# Orthonormal bases for product measures

Jordan Bell
October 22, 2015

## 1 Measure and integration theory

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

Let $(X,\mathscr{A},\mu)$ be a measure space. It is a fact that if $f_{n}$ is a sequence of $\mathscr{A}\to\overline{\mathscr{B}}$ measurable functions then $\sup_{n}f_{n}$ and $\inf_{n}f_{n}$ are $\mathscr{A}\to\overline{\mathscr{B}}$ measurable, and thus if $f_{n}$ is a sequence of $\mathscr{A}\to\overline{\mathscr{B}}$ measurable functions that converge pointwise to a function $f:X\to\overline{\mathbb{R}}$, then $f$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable.11 1 Heinz Bauer, Measure and Integration Theory, p. 52, Corollary 9.7. If $f_{1},\ldots,f_{n}$ are $\mathscr{A}\to\overline{\mathscr{B}}$ measurable, then so are $f_{1}\vee\cdots\vee f_{n}$ and $f_{1}\wedge\cdots\wedge f_{n}$, and a function $f:X\to\overline{\mathbb{R}}$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable if and only if both $f^{+}=f\vee 0$ and $f^{-}=-(f\wedge 0)$ are $\mathscr{A}\to\overline{\mathscr{B}}$ measurable. In particular, if $f$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable then so is $|f|=f^{+}+f^{-}$.

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

 $u,v\in E,\;\alpha\geq 0\quad\Rightarrow\quad\alpha u,\;u+v,\;u\cdot v,\;u\vee v% ,\;u\wedge v\in E.$

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

 $I_{\mu}u=\sum_{i=1}^{n}a_{i}\mu(A_{i}),$

where $u$ has range $\{a_{1},\ldots,a_{n}\}$ and $A_{i}=u^{-1}(a_{i})$. One proves that $I_{\mu}:E\to[0,\infty]$ is positive homogeneous, additive, and order preserving.22 2 Heinz Bauer, Measure and Integration Theory, pp. 55–56, §10.

It is a fact33 3 Heinz Bauer, Measure and Integration Theory, p. 57, Theorem 11.1. that if $u_{n}$ is a nondecreasing sequence in $E$ and $u\in E$ then

 $u\leq\sup_{n}u_{n}\quad\Rightarrow\quad I_{\mu}u\leq\sup_{n}I_{\mu}u_{n}.$

It follows that if $u_{n}$ and $v_{n}$ are sequences in $E$ then

 $\sup_{n}u_{n}=\sup_{n}v_{n}\quad\Rightarrow\quad\sup_{n}I_{\mu}u_{n}=\sup_{n}I% _{\mu}v_{n}.$ (1)

Define $E^{*}=E^{*}(\mathscr{A})$ to be the set of all functions $f:X\to[0,\infty]$ 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 $E\to[0,\infty]$ to $E^{*}\to[0,\infty]$ by defining $I_{\mu}f=\sup_{n}I_{\mu}u_{n}$. One proves44 4 Heinz Bauer, Measure and Integration Theory, pp. 58–59, §11. that

 $f,g\in E^{*},\;\alpha\geq 0\quad\Rightarrow\quad\alpha f,\;f+g,\;f\cdot g,\;f% \vee g,\;f\wedge g\in E^{*}$

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

The monotone convergence theorem55 5 Heinz Bauer, Measure and Integration Theory, p. 59, Theorem 11.4. states that if $f_{n}$ is a sequence in $E^{*}$ then $\sup_{n}f_{n}\in E^{*}$ and

 $I_{\mu}\left(\sup_{n}f_{n}\right)=\sup_{n}I_{\mu}f_{n}.$

We now prove a characterization of $E^{*}$.66 6 Heinz Bauer, Measure and Integration Theory, p. 61, Theorem 11.6.

###### Theorem 1.

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

Now suppose that $f:X\to[0,\infty]$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable. For $n\geq 1$ and $0\leq i\leq n2^{n}-1$ let

 $A_{i,n}=\{f\geq i2^{-n}\}\cap\{f<(i+1)2^{-n}\}=\{i2^{-n}\leq f<(i+1)2^{-n}\},$

and for $i=n2^{n}$ let

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

Because $f$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable, the sets $A_{i,n}$ belong to $\mathscr{A}$. For each $n$, the sets $A_{0,n},\ldots A_{n2^{n}-1,n},A_{n2^{n},n}$ are pairwise disjoint and their union is equal to $X$. It is apparent that

