (K_ j)$ is finite. $\square$

Proof. (1) The integrals $\int f_ j\, \mathrm d\mu $ increase and are bounded by $\int f\, \mathrm d\mu $, so we must show that their limit is at least $\int f\, \mathrm d\mu $. Let $s\leq f$ be a nonnegative simple function and $0<t<1$. The measurable sets $E_ j=\{ f_ j\geq ts\} $ increase with $j$, and their union contains $\{ s>0\} $, since $f(x)\geq s(x)>ts(x)$ there. Applying continuity from below to the sets $E_ j\cap \{ s=a\} $ for the finitely many values $a$ of $s$, we obtain

Taking the supremum over $s$ and then letting $t\uparrow 1$ proves (1); infinite integrals are allowed.

(2) The functions $g_ j=\inf _{k\geq j}f_ k$ are measurable, they increase to $\varliminf _ jf_ j$, and $\int g_ j\, \mathrm d\mu \leq \int f_ k\, \mathrm d\mu $ for every $k\geq j$. Applying (1) to $(g_ j)$ gives (2).

(3) Letting $j\to \infty $ in $|f_ j|\leq g$ gives $|f|\leq g$ almost everywhere, so $f$ is integrable. The functions $2g-|f_ j-f|$ are nonnegative almost everywhere, and (2), applied after a modification on a null set, gives

Since $\int 2g\, \mathrm d\mu <\infty $, it follows that $\int |f_ j-f|\, \mathrm d\mu \to 0$. $\square$

Proof. For $u<v$ in $I$ put $s(u,v)=(\varphi (v)-\varphi (u))/(v-u)$. If $u<v<w$ lie in $I$, then $v=\lambda u+(1-\lambda )w$ with $\lambda =(w-v)/(w-u)\in (0,1)$, and the convexity inequality $\varphi (v)\leq \lambda \varphi (u)+(1-\lambda )\varphi (w)$ is equivalent to each of the inequalities $s(u,v)\leq s(u,w)$ and $s(u,w)\leq s(v,w)$. Hence $s(u,v)$ is nondecreasing in each of its two arguments. Consequently, if $[x,y]\subset I$ and $x'<x$, $y'>y$ are points of $I$, then $s(x',x)\leq s(u,v)\leq s(y,y')$ whenever $x\leq u<v\leq y$, so $\varphi $ is Lipschitz on $[x,y]$. In particular, $\varphi $ is continuous, and $\varphi \circ f$ is measurable.

Let $a=\int f\, \mathrm dP$. If $I$ has a finite left endpoint $\alpha $, then $f-\alpha >0$ on $X$, so one of the sets $\{ f-\alpha >1/k\} $, which increase to $X$, has positive measure, and therefore $a-\alpha =\int (f-\alpha )\, \mathrm dP>0$. In the same way, $a$ is smaller than a finite right endpoint of $I$; thus $a\in I$. For $u<a<w$ in $I$ we have $s(u,a)\leq s(u,w)\leq s(a,w)$, so $c=\sup _{u<a}s(u,a)$ is finite and $c\leq s(a,w)$ for every $w>a$. Hence $\varphi (y)\geq \varphi (a)+c(y-a)$ for every $y\in I$: for $y<a$ this is the inequality $s(y,a)\leq c$, and for $y>a$ it is $s(a,y)\geq c$. It follows that $\varphi \circ f\geq \varphi (a)+c(f-a)$, where the right side is integrable. Therefore the negative part of $\varphi \circ f$ is integrable, and since $\int (f-a)\, \mathrm dP=0$, integration gives $\int \varphi \circ f\, \mathrm dP\geq \varphi (a)$. $\square$

Proof. Write $X=\bigcup _ mX_ m$ with measurable sets $X_ m$ of finite measure. Every member $E$ of the family satisfies $\mu (E\cap X_ m)\geq 1/k$ for some $m$ and some integer $k\geq 1$. For fixed $m$ and $k$, at most $k\mu (X_ m)$ members have this property, because the corresponding sets $E\cap X_ m$ are pairwise disjoint subsets of $X_ m$. Hence the family is a countable union of finite subfamilies. $\square$

