Orthonormal bases for product measures

1 Measure and integration theory

Let $ℬ$ be the Borel $\sigma$-algebra of $ℝ$, and let $\overline{ℬ}$ be the Borel $\sigma$-algebra of $[-\mathrm{\infty },\mathrm{\infty }]=ℝ\cup \{-\mathrm{\infty },\mathrm{\infty }\}$: the elements of $\overline{ℬ}$ are those subsets of $\overline{ℝ}$ of the form $B,B\cup \{-\mathrm{\infty }\},B\cup \{\mathrm{\infty }\},B\cup \{-\mathrm{\infty },\mathrm{\infty }\}$, with $B\in ℬ$.

Let $(X,𝒜,\mu )$ be a measure space. It is a fact that if $f_n$ is a sequence of $𝒜\to \overline{ℬ}$ measurable functions then ${sup}_n{f}_n$ and ${inf}_n{f}_n$ are $𝒜\to \overline{ℬ}$ measurable, and thus if $f_n$ is a sequence of $𝒜\to \overline{ℬ}$ measurable functions that converge pointwise to a function $f:X\to \overline{ℝ}$, then $f$ is $𝒜\to \overline{ℬ}$ measurable. If $f_1,\mathrm{\dots },f_n$ are $𝒜\to \overline{ℬ}$ measurable, then so are $f_1\vee \mathrm{\cdots }\vee f_n$ and $f_1\wedge \mathrm{\cdots }\wedge f_n$, and a function $f:X\to \overline{ℝ}$ is $𝒜\to \overline{ℬ}$ measurable if and only if both $f^{+}=f\vee 0$ and $f^{-}=-(f\wedge 0)$ are $𝒜\to \overline{ℬ}$ measurable. In particular, if $f$ is $𝒜\to \overline{ℬ}$ measurable then so is $|f|=f^{+}+f^{-}$.

A simple function is a function $f:X\to ℝ$ that is $𝒜\to ℬ$ measurable and whose range is finite. Let $E=E(𝒜)$ be the collection of nonnegative simple functions. It is straightforward to prove that

$$u,v\in E,\alpha \ge 0\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}\alpha u,u+v,u\cdot v,u\vee v,u\wedge v\in E.$$

Define $I_{\mu }:E\to [0,\mathrm{\infty }]$ by

$$I_{\mu }u=\sum _{i=1}^n{a}_i\mu (A_i),$$

where $u$ has range $\{a_1,\mathrm{\dots },a_n\}$ and $A_i=u^{-1}(a_i)$. One proves that $I_{\mu }:E\to [0,\mathrm{\infty }]$ is positive homogeneous, additive, and order preserving.

It is a fact that if $u_n$ is a nondecreasing sequence in $E$ and $u\in E$ then

$$u\le \underset{n}{sup}{u}_n\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}{I}_{\mu }u\le \underset{n}{sup}{I}_{\mu }{u}_n.$$

It follows that if $u_n$ and $v_n$ are sequences in $E$ then

$$\underset{n}{sup}{u}_n=\underset{n}{sup}{v}_n\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}\underset{n}{sup}{I}_{\mu }{u}_n=\underset{n}{sup}{I}_{\mu }{v}_n.$$ (1)

Define $E^{*}=E^{*}(𝒜)$ to be the set of all functions $f:X\to [0,\mathrm{\infty }]$ for which there is a nondecreasing sequence $u_n$ in $E$ satisfying ${sup}_n{u}_n=f$, in other words, there is a sequence $u_n$ in $E$ satisfying $u_n\uparrow f$. From (1), for $f\in E^{*}$ and sequences $u_n,v_n\in E$ with ${sup}_n{u}_n=f$ and ${sup}_n{v}_n=f$, it holds that ${sup}_n{I}_{\mu }{u}_n={sup}_n{I}_{\mu }{v}_n$. Also, if $u\in E$ then $u_n=u$ is a nondecreasing sequence in $E$ with $u={sup}_n{u}_n$, so $u\in E^{*}$. Then it makes sense to extend $I_{\mu }$ from $E\to [0,\mathrm{\infty }]$ to $E^{*}\to [0,\mathrm{\infty }]$ by defining $I_{\mu }f={sup}_n{I}_{\mu }{u}_n$. One proves that

$$f,g\in E^{*},\alpha \ge 0\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}\alpha f,f+g,f\cdot g,f\vee g,f\wedge g\in E^{*}$$

and that $I_{\mu }:E^{*}\to [0,\mathrm{\infty }]$ is positive homogeneous, additive, and order preserving.