 $A_{i,n}=A_{2i,n+1}\cup A_{2i+1,n+1},\qquad 0\leq i\leq n2^{n}-1.$ (2)

Define

 $u_{n}=\sum_{i=0}^{n2^{n}}i2^{-n}1_{A_{i,n}},$

which belongs to $E$. For $x\in X$, either $f(x)=\infty$ or $0\leq f(x)<\infty$. In the first case, $u_{n}(x)=n$ for all $n\geq 1$. In the second case, $u_{n}(x)\leq f(x) for all $n>f(x)$. Therefore $u_{n}(x)\uparrow f(x)$ as $n\to\infty$, 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 $I_{\mu}:E^{*}\to[0,\infty]$. Suppose that $f:X\to\overline{\mathbb{R}}$ is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable. Then $f^{+},f^{-}:X\to[0,\infty]$ are $\mathscr{A}\to\overline{\mathscr{B}}$ measurable so by Theorem 1, $f^{+},f^{-}\in E^{*}$. Then $I_{\mu}f^{+},I_{\mu}f^{-}\in[0,\infty]$. We say that a function $f:X\to\overline{\mathbb{R}}$ is $\mu$-integrable if it is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable and $I_{\mu}f^{+}<\infty$ and $I_{\mu}f^{-}<\infty$. One checks that a function $f:X\to\overline{\mathbb{R}}$ is $\mu$-integrable if and only if it is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable and $I_{\mu}|f|<\infty$. If $f:X\to\overline{\mathbb{R}}$ is $\mu$-integrable, we now define $I_{\mu}f\in\mathbb{R}$ by

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

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

One proves that if $f,g:X\to\overline{\mathbb{R}}$ are $\mu$-integrable and $\alpha\in\mathbb{R}$ 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.77 7 Heinz Bauer, Measure and Integration Theory, p. 65, Theorem 12.3. Furthermore, $I_{\mu}$ is order preserving.

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

 $I_{\mu}f=I_{\mu}u+iI_{\mu}v.$

## 2 ℒ²

Let $(X,\mathscr{A},\mu)$ be a measure space and for $1\leq p<\infty$ let $\mathscr{L}^{p}(\mu)$ be the collection of Borel measurable functions $f:X\to\mathbb{C}$ such that $|f|^{p}$ is $\mu$-integrable. For complex $a,b$, because $x\mapsto x^{p}$ is convex we have by Jensen’s inequality

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

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

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

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

For Borel measurable $f:X\to\mathbb{C}$ define

 $\left\|f\right\|_{L^{p}}=\left(\int_{X}|f|^{p}d\mu\right)^{1/p}.$

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

 $\displaystyle\left\|f+g\right\|_{L^{p}}^{p}$ $\displaystyle\leq\int_{X}|f||f+g|^{p-1}d\mu+\int_{X}|g||f+g|^{p-1}d\mu$ $\displaystyle\leq\left\|f\right\|_{L^{p}}\left\||f+g|^{p-1}\right\|_{L^{p^{% \prime}}}+\left\|g\right\|_{L^{p}}\left\||f+g|^{p-1}\right\|_{L^{p^{\prime}}}$ $\displaystyle=\left\|f\right\|_{L^{p}}\left\|f+g\right\|_{L^{p}}^{p-1}+\left\|% g\right\|_{L^{p}}\left\|f+g\right\|_{L^{p}}^{p-1},$

which implies that $\left\|f+g\right\|_{L^{p}}\leq\left\|f\right\|_{L^{p}}+\left\|g\right\|_{L^{p}}$, and hence $\left\|\cdot\right\|_{L^{p}}$ is a seminorm on $\mathscr{L}^{p}(\mu)$.

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

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

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

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 $\mathscr{L}^{p}(\mu)$ with $\sum_{k=1}^{\infty}\left\|f\right\|_{L^{p}}<\infty$. For $n\geq 1$ let $g_{n}(x)=\left(\sum_{k=1}^{n}|f_{k}(x)|\right)^{p}$ and define $g:X\to[0,\infty]$ by

 $g(x)=\left(\sum_{k=1}^{\infty}|f_{k}(x)|\right)^{p}=\lim_{n\to\infty}g_{n}(x),$

which is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable, being the pointwise limit of a sequence of functions each of which is $\mathscr{A}\to\overline{\mathscr{B}}$ measurable. Because $g_{1}\leq g_{2}\leq\cdots$, by the monotone convergence theorem,