Proof. (1) Let $\mathcal L$ be the intersection of all families of subsets of $Z$ which contain $\mathcal P$ and are closed under complements and countable disjoint unions; it is the smallest such family. It contains $Z=A\cup (Z\setminus A)$ for any $A\in \mathcal P$. If $B\subset A$ belong to $\mathcal L$, so does $A\setminus B$, the complement of the disjoint union of $Z\setminus A$ and $B$. For $A\in \mathcal P$, the sets $B\in \mathcal L$ with $A\cap B\in \mathcal L$ form a family which contains $\mathcal P$ and is closed under countable disjoint unions and under complements, because $A\cap (Z\setminus B)=A\setminus (A\cap B)$; hence it equals $\mathcal L$. Thus $A\cap B\in \mathcal L$ whenever $A\in \mathcal P$ and $B\in \mathcal L$. Repeating the argument with $A\in \mathcal L$ shows that $\mathcal L$ is closed under finite intersections. Writing a countable union $\bigcup _ jA_ j$ of members of $\mathcal L$ as the disjoint union of the sets $A_ j\cap (Z\setminus A_1)\cap \cdots \cap (Z\setminus A_{j-1})$, we see that $\mathcal L$ is a $\sigma $-algebra. Hence $\sigma (\mathcal P)\subset \mathcal L$, and $\mathcal L$ is contained in every family as in (1).

(2) The sets $E\in \sigma (\mathcal P)$ with $\lambda (E)=\lambda '(E)$ form a family which contains $\mathcal P$, is closed under countable disjoint unions by countable additivity, and is closed under complements because $\lambda (Z\setminus E)=\lambda (Z)-\lambda (E)$ and $\lambda '(Z\setminus E)=\lambda '(Z)-\lambda '(E)$ are differences of finite numbers. By (1), this family is all of $\sigma (\mathcal P)$. $\square$

Proof. The product $\sigma $-algebra is generated by the measurable rectangles $A\times B$, and the intersection of two rectangles is a rectangle, so lemma 2.2.6 applies to the family of rectangles.

Step 1: finite measures. Suppose that $\mu $ and $\nu $ are finite. The sets $E\subset X\times Y$ whose sections $E_ x=\{ y:(x,y)\in E\} $ are all measurable form a $\sigma $-algebra containing the rectangles, so sections of product-measurable sets are measurable. The product-measurable sets $E$ for which $x\mapsto \nu (E_ x)$ is measurable include the rectangles, since $\nu ((A\times B)_ x)=\mathbf1_ A(x)\nu (B)$. They are closed under complements, as $\nu ((X\times Y\setminus E)_ x)=\nu (Y)-\nu (E_ x)$, and under countable disjoint unions, so by lemma 2.2.6(1) they are all product-measurable sets.

By monotone convergence (proposition 2.2.3), $\lambda (E)=\int _ X\nu (E_ x)\, \mathrm d\mu (x)$ is a measure with $\lambda (A\times B)=\mu (A)\nu (B)$. If $\lambda '$ is another such measure, then $\lambda '(X\times Y)=\mu (X)\nu (Y)=\lambda (X\times Y)$, so $\lambda '=\lambda $ by lemma 2.2.6(2). In particular, $\lambda $ coincides with the measure $E\mapsto \int _ Y\mu (E^ y)\, \mathrm d\nu (y)$, where $E^ y=\{ x:(x,y)\in E\} $, which is obtained by exchanging the roles of $X$ and $Y$. This proves uniqueness and both integral identities for indicator functions. If $f\geq 0$ is product-measurable, its sections are measurable, because $\{ y:f(x,y)>t\} $ is a section of $\{ f>t\} $. Linearity extends the identities, together with the measurability of the inner integrals, to nonnegative simple functions, and monotone convergence, applied to an increasing sequence of such functions converging to $f$, extends them to $f$.