The monotone convergence theorem states that if $f_n$ is a sequence in $E^{*}$ then ${sup}_n{f}_n\in E^{*}$ and

$$I_{\mu }\left(\underset{n}{sup}{f}_n\right)=\underset{n}{sup}{I}_{\mu }{f}_n.$$

We now prove a characterization of $E^{*}$.

Theorem 1.

$E^{*}$ is equal to the set of functions $X\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{]}$ that are $\mathrm{A}\mathrm{\to }\overline{\mathrm{B}}$ measurable.

Proof.

If $f\in E^{*}$, then there is a sequence $u_n$ in $E$ with $u_n\uparrow f$. Because each $u_n$ is measurable $𝒜\to \overline{ℬ}$, so is $f$.

Now suppose that $f:X\to [0,\mathrm{\infty }]$ is $𝒜\to \overline{ℬ}$ measurable. For $n\ge 1$ and $0\le i\le n{2}^n-1$ let

and for $i=n{2}^n$ let

$$A_{i,n}=\{f\ge n\}.$$

Because $f$ is $𝒜\to \overline{ℬ}$ measurable, the sets $A_{i,n}$ belong to $𝒜$. For each $n$, the sets $A_{0,n},\mathrm{\dots }{A}_{n{2}^n-1,n},A_{n{2}^n,n}$ are pairwise disjoint and their union is equal to $X$. It is apparent that

$$A_{i,n}=A_{2i,n+1}\cup A_{2i+1,n+1},0\le i\le n{2}^n-1.$$ (2)

Define

$$u_n=\sum _{i=0}^{n{2}^n}i{2}^{-n}{1}_{A_{i,n}},$$

which belongs to $E$. For $x\in X$, either $f(x)=\mathrm{\infty }$ or . In the first case, $u_n(x)=n$ for all $n\ge 1$. In the second case, for all $n>f(x)$. Therefore $u_n(x)\uparrow f(x)$ as $n\to \mathrm{\infty }$, and because this is true for each $x\in X$, this means $u_n\uparrow f$ and so $f\in E^{*}$. ∎

So far we have defined $I_{\mu }:E^{*}\to [0,\mathrm{\infty }]$. Suppose that $f:X\to \overline{ℝ}$ is $𝒜\to \overline{ℬ}$ measurable. Then $f^{+},f^{-}:X\to [0,\mathrm{\infty }]$ are $𝒜\to \overline{ℬ}$ measurable so by Theorem 1, $f^{+},f^{-}\in E^{*}$. Then $I_{\mu }{f}^{+},I_{\mu }{f}^{-}\in [0,\mathrm{\infty }]$. We say that a function $f:X\to \overline{ℝ}$ is $\mu$-integrable if it is $𝒜\to \overline{ℬ}$ measurable and and . One checks that a function $f:X\to \overline{ℝ}$ is $\mu$-integrable if and only if it is $𝒜\to \overline{ℬ}$ measurable and . If $f:X\to \overline{ℝ}$ is $\mu$-integrable, we now define $I_{\mu }f\in ℝ$ by

$$I_{\mu }f=I_{\mu }{f}^{+}-I_{\mu }{f}^{-}.$$

For example, if and $S$ is a subset of $X$ that does not belong to $𝒜$, define $f:X\to ℝ$ by $f=1_S-1_{X\setminus S}$. Then $f^{+}=1_S$ and $f^{-}=1_{X\setminus S}$, and thus $f$ is not $𝒜\to \overline{ℬ}$ measurable, so it is not $\mu$-integrable. But $|f|=1$ belongs to $E$, and by hypothesis, showing that $|f|$ is $\mu$-integrable while $f$ is not.

