11 Product measures, Tonelli, Fubini theorems

product measure

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces, and, let $X\times Y$ be the Cartesian product of the sets $X$ and $Y$. A subset of $X\times Y$ is a measurable rectangle if it has the form $A\times B$ for some $A\in \mathcal{A}$ and some $B\in \mathcal{B}$; the $\sigma$-algebra on $X\times Y$ generated by the collection of all measurable rectangles is called the product of the $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ and is denoted by $\mathcal{A}\otimes \mathcal{B}$.

Lemma

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces.

• If $E$ is a subset of $X\times Y$ that belongs to $\mathcal{A}\otimes \mathcal{B}$, then each section $E_x$ belongs to $\mathcal{B}$ and each section $E^y$ belongs to $\mathcal{A}$.
• If $f$ is an extended real-valued (or a complex-valued) $A\otimes B$-measurable function on $X\times Y$, then each section $f_x$ is $\mathcal{B}$-measurable and each section $f^y$ is $\mathcal{A}$-measurable.

\begin{proof} See here. It is similar as the proof in 3 the Monotone Class Theorem. \end{proof}

Lemma

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces. If $E\in \mathcal{A}\otimes \mathcal{B}$, then the function $x\to \nu \left(E_x\right)$ is $\mathcal{A}$-measurable and the function $y\to$ $\mu \left(E^y\right)$ is $\mathcal{B}$-measurable.

\begin{proof} See here. It is also similar as the proof in 3 the Monotone Class Theorem. \end{proof}

Theorem

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces. Then there is a unique measure $\mu \times \nu$ on the $\sigma$-algebra $\mathcal{A}\otimes \mathcal{B}$ such that $(\mu \times \nu )(A\times B)=$ $\mu (A)\nu (B)$ holds for each $A\in \mathcal{A}$ and $B\in \mathcal{B}$. Furthermore, for any $E\in \mathcal{A}\otimes \mathcal{B}$, we have

$$(\mu \times \nu )(E)={\int }_X\nu \left(E_x\right)d\mu ={\int }_Y\mu \left(E^y\right)d\nu$$

The measure $\mu \times \nu$ is called the product of $\mu$ and $\nu$.

\begin{proof} We define $(\mu \times \nu {)}_1:={\int }_X\nu (E_x)d\mu$ and $(\mu \times \nu {)}_2:={\int }_Y\mu (E^y)d\nu$. Then it suffices to show:

• $(\mu \times \nu {)}_i$ are measures;
• these two measures coincide on all measurable rectangles;
• the set of measurable rectangles form a $\pi$-system;
• use ^ljjfjj.

Also see here. \end{proof}

Tonelli's Theorem

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces, and let $f:X\times Y\to [0,\infty ]$ be $\mathcal{A}\otimes \mathcal{B}$-measurable. Then

• the function $x\to {\int }_Y{f}_{x}d\nu$ is $\mathcal{A}$-measurable and the function $y\to {\int }_X{f}^{y}d\mu$ is $\mathcal{B}$-measurable, and
• $f$ satisfies

$${\int }_{X\times Y}fd(\mu \times \nu )={\int }_X\left({\int }_Y{f}_{x}d\nu \right)d\mu ={\int }_Y\left({\int }_X{f}^{y}d\mu \right)d\nu .$$

Note that the functions $f_x$ and $f^y$ are nonnegative and measurable; thus the expression ${\int }_Y{f}_{x}d\nu$ is defined for each $x$ in $X$ and the expression ${\int }_X{f}^{y}d\mu$ is defined for each $y\in Y$.

\begin{proof} It suffices to show these statements hold for character functions by ^db74fa and ^b32528 and so for simple function. Then use simple function to approximate measurable function and by MCT we finish the proof.

Also see here. \end{proof}

Fubini's Theorem

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces, and let $f:X\times Y\to [-\infty ,\infty ]$ be $\mathcal{A}\otimes \mathcal{B}$-measurable and ${\int }_{X\times Y}|f|d(\mu \times \nu )<\infty$. Then

$$\begin{array}{rl}& {\int }_Y|f(x,y)|d\nu <\infty \text{ for a.e. }x\in X,\\ & {\int }_X|f(x,y)|d\mu <\infty \text{ for a.e. }y\in Y.\end{array}$$

Furthermore, $x\to {\int }_{Y}f(x,y)d\nu$ is an $\mathcal{A}$-measurable function on $X$ and $y\to {\int }_{X}f(x,y)d\mu$ is a $\mathcal{B}$-measurable function on $Y$, and

$${\int }_{X\times Y}fd(\mu \times \nu )={\int }_X\left({\int }_Y{f}_{x}d\nu \right)d\mu ={\int }_Y\left({\int }_X{f}^{y}d\mu \right)d\nu .$$

\begin{proof} By ${\int }_{X\times Y}|f|<\infty$, we have ${\int }_Y|f|$, ${\int }_X|f|$, ${\int }_Y{f}_x^{+}$, ${\int }_Y{f}_x^{-}$ are finite almost everywhere. Then use ^67db20.

Also see here. \end{proof}

Remark. $\mathcal{A},\mathcal{B}$ being $\sigma$-finite is necessary. As a counterexample, consider Lebesgue measure $(\mathbb{R},\mathcal{A},\mu )$, counting measure $(\mathbb{R},\mathcal{B},\nu )$ and $f(x,y)={\delta }_{xy}$ with $0⩽x⩽1$.