Step 2: $\sigma $-finite measures. Partition $X$ and $Y$ into countably many measurable sets $X_ m$ and $Y_ n$ of finite measure, and consider the finite measures $\mu _ m(A)=\mu (A\cap X_ m)$ on $X$ and $\nu _ n(B)=\nu (B\cap Y_ n)$ on $Y$. Let $\lambda _{mn}$ be the product of $\mu _ m$ and $\nu _ n$ given by Step 1, and put $\mu \times \nu =\sum _{m,n}\lambda _{mn}$. This is a measure on the product $\sigma $-algebra, and $(\mu \times \nu )(A\times B)=\sum _{m,n}\mu (A\cap X_ m)\nu (B\cap Y_ n)=\mu (A)\nu (B)$. If $\lambda '$ is another measure with this property, then $\lambda '_{mn}(E)=\lambda '(E\cap (X_ m\times Y_ n))$ is a finite measure with $\lambda '_{mn}(A\times B)=\mu _ m(A)\nu _ n(B)$, so $\lambda '_{mn}=\lambda _{mn}$ by Step 1, and $\lambda '=\sum _{m,n}\lambda '_{mn}=\mu \times \nu $.

The integral of a nonnegative measurable function with respect to a countable sum of measures is the sum of the integrals. This holds for indicator functions by definition and for simple functions by linearity; in general it follows by monotone convergence, which also allows us to interchange a limit with a series of nonnegative terms. Applying this to $\mu \times \nu =\sum _{m,n}\lambda _{mn}$, to $\nu =\sum _ n\nu _ n$ and to $\mu =\sum _ m\mu _ m$, and using Step 1 for each $\lambda _{mn}$, we obtain for every nonnegative product-measurable $f$

where $x\mapsto \int _ Yf(x,y)\, \mathrm d\nu (y)=\sum _ n\int _ Yf(x,y)\, \mathrm d\nu _ n(y)$ is measurable as a countable sum of measurable functions. The other iterated integral is treated in the same way, with the roles of $X$ and $Y$ exchanged.

Now let $\int |f|\, \mathrm d(\mu \times \nu )<\infty $. By the nonnegative case applied to $|f|$, the measurable function $x\mapsto \int _ Y|f(x,y)|\, \mathrm d\nu (y)$ is integrable, so it is finite outside a measurable $\mu $-null set $N$. For $x\notin N$ the section $f(x,\cdot )$ is integrable, and its integral is the corresponding combination of the integrals of the sections of the positive and negative parts of $\operatorname{Re}f$ and $\operatorname{Im}f$. These four functions have finite integrals with respect to $\mu \times \nu $; defining the inner integral to be zero on $N$ and applying the nonnegative case to each of them, we obtain the first equality for $f$. The second follows by exchanging the roles of $X$ and $Y$. $\square$

Proof. Let $p_1$ and $p_2$ be the projections of $Z_1\times Z_2$ onto its factors. For $i=1,2$, the sets $E\subset Z_ i$ with $p_ i^{-1}(E)$ Borel form a $\sigma $-algebra which contains the open sets, because $p_ i$ is continuous; hence it contains all Borel sets. Thus $A_1\times A_2=p_1^{-1}(A_1)\cap p_2^{-1}(A_2)$ is Borel whenever $A_1$ and $A_2$ are Borel. Conversely, we may assume that $Z_1$ and $Z_2$ are nonempty, and we choose countable dense subsets $D_1\subset Z_1$ and $D_2\subset Z_2$. Let $W\subset Z_1\times Z_2$ be open and $(z_1,z_2)\in W$. There is $r>0$ with $B(z_1,r)\times B(z_2,r)\subset W$, where $B(z_ i,r)$ is the open ball in $Z_ i$. Choose $a_ i\in D_ i$ with $\operatorname {dist}(a_ i,z_ i)<r/3$ and a rational number $q\in (r/3,2r/3)$; then $z_ i\in B(a_ i,q)\subset B(z_ i,r)$ by the triangle inequality. Hence $W$ is the union of those of the countably many products $B(a_1,q_1)\times B(a_2,q_2)$, with $a_ i\in D_ i$ and positive rational $q_ i$, that are contained in $W$. Each of them is a product of open sets, so every open subset of $Z_1\times Z_2$, and therefore every Borel subset, belongs to the $\sigma $-algebra generated by the products of Borel sets. $\square$