One proves that if $f,g:X\to \overline{ℝ}$ are $\mu$-integrable and $\alpha \in ℝ$ then $\alpha f$ is $\mu$-integrable and

$$I_{\mu }(\alpha f)=\alpha I_{\mu }f,$$

if $f+g$ is defined on all $X$ then $f+g$ is $\mu$-integrable and

$$I_{\mu }(f+g)=I_{\mu }f+I_{\mu }g,$$

and $f\vee g,f\wedge g$ are $\mu$-integrable. Furthermore, $I_{\mu }$ is order preserving.

Let $f:X\to ℂ$ be a function and write $f=u+iv$. One proves that $f$ is Borel measurable (i.e. $𝒜\to {ℬ}_{ℂ}$ measurable), if and only if $u$ and $v$ are measurable $𝒜\to ℬ$. We define $f$ to be $\mu$-integrable if both $u$ and $v$ are $\mu$-integrable, and define

$$I_{\mu }f=I_{\mu }u+i{I}_{\mu }v.$$

2 ℒ²

Let $(X,𝒜,\mu )$ be a measure space and for let ${ℒ}^p(\mu )$ be the collection of Borel measurable functions $f:X\to ℂ$ such that ${|f|}^p$ is $\mu$-integrable. For complex $a,b$, because $x\mapsto x^p$ is convex we have by Jensen’s inequality

$${\left|\frac{a+b}{2}\right|}^p\le {\left(\frac{1}{2}|a|+\frac{1}{2}|b|\right)}^p\le \frac{1}{2}{|a|}^p+\frac{1}{2}{|b|}^p=\frac{1}{2}({|a|}^p+{|b|}^p),$$

so ${|a+b|}^p\le 2^{p-1}({|a|}^p+{|b|}^p)$. Thus if $f,g\in {ℒ}^p(\mu )$ then

$${|f+g|}^p\le 2^{p-1}({|f|}^p+{|g|}^p),$$

which implies that ${ℒ}^p(\mu )$ is a linear space.

For Borel measurable $f:X\to ℂ$ define

$${\parallel f\parallel }_{L^p}={\left({\int }_X{|f|}^p𝑑\mu \right)}^{1/p}.$$

For $f,g\in {ℒ}^p(\mu )$, by Hölder’s inequality, with $\frac{1}{p}+\frac{1}{p^{\prime }}=1$ (for which $p^{\prime }=\frac{p}{p-1}$),

${\parallel f+g\parallel }_{L^p}^p$ $\le {\displaystyle {\int }_X}|f|{|f+g|}^{p-1}𝑑\mu +{\displaystyle {\int }_X}|g|{|f+g|}^{p-1}𝑑\mu$

$\le {\parallel f\parallel }_{L^p}{\parallel {|f+g|}^{p-1}\parallel }_{L^{p^{\prime }}}+{\parallel g\parallel }_{L^p}{\parallel {|f+g|}^{p-1}\parallel }_{L^{p^{\prime }}}$

$={\parallel f\parallel }_{L^p}{\parallel f+g\parallel }_{L^p}^{p-1}+{\parallel g\parallel }_{L^p}{\parallel f+g\parallel }_{L^p}^{p-1},$

which implies that ${\parallel f+g\parallel }_{L^p}\le {\parallel f\parallel }_{L^p}+{\parallel g\parallel }_{L^p}$, and hence $\parallel \cdot {\parallel }_{L^p}$ is a seminorm on ${ℒ}^p(\mu )$.