 $\int_{X}gd\mu=\lim_{n\to\infty}\int_{X}g_{n}d\mu.$

But

 $\left(\int_{X}g_{n}d\mu\right)^{1/p}=\left\|\sum_{k=1}^{n}|f_{k}|\right\|_{L^{% p}}\leq\sum_{k=1}^{n}\left\|f_{k}\right\|_{L^{p}}\leq\sum_{k=1}^{\infty}\left% \|f_{k}\right\|_{L^{p}},$

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

 $f(x)=1_{X\setminus E}(x)\sum_{k=1}^{\infty}f_{k}(x),$

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

 $\lim_{n\to\infty}\left|\sum_{k=1}^{n}f_{k}(x)-f(x)\right|^{p}=0$

and

 $\left|\sum_{k=1}^{n}f_{k}(x)-f(x)\right|^{p}\leq g(x),$

so by the dominated convergence theorem,88 8 Heinz Bauer, Measure and Integration Theory, p. 83, Theorem 15.6.

 $\lim_{n\to\infty}\int_{X}\left|\sum_{k=1}^{n}f_{k}(x)-f(x)\right|^{p}d\mu=0.$

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

 $\lim_{n\to\infty}\left\|\sum_{k=1}^{n}f_{k}-f\right\|_{L^{p}}=0.$

Hence, if $f_{k}$ is a sequence in $L^{p}(\mu)$ such that $\sum_{k=1}^{\infty}\left\|f_{k}\right\|_{L^{p}}<\infty$ then there is some $f\in L^{p}(\mu)$ such that $\sum_{k=1}^{n}f_{k}\to f$ in the norm $\left\|\cdot\right\|_{L^{p}}$. This implies that $L^{p}(\mu)$ is a Banach space.

We say that the $\sigma$-algebra $\mathscr{A}$ is countably generated if there is a countable subset $\mathscr{C}$ of $\mathscr{A}$ such that $\mathscr{A}=\sigma(\mathscr{C})$ and we say that a topological space is separable if there exists a countable dense subset of it. It can be proved that if $\mathscr{A}$ is countably generated and $\mu$ is $\sigma$-finite, then for $1\leq p<\infty$ there is a countable collection of simple functions that is dense in $L^{p}(\mu)$, showing that $L^{p}(\mu)$ is separable.99 9 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition 3.4.5.

###### Theorem 2.

Let $(X,\mathscr{A},\mu)$ be a measure space and let $1\leq p<\infty$. $L^{p}(\mu)$ with the norm $\left\|\cdot\right\|_{L^{p}}$ is a Banach space, and if $\mathscr{A}$ is countably generated and $\mu$ is $\sigma$-finite then $L^{p}(\mu)$ is separable.

For $f,g\in\mathscr{L}^{2}(\mu)$, let

 $\left\langle f,g\right\rangle_{L^{2}(\mu)}=\int_{X}f\cdot\overline{g}d\mu.$

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

## 3 Product measures

Let $(X_{1},\mathscr{A}_{1},\mu_{1})$ and $(X_{1},\mathscr{A}_{1},\mu_{1})$ be measure spaces and let $\mathscr{A}_{1}\otimes\mathscr{A}_{2}$ be the product $\sigma$-algebra. For $Q\subset X_{1}\times X_{2}$, write

 $Q_{x_{1}}=\{x_{2}\in X_{2}:(x_{1},x_{2})\in Q\},\qquad Q_{x_{2}}=\{x_{1}\in X_% {1}:(x_{1},x_{2})\in Q\}.$

One proves that if $\mu_{1}$ and $\mu_{2}$ are $\sigma$-finite, then for each $Q\in\mathscr{A}_{1}\otimes\mathscr{A}_{2}$ the function $x_{1}\mapsto\mu_{2}(Q_{x_{1}})$ is $\mathscr{A}_{1}\to\overline{\mathscr{B}}$ measurable and the function $x_{2}\mapsto\mu_{1}(Q_{x_{2}})$ is $\mathscr{A}_{2}\to\overline{\mathscr{B}}$ measurable.1010 10 Heinz Bauer, Measure and Integration Theory, p. 135, Lemma 23.2. If $\mu_{1}$ and $\mu_{2}$ are $\sigma$-finite, one proves1111 11 Heinz Bauer, Measure and Integration Theory, p. 136, Theorem 23.3. that there is a unique measure $\mu:\mathscr{A}_{1}\otimes\mathscr{A}_{2}\to[0,\infty]$ that satisfies