Remark of the remark above. 这一节几乎所有结果都需要 $\sigma$-finite，主要是因为我们需要 $\nu (E_x)$ 和 $\mu (E^y)$ 可测来定义 $\mu \times \nu$ 上的积分, which is defined as ${\int }_{X\times Y}fd(\mu \times \nu )={\int }_X\left({\int }_Y{f}_{x}d\nu \right)d\mu$.

• ^db74fa
  • we prove the finite measure case and then the $\sigma$-finite case;
  • it ensures that $\nu (E_x)$ and $\mu (E^y)$ are measurable sets, which is important for ^b32528.
• ^b32528 is a proof of Tonelli’s theorem for character functions. Only with this conclusion, Tonelli’s theorem holds for simple functions and so measurable functions.

Example

Assume that $f:X\to [0,\infty ]$ is an $\mathcal{A}$-measurable function. We have the often useful relation

$${\int }_{X}fd\mu ={\int }_0^{\infty }\mu (\{x\in X:f(x)>t\})dt.$$

\begin{proof} Define $E=\{(x,t):0⩽t<f(x)\}$, which is the region under the graph of $f$. Since $f$ is measurable, there exists $s_n\uparrow f$ and $E_{s_n}=\{(x.t):0⩽t<s_n(x)\}$ is measurable. Then we can compute $\mu \times m(E)$ by ^67db20. Also see here. \end{proof}

Theorem

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces, and let $K$ be a $\mathcal{A}\otimes \mathcal{B}$-measurable function on $X\times Y$. Suppose that there exists $C>0$ such that ${\int }_X|K(x,y)|d\mu ⩽C$ for a.e. $y\in Y$ and ${\int }_Y|K(x,y)|d\nu ⩽C$ for a.e. $x\in X$, and that $1<p<\infty$. If $f\in L^p(\nu )$, the integral

$$(Tf)(x)={\int }_{Y}K(x,y)f(y)d\nu$$

converges absolutely for a.e. $x\in X$, the function $Tf$ thus defined is in $L^p(\mu )$, and $\Vert Tf{\Vert }_p⩽C\Vert f{\Vert }_p$.

\begin{proof} Write $|K(x,y)|=|K(x,y){|}^{1/q}|K(x,y){|}^{1/p}$, then we have

$$\begin{array}{rl}{\int }_Y|K(x,y)f(y)|d\nu & ={\int }_Y|K(x,y){|}^{1/q}(|K(x,y{)}^{1/p}|f(y)||)\\ & ⩽{\left({\int }_Y|K(x,y)|d\nu \right)}^{1/q}{\left({\int }_Y|K(x,y)||f(y){|}^{p}d\nu \right)}^{1/p}\\ & ⩽C^{1/q}{\left({\int }_Y|K(x,y)||f(y){|}^{p}d\nu \right)}^{1/p}.\end{array}$$

By ^67db20 we have

$$\begin{array}{rl}{\int }_X{\left({\int }_Y|K(x,y)f(y)|d\nu \right)}^{p}d\mu & ⩽C^{p/q}{\int }_X{\int }_Y|K(x,y)||f(y){|}^{p}d\nu d\mu \\ & ⩽C^{(p/q)+1}{\int }_Y|f(y){|}^{p}d\nu \end{array}$$

and we finish the proof. Also see here. \end{proof}

The following theorem is a generalization of Minkowski’s inequality, which replace $\sum$ as $\int$.

Theorem

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces, and let $f$ be a nonnegative $\mathcal{A}\otimes \mathcal{B}$-measurable function on $X\times Y$. If $1\le p<\infty$, then

$${\left[{\int }_X{\left({\int }_{Y}f(x,y)d\nu \right)}^{p}d\mu \right]}^{1/p}\le {\int }_Y{\left({\int }_{X}f(x,y{)}^{p}d\mu \right)}^{1/p}d\nu .$$

\begin{proof} If $p=1$, it follows from Tonelli’s theorem. If $1<p<\infty$, let $q$ be the conjugate exponent to $p$ and suppose $g\in L^q(\mu )$. WLOG assume the right hand side is finite. Then by Tonelli’s theorem and Holder’s inequality, we have

$$\begin{array}{rl}{\int }_X\left({\int }_{Y}f(x,y)d\nu \right)|g(x)|d\mu & ={\int }_Y\left({\int }_{X}f(x,y)|g(x)|d\mu \right)d\nu \\ & ⩽\Vert g{\Vert }_q{\int }_Y{\left({\int }_{X}f(x,y{)}^{p}d\mu \right)}^{1/p}d\nu <\infty \end{array}$$

So the map $T:L^{\ast }(\mu )\to \mathbb{R}$ defined by

$$T(g)={\int }_X\left({\int }_{Y}f(x,y)d\nu \right)g(x)d\mu$$

is a well-defined (i.e., the integral over with respect to $d\mu$ is defined), bounded linear functional on $L^9(\mu )$, and

$$\Vert T\Vert ⩽{\int }_Y{\left({\int }_{X}f(x,y{)}^{P}d\mu \right)}^{1/p}d\nu .$$

On the other hand, it follows from the Riesz representation theorem that

$$\Vert T\Vert ={\left({\int }_X{\left({\int }_{Y}f(x,y)d\nu \right)}^{p}d\mu \right)}^{1/p}$$

Now we finish the proof. \end{proof}

Remark. 用了 Riesz representation theorem. 右边凑一个 Holder 比较容易想到，左边是线性变换的模长就有点变态了。