Let ${𝒩}^p(\mu )$ be the set of those $f\in {ℒ}^p(\mu )$ such that ${\parallel f\parallel }_{L^p}=0$. ${𝒩}^p(\mu )$ is a linear subspace of ${ℒ}^p(\mu )$, and we define

$$L^p(\mu )={ℒ}^p(\mu )/{𝒩}^p(\mu )=\{f+{𝒩}^p(\mu ):f\in {ℒ}^p(\mu )\}.$$

$L^p(\mu )$ is a normed linear space with the norm $\parallel \cdot {\parallel }_{L^p}$.

It is a fact that if $V$ is a normed linear space then $V$ is complete if and only if each absolutely convergent series in $V$ converges in $V$. Suppose that $f_k$ is a sequence in ${ℒ}^p(\mu )$ with . For $n\ge 1$ let $g_n(x)={\left({\sum }_{k=1}^n|f_k(x)|\right)}^p$ and define $g:X\to [0,\mathrm{\infty }]$ by

$$g(x)={\left(\sum _{k=1}^{\mathrm{\infty }}|f_k(x)|\right)}^p=\underset{n\to \mathrm{\infty }}{lim}{g}_n(x),$$

which is $𝒜\to \overline{ℬ}$ measurable, being the pointwise limit of a sequence of functions each of which is $𝒜\to \overline{ℬ}$ measurable. Because $g_1\le g_2\le \mathrm{\cdots }$, by the monotone convergence theorem,

$${\int }_{X}g𝑑\mu =\underset{n\to \mathrm{\infty }}{lim}{\int }_X{g}_n𝑑\mu .$$

But

$${\left({\int }_X{g}_n𝑑\mu \right)}^{1/p}={\parallel \sum _{k=1}^n|f_k|\parallel }_{L^p}\le \sum _{k=1}^n{\parallel f_k\parallel }_{L^p}\le \sum _{k=1}^{\mathrm{\infty }}{\parallel f_k\parallel }_{L^p},$$

which implies that , meaning that $g:X\to [0,\mathrm{\infty }]$ is integrable. The fact that $g$ is integrable implies $\mu (E)=0$, where $E=\{x\in X:g(x)=\mathrm{\infty }\}\in 𝒜$. For $x\in X\setminus E$, and because $ℂ$ is complete this implies that ${\sum }_{k=1}^{\mathrm{\infty }}{f}_k(x)\in ℂ$, and so it makes sense to define $f:X\to ℂ$ by

$$f(x)=1_{X\setminus E}(x)\sum _{k=1}^{\mathrm{\infty }}{f}_k(x),$$

which is Borel measurable. Furthermore, ${|f|}^p\le g$, and because $g$ is integrable this implies that $f\in {ℒ}^p(\mu )$. For $x\in X\setminus E$,

$$\underset{n\to \mathrm{\infty }}{lim}{\left|\sum _{k=1}^n{f}_k(x)-f(x)\right|}^p=0$$

and

$${\left|\sum _{k=1}^n{f}_k(x)-f(x)\right|}^p\le g(x),$$

so by the dominated convergence theorem,

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_X{\left|\sum _{k=1}^n{f}_k(x)-f(x)\right|}^p𝑑\mu =0.$$

Because $x\mapsto x^{1/p}$ is continuous this implies

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel \sum _{k=1}^n{f}_k-f\parallel }_{L^p}=0.$$

Hence, if $f_k$ is a sequence in $L^p(\mu )$ such that then there is some $f\in L^p(\mu )$ such that ${\sum }_{k=1}^n{f}_k\to f$ in the norm $\parallel \cdot {\parallel }_{L^p}$. This implies that $L^p(\mu )$ is a Banach space.

We say that the $\sigma$-algebra $𝒜$ is countably generated if there is a countable subset $𝒞$ of $𝒜$ such that $𝒜=\sigma (𝒞)$ and we say that a topological space is separable if there exists a countable dense subset of it. It can be proved that if $𝒜$ is countably generated and $\mu$ is $\sigma$-finite, then for there is a countable collection of simple functions that is dense in $L^p(\mu )$, showing that $L^p(\mu )$ is separable.

Theorem 2.