 $\mu(A_{1}\times A_{2})=\mu_{1}(A_{1})\mu_{2}(A_{2}),\qquad A_{1}\in\mathscr{A}% _{1},A_{2}\in\mathscr{A}_{2}.$

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

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

 $f_{x_{1}}(x_{2})=f(x_{1},x_{2}),\qquad x_{2}\in X_{2}$

and for $x_{2}\in X_{2}$, define $f_{x_{2}}:X_{1}\to X^{\prime}$ by

 $f_{x_{2}}(x_{1})=f(x_{1},x_{2}),\qquad x_{1}\in X_{1}.$

For $Q\subset X_{1}\times X_{2}$,

 $(1_{Q})_{x_{1}}=1_{Q_{x_{1}}},\qquad(1_{Q})_{x_{2}}=1_{Q_{x_{2}}}.$

It is straightforward to prove that if $(X^{\prime},\mathscr{A}^{\prime})$ is a measurable space and $f:(X_{1}\times X_{2},\mathscr{A}_{1}\otimes\mathscr{A}_{2})\to(X^{\prime},% \mathscr{A}^{\prime})$ is measurable, then for each $x_{1}\in X_{1}$ the function $f_{x_{1}}:X_{2}\to X^{\prime}$ is measurable $\mathscr{A}_{2}\to\mathscr{A}^{\prime}$ and for each $x_{2}\in X_{2}$ the function $f_{x_{2}}:X_{1}\to X^{\prime}$ is measurable $\mathscr{A}_{1}\to\mathscr{A}^{\prime}$.1212 12 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma 23.5.

Tonelli’s theorem1313 13 Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6. states that if $(X_{1},\mathscr{A}_{1},\mu_{1})$ and $(X_{1},\mathscr{A}_{1},\mu_{1})$ are $\sigma$-finite measure spaces and $f:X_{1}\times X_{2}\to[0,\infty]$ is $\mathscr{A}_{1}\otimes\mathscr{A}_{2}\to\overline{\mathscr{B}}$ measurable, then the functions

 $x_{2}\mapsto\int_{X_{1}}f_{x_{2}}d\mu_{1},\qquad x_{1}\mapsto\int_{X_{2}}f_{x_% {1}}d\mu_{2}$

are $\mathscr{A}_{2}\to\overline{\mathscr{B}}$ measurable and $\mathscr{A}_{1}\to\overline{\mathscr{B}}$ measurable respectively, and

 $\begin{split}&\displaystyle\int_{X_{1}\times X_{2}}fd(\mu_{1}\otimes\mu_{2})\\ \displaystyle=&\displaystyle\int_{X_{2}}\left(\int_{X_{1}}f_{x_{2}}d\mu_{1}% \right)d\mu_{2}(x_{2})\\ \displaystyle=&\displaystyle\int_{X_{1}}\left(\int_{X_{2}}f_{x_{1}}d\mu_{2}% \right)d\mu_{1}(x_{1}).\end{split}$ (3)

Fubini’s theorem1414 14 Heinz Bauer, Measure and Integration Theory, p. 139, Corollary 23.7. states that if $(X_{1},\mathscr{A}_{1},\mu_{1})$ and $(X_{2},\mathscr{A}_{2},\mu_{2})$ are $\sigma$-finite measure spaces and $f:X_{1}\times X_{2}\to\overline{\mathbb{R}}$ is $\mu_{1}\otimes\mu_{2}$-integrable then there is some $A_{1}\in\mathscr{A}_{1}$ with $\mu_{1}(A_{1})=0$ such that for $x_{1}\in X_{1}\setminus A_{1}$ the function $f_{x_{1}}:X_{2}\to\overline{\mathbb{R}}$ is $\mu_{2}$-integrable, and there is some $A_{2}\in\mathscr{A}_{2}$ with $\mu_{2}(A_{2})=0$ such that for $x_{2}\in X_{2}\setminus A_{2}$ the function $f_{x_{2}}:X_{1}\to\overline{\mathbb{R}}$ is $\mu_{1}$-integrable. Furthermore, define $F_{1}:X_{1}\to\mathbb{R}$ 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 $F_{2}:X_{2}\to\mathbb{R}$ 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

 $\int_{X_{1}\times X_{2}}fd(\mu_{1}\otimes\mu_{2})=\int_{X_{1}}F_{1}d\mu_{1}=% \int_{X_{2}}F_{2}d\mu_{2}.$

