Basis functions on two Hilbert spaces - showing the dot product is the basis of the product Hilbert space Question: Let $(X,\mathcal M, \mu)$ and $(Y, \mathcal N, \nu)$ be two $\sigma$-finite measure spaces. If functions $e_1, e_2 \ldots $ are a Hilbert space basis of $L^2(X)$ and functions $f_1, f_2, \ldots $ are a Hilbert space basis of $L^2(Y)$, prove that $e_i\cdot f_j$ is a basis of $L^2(X\times Y)$.
(Here $e_i\cdot f_j$ is a function on $X\times Y$ such that its value at $(x,y)\in X\times Y$ is $e_i(x)\cdot f_j(y)$.) 
My Attempt: Assume that the basis functions are all orthonormal because if they are not, I could just use Graham Schmidt. I know that for any $x \in X$, $$x= \sum_n \langle x, e_n\rangle e_n$$ and for any $y \in Y$, $$y= \sum_m \langle y, f_m\rangle f_m$$
Now, for any element $z \in X \times Y$, I want to show that we can write it using the $e_n$ and $f_n$ basis functions. This is where I am having a tough time and was hoping to get some help.
Thank you.  
 A: To simplify things slightly, I will only present the result for separable Hilbert spaces (i.e. the ones with a countable basis, which suffices in your case). 

Let $\mathcal H$ be a Hilbert space and $\{e_n\}_{n=1}^\infty \subset \mathcal H$ an orthonormal system. Then
  $$\{e_n\}_{n=1}^\infty \text{ is a basis for }\mathcal H \quad\iff\quad \langle x,e_n\rangle=0 \;\text{ for all } n\geq1 \text{ implies that } x = 0. $$

Since you said that you found a proof for $(\Rightarrow)$ in the literature, I will only prove $(\Leftarrow)$. So suppose that $\langle x,e_n\rangle=0$ for all $n\geq 1$ implies that $x=0$. Let $x \in \mathcal H$ and consider
$$ x_n = \sum_{j=1}^n \langle x, e_j \rangle e_j. $$
For $n > m$, we then have
\begin{align}
\Vert x_n-x_m \Vert^2 &= \left\Vert \sum_{j=m+1}^n \langle x, e_j \rangle e_j \right\Vert^2 = \left\langle \sum_{j=m+1}^n \langle x, e_j \rangle e_j, \sum_{j=m+1}^n \langle x, e_j \rangle e_j \right\rangle\\
&= \sum_{j=m+1}^n\sum_{k=m+1}^n \langle x, e_j \rangle \overline{\langle x, e_k\rangle} \langle e_j, e_k \rangle = \sum_{j=m+1}^n |\langle x, e_j \rangle|^2,
\end{align}
since $\langle e_j, e_k \rangle = 0$ whenever $j\not=k$. This in fact shows that $(x_n)_{n=1}^\infty$ is a Cauchy sequence, since
$$ \left(\sum_{j=1}^n |\langle x, e_j \rangle|^2\right)_{n=1}^\infty $$
is an increasing sequence which is upper bounded by $\Vert x\Vert^2$ (Bessel's inequality), and hence convergent, and in particular Cauchy.
Since $\mathcal H$ is a Hilbert space, it is complete, so $x_n$ converges:
$$ x_n = \sum_{j=1}^n \langle x, e_j \rangle e_j \to \tilde x = \sum_{j=1}^\infty \langle x, e_j \rangle e_j \in \mathcal H $$
It is easily checked that this limit is indeed equal to $x$, by proving that
$$ \langle x-\tilde x, e_n \rangle = \lim_{k\to\infty} \langle x-x_k, e_n \rangle = 0 $$
for all $n\geq 1$, and then using the assumption. Can you do that?
Please let me know if that answers your question.