Let $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{,}\mu \mathrm{)}$ be a measure space and let . $L^p\mathit{}\mathrm{(}\mu \mathrm{)}$ with the norm $\mathrm{\parallel }\mathrm{\cdot }{\mathrm{\parallel }}_{L^p}$ is a Banach space, and if $\mathrm{A}$ is countably generated and $\mu$ is $\sigma$-finite then $L^p\mathit{}\mathrm{(}\mu \mathrm{)}$ is separable.

For $f,g\in {ℒ}^2(\mu )$, let

$${⟨f,g⟩}_{L^2(\mu )}={\int }_{X}f\cdot \overline{g}𝑑\mu .$$

This is an inner product on $L^2(\mu )$, and thus $L^2(\mu )$ is a Hilbert space.

3 Product measures

Let $(X_1,{𝒜}_1,{\mu }_1)$ and $(X_1,{𝒜}_1,{\mu }_1)$ be measure spaces and let ${𝒜}_1\otimes {𝒜}_2$ be the product $\sigma$-algebra. For $Q\subset X_1\times X_2$, write

$$Q_{x_1}=\{x_2\in X_2:(x_1,x_2)\in Q\},Q_{x_2}=\{x_1\in X_1:(x_1,x_2)\in Q\}.$$

One proves that if ${\mu }_1$ and ${\mu }_2$ are $\sigma$-finite, then for each $Q\in {𝒜}_1\otimes {𝒜}_2$ the function $x_1\mapsto {\mu }_2(Q_{x_1})$ is ${𝒜}_1\to \overline{ℬ}$ measurable and the function $x_2\mapsto {\mu }_1(Q_{x_2})$ is ${𝒜}_2\to \overline{ℬ}$ measurable. If ${\mu }_1$ and ${\mu }_2$ are $\sigma$-finite, one proves that there is a unique measure $\mu :{𝒜}_1\otimes {𝒜}_2\to [0,\mathrm{\infty }]$ that satisfies

$$\mu (A_1\times A_2)={\mu }_1(A_1){\mu }_2(A_2),A_1\in {𝒜}_1,A_2\in {𝒜}_2.$$

The measure $\mu$ satisfies

$$\mu (Q)={\int }_{X_1}{\mu }_2(Q_{x_1})𝑑{\mu }_1(x_1)={\int }_{X_2}{\mu }_1(Q_{x_2})𝑑{\mu }_2(x_2)$$

for $Q\in {𝒜}_1\otimes {𝒜}_2$, and is itself $\sigma$-finite. We write $\mu ={\mu }_1\otimes {\mu }_2$, and call $\mu$ the product measure of ${\mu }_{\mathrm{1}}$ and ${\mu }_{\mathrm{2}}$.

Let $X^{\prime }$ be a set and let $f:X_1\times X_2\to X^{\prime }$ be a function. For $x_1\in X_1$, define $f_{x_1}:X_2\to X^{\prime }$ by

$$f_{x_1}(x_2)=f(x_1,x_2),x_2\in X_2$$

and for $x_2\in X_2$, define $f_{x_2}:X_1\to X^{\prime }$ by

$$f_{x_2}(x_1)=f(x_1,x_2),x_1\in X_1.$$

For $Q\subset X_1\times X_2$,

$${(1_Q)}_{x_1}=1_{Q_{x_1}},{(1_Q)}_{x_2}=1_{Q_{x_2}}.$$

It is straightforward to prove that if $(X^{\prime },{𝒜}^{\prime })$ is a measurable space and $f:(X_1\times X_2,{𝒜}_1\otimes {𝒜}_2)\to (X^{\prime },{𝒜}^{\prime })$ is measurable, then for each $x_1\in X_1$ the function $f_{x_1}:X_2\to X^{\prime }$ is measurable ${𝒜}_2\to {𝒜}^{\prime }$ and for each $x_2\in X_2$ the function $f_{x_2}:X_1\to X^{\prime }$ is measurable ${𝒜}_1\to {𝒜}^{\prime }$.