Suppose that $(X_{1},\mathscr{A}_{1},\mu_{1})$ and $(X_{2},\mathscr{A}_{2},\mu_{2})$ are $\sigma$-finite measure spaces. For $e:X_{1}\to\mathbb{C}$ and $f:X_{2}\to\mathbb{C}$, define $e\otimes f:X_{1}\times X_{2}\to\mathbb{C}$ by

 $(e\otimes f)(x_{1},x_{2})=e(x_{1})f(x_{2}),$

which is Borel measurable $X_{1}\times X_{2}\to\mathbb{C}$ if $e$ and $f$ are Borel measurable. If $e\in\mathscr{L}^{2}(\mu_{1})$ and $f\in\mathscr{L}^{2}(\mu_{2})$, then by Tonelli’s theorem $e\otimes f:X_{1}\times X_{2}\to\mathbb{C}$ belongs to $\mathscr{L}^{2}(\mu_{1}\otimes\mu_{2})$. For $e,e^{\prime}\in\mathscr{L}^{2}(\mu_{1})$ and $f,f^{\prime}\in\mathscr{L}^{2}(\mu_{2})$, by Fubini’s theorem,

 $\begin{split}&\displaystyle\left\langle e\otimes f,e^{\prime}\otimes f^{\prime% }\right\rangle_{L^{2}(\mu_{1}\otimes\mu_{2})}\\ \displaystyle=&\displaystyle\int_{X_{1}\times X_{2}}e(x_{1})f(x_{2})\overline{% e^{\prime}(x_{1})f^{\prime}(x_{2})}d(\mu_{1}\otimes\mu_{2})(x_{1},x_{2})\\ \displaystyle=&\displaystyle\int_{X_{2}}\left(\int_{X_{1}}e(x_{1})\overline{e^% {\prime}(x_{1})}d\mu_{1}(x_{1})\right)f(x_{2})\overline{f^{\prime}(x_{2})}d\mu% _{2}(x_{2})\\ \displaystyle=&\displaystyle\left\langle e,e^{\prime}\right\rangle_{L^{2}(\mu_% {1})}\cdot\left\langle f,f^{\prime}\right\rangle_{L^{2}(\mu_{2})}.\end{split}$

Therefore, if $E\subset\mathscr{L}^{2}(\mu_{1})$ is an orthonormal set in $L^{2}(\mu_{1})$ and $F\subset\mathscr{L}^{2}(\mu_{2})$ is an orthonormal set in $L^{2}(\mu_{2})$, then $\{e\otimes f:e\in E,f\in F\}\subset\mathscr{L}^{2}(\mu_{1}\otimes\mu_{2})$ is an orthonormal set in $L^{2}(\mu_{1}\otimes\mu_{2})$.

###### Theorem 3.

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

###### Proof.

To show that $\Phi$ is an orthonormal basis for $L^{2}(\mu_{1}\otimes\mu_{2})$ it suffices to prove that if $h\in\mathscr{L}^{2}(\mu_{1}\otimes\mu_{2})$ belongs to the orthogonal complement of $\Phi$ then $h\in\mathscr{N}^{2}(\mu_{1}\otimes\mu_{2})$. Thus, suppose that $h\in\mathscr{L}^{2}(\mu_{1}\otimes\mu_{2})$ and that $\left\langle h,e\otimes f\right\rangle_{L^{2}(\mu_{1}\otimes\mu_{2})}=0$ 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 $A_{f}\in\mathscr{A}_{1}$ with $\mu_{1}(A_{f})=0$ such that $\int_{X_{2}}h_{x_{1}}fd\mu_{2}=0$ for $x_{1}\not\in A_{f}$. Let $A_{1}=\bigcup_{f\in F}A_{f}$, for which $\mu_{1}(A_{1})=0$. If $x_{1}\not\in A_{1}$ 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

 $\displaystyle\int_{X_{1}\times X_{2}}|h|^{2}d(\mu_{1}\otimes\mu_{2})$ $\displaystyle=\int_{X_{1}}\left(\int_{X_{2}}|h_{x_{1}}|^{2}d\mu_{2}\right)d\mu% _{1}(x_{1})$ $\displaystyle=\int_{X_{1}\setminus A_{1}}\left(\int_{X_{2}}|h_{x_{1}}|^{2}d\mu% _{2}\right)d\mu_{1}(x_{1})$ $\displaystyle=0,$

which implies that $h=0$ $\mu_{1}\otimes\mu_{2}$-almost everywhere. ∎