Let $X$ and $Y$ be measure spaces and $X\times Y$ the product measure space.
I've always heard people saying that $L^2(X\times Y) \cong L^2(X)\otimes L^2(Y)$, with a cannonical isomorfism given by $f(x)\otimes g(y)\mapsto f(x)g(y)$, but I've never found a proof of this fact.
I'd be grateful if someone could either help me with the proof or provide a good reference for it.
On
note that by Fubini we have for a function $f:X \times Y \rightarrow \mathbb{R}$ in $L^2$ the property $$\infty>\int_{X \times Y} f^2\: d\mu(X \times Y) = \int_X\int_Yf^2 \: d\mu(y)d\mu(X)=\int_Xg^2\:d\mu(X)\int_Yh^2\:d\mu(Y)$$ where $g,h$ denote the ($L^2$ integrable) slices of $f$ along $X$,$Y$ respectively.
I manage to prove only under Euclidean settings:
If $H_{1}=L^{2}({\bf{R}}^{n})$ and $H_{2}=L^{2}({\bf{R}}^{m})$ and consider $H_{3}=L^{2}({\bf{R}}^{n}\times{\bf{R}}^{m})$ all equipped with standard Lebesgue measures.
Now let the map $\tau: H_{1}\otimes H_{2}\rightarrow H_{3}$ be such that $\tau(f\otimes g):(x,y)\rightarrow f(x)g(y)$. If $\tau(f\otimes g)=0$, then either $f=0$ or $g=0$ a.e. in $H_{1}$ or $H_{2}$ respectively. In either case, we have $f\otimes g=0\otimes g=0$ or $f\otimes g=f\otimes 0=0$, so $\tau$ is injective.
We equip $H_{1}\otimes H_{2}$ with the canonical $L^{2}$ inner product.
For all $f\in H_{3}$, $f$ can be $L^{2}$ approximated by a sequence of functions of the form $\displaystyle\sum_{i=1}^{n}\varphi_{i}(x)\psi_{i}(y)$, where $\varphi,\psi$ are simple functions in $H_{1}$ and $H_{2}$ respectively.
Therefore with $\xi_{i}:(x,y)\rightarrow\varphi_{i}(x)\psi_{i}(y)$, $\tau^{-1}(\xi_{i})\in H_{1}\otimes H_{2}$ are such that $\tau^{-1}(\xi_{i})\rightarrow \eta$ in the completion of $H_{1}\widehat{\otimes}H_{2}$ of $H_{1}\otimes H_{2}$. Now $\tau(\eta)=f$, so $\eta$ is surjective. One can also show that $\tau$ is topological isomorphism.
The above reasoning is rough, for more information, see Linear Operators in Hilbert Spaces, Joachim Weidmann, page 48.