Tonelli’s theorem states that if $(X_1,{𝒜}_1,{\mu }_1)$ and $(X_1,{𝒜}_1,{\mu }_1)$ are $\sigma$-finite measure spaces and $f:X_1\times X_2\to [0,\mathrm{\infty }]$ is ${𝒜}_1\otimes {𝒜}_2\to \overline{ℬ}$ measurable, then the functions

$$x_2\mapsto {\int }_{X_1}{f}_{x_2}𝑑{\mu }_1,x_1\mapsto {\int }_{X_2}{f}_{x_1}𝑑{\mu }_2$$

are ${𝒜}_2\to \overline{ℬ}$ measurable and ${𝒜}_1\to \overline{ℬ}$ measurable respectively, and

(3)

Fubini’s theorem states that if $(X_1,{𝒜}_1,{\mu }_1)$ and $(X_2,{𝒜}_2,{\mu }_2)$ are $\sigma$-finite measure spaces and $f:X_1\times X_2\to \overline{ℝ}$ is ${\mu }_1\otimes {\mu }_2$-integrable then there is some $A_1\in {𝒜}_1$ with ${\mu }_1(A_1)=0$ such that for $x_1\in X_1\setminus A_1$ the function $f_{x_1}:X_2\to \overline{ℝ}$ is ${\mu }_2$-integrable, and there is some $A_2\in {𝒜}_2$ with ${\mu }_2(A_2)=0$ such that for $x_2\in X_2\setminus A_2$ the function $f_{x_2}:X_1\to \overline{ℝ}$ is ${\mu }_1$-integrable. Furthermore, define $F_1:X_1\to ℝ$ by $F_1(x_1)={\int }_{X_2}{f}_{x_1}𝑑{\mu }_2$ for $x_1\in X_1\setminus A_1$ and $F_1(x_1)=0$ for $x_1\in A_1$, and define $F_2:X_2\to ℝ$ by $F_2(x_2)={\int }_{X_1}{f}_{x_2}𝑑{\mu }_1$ for $x_2\in X_2\setminus A_2$ and $F_2(x_2)=0$ for $x_2\in A_2$. The functions $F_1$ and $F_2$ are ${\mu }_1$-integrable and ${\mu }_2$-integrable respectively, and

$${\int }_{X_1\times X_2}fd({\mu }_1\otimes {\mu }_2)={\int }_{X_1}{F}_1𝑑{\mu }_1={\int }_{X_2}{F}_2𝑑{\mu }_2.$$

Suppose that $(X_1,{𝒜}_1,{\mu }_1)$ and $(X_2,{𝒜}_2,{\mu }_2)$ are $\sigma$-finite measure spaces. For $e:X_1\to ℂ$ and $f:X_2\to ℂ$, define $e\otimes f:X_1\times X_2\to ℂ$ by

$$(e\otimes f)(x_1,x_2)=e(x_1)f(x_2),$$

which is Borel measurable $X_1\times X_2\to ℂ$ if $e$ and $f$ are Borel measurable. If $e\in {ℒ}^2({\mu }_1)$ and $f\in {ℒ}^2({\mu }_2)$, then by Tonelli’s theorem $e\otimes f:X_1\times X_2\to ℂ$ belongs to ${ℒ}^2({\mu }_1\otimes {\mu }_2)$. For $e,e^{\prime }\in {ℒ}^2({\mu }_1)$ and $f,f^{\prime }\in {ℒ}^2({\mu }_2)$, by Fubini’s theorem,

Therefore, if $E\subset {ℒ}^2({\mu }_1)$ is an orthonormal set in $L^2({\mu }_1)$ and $F\subset {ℒ}^2({\mu }_2)$ is an orthonormal set in $L^2({\mu }_2)$, then $\{e\otimes f:e\in E,f\in F\}\subset {ℒ}^2({\mu }_1\otimes {\mu }_2)$ is an orthonormal set in $L^2({\mu }_1\otimes {\mu }_2)$.

