The product of square integrable functions is the space of square integrable functions on the product

Theorem : Given two separable Hilbert spaces $\mathcal{H}_1 = L^2(X, \mu)$ and $\mathcal{H}_2 = L^2(Y, \lambda)$, there exists a unique isometric isomorphism

\begin{equation} L^2(X, \mu) \otimes L^2(Y, \lambda) \cong L^2(X \times Y, \mu \times \lambda) \end{equation}

which maps $f \otimes g$ to $f(x) g(y)$.

Proof : Being separable Hilbert spaces, $\mathcal{H}_1$ and $\mathcal{H}_2$ can be given a basis, $\{ \psi^1_i(x) \}$ and $\{ \psi^2_i(y) \}$. We need to show first that $\{ \psi^1_i(x) \psi^2_j(y) \}$ is a basis of $L^2(X \times Y, \mu \times \lambda)$. This is true if this set obeys Parseval's identity : for any $v \in L^2(X \times Y, \mu \times \lambda)$,

\begin{equation} \sum_{i,j} |\langle v, \psi^1_i \psi^2_j \rangle|^2 = \langle v, v \rangle \end{equation}

Consider our function $v(x,y)$. Then :

\begin{eqnarray} \langle v, \psi^1_i \psi^2_j \rangle &=& \int v^*(x,y) \psi^1_i(x) \psi^2_j(y) d\mu(x) d\lambda(y) \end{eqnarray}