Product of Hilbert bases of $L^2(\mathbb{R}^p)$ and $L^2(\mathbb{R}^q)$ is a Hilbert basis for $L^2(\mathbb{R}^{p+q})$ Let $(\alpha_n)_n$ be a Hilbert basis of $L^2(\mathbb{R}^p)$ and let $(\beta_k)_k$ be a Hilbert basis for $L^2(\mathbb{R}^q)$. I need to show that $(\alpha_n \beta_k)_{(n,k) \in \mathbb{Z}}$ is also a Hilbert basis for $L^2(\mathbb{R}^{p+q})$.
My attempt would be, given $f(x,y) \in L^2(\mathbb{R}^{p+q})$ to first approximate $f$ by the first variable, namely there exist coefficients $c_n(y)$ such that $f_y(x) = \sum c_n(y) \alpha_n(x)$. Now $c_n(y)$ can also be represented as $c_n(y) = \sum d_k \beta_k(y)$ and using the Fubini theorem for series, we can conclude the result. However there seems to be something missing regarding the measurability of $c_n(y)$, in order to be able to show that it is indeed in $L^2$. Is there a way to salvage this approach?
P.S. Is it also possible to maybe use the result that all separable Hilbert spaces are isomorphic to $l^2(\mathbb{Z})$?
 A: To show that $\{ \alpha_n(x)\beta_m(y) \}_{