Theorem 3.

Let $\mathrm{(}{X}_{\mathrm{1}}\mathrm{,}{\mathrm{A}}_{\mathrm{1}}\mathrm{,}{\mu }_{\mathrm{1}}\mathrm{)}$ and $\mathrm{(}{X}_{\mathrm{2}}\mathrm{,}{\mathrm{A}}_{\mathrm{2}}\mathrm{,}{\mu }_{\mathrm{2}}\mathrm{)}$ be $\sigma$-finite measure spaces and suppose that $L^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{1}}\mathrm{)}$ and $L^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{2}}\mathrm{)}$ are separable. If $E\mathrm{\subset }{\mathrm{L}}^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{1}}\mathrm{)}$ is an orthonormal basis for $L^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{1}}\mathrm{)}$ and $F\mathrm{\subset }{\mathrm{L}}^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{2}}\mathrm{)}$ is an orthonormal basis for $L^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{2}}\mathrm{)}$, then $\mathrm{\Phi }\mathrm{=}\mathrm{\{}e\mathrm{\otimes }f\mathrm{:}e\mathrm{\in }E\mathrm{,}f\mathrm{\in }F\mathrm{\}}\mathrm{\subset }{\mathrm{L}}^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{1}}\mathrm{\otimes }{\mu }_{\mathrm{2}}\mathrm{)}$ is an orthonormal basis for $L^{\mathrm{2}}\mathit{}\mathrm{(}{\mu }_{\mathrm{1}}\mathrm{\otimes }{\mu }_{\mathrm{2}}\mathrm{)}$.

Proof.

To show that $\mathrm{\Phi }$ is an orthonormal basis for $L^2({\mu }_1\otimes {\mu }_2)$ it suffices to prove that if $h\in {ℒ}^2({\mu }_1\otimes {\mu }_2)$ belongs to the orthogonal complement of $\mathrm{\Phi }$ then $h\in {𝒩}^2({\mu }_1\otimes {\mu }_2)$. Thus, suppose that $h\in {ℒ}^2({\mu }_1\otimes {\mu }_2)$ and that ${⟨h,e\otimes f⟩}_{L^2({\mu }_1\otimes {\mu }_2)}=0$ for all $e\in E,f\in F$. Using Fubini’s theorem,

$${\int }_{X_1}e(x_1)\left({\int }_{X_2}{h}_{x_1}(x_2)f(x_2)𝑑{\mu }_2(x_2)\right)𝑑{\mu }_1(x_1)=0.$$

Because this is true for all $e\in E$ and $E$ is dense in $L^2({\mu }_1)$, it follows that there is some $A_f\in {𝒜}_1$ with ${\mu }_1(A_f)=0$ such that ${\int }_{X_2}{h}_{x_1}f𝑑{\mu }_2=0$ for $x_1\notin A_f$. Let $A_1={\bigcup }_{f\in F}{A}_f$, for which ${\mu }_1(A_1)=0$. If $x_1\notin A_1$ then ${\int }_{X_2}{h}_{x_1}f𝑑{\mu }_2=0$ for all $f\in F$, and because $F$ is dense in $L^2({\mu }_2)$ this implies that $h_{x_1}=0$ ${\mu }_2$-almost everywhere. Then

${\displaystyle {\int }_{X_1\times X_2}}{|h|}^{2}d({\mu }_1\otimes {\mu }_2)$ $={\displaystyle {\int }_{X_1}}\left({\displaystyle {\int }_{X_2}}{|h_{x_1}|}^2𝑑{\mu }_2\right)𝑑{\mu }_1(x_1)$

$={\displaystyle {\int }_{X_1\setminus A_1}}\left({\displaystyle {\int }_{X_2}}{|h_{x_1}|}^2𝑑{\mu }_2\right)𝑑{\mu }_1(x_1)$

$=0,$

which implies that $h=0$ ${\mu }_1\otimes {\mu }_2$-almost everywhere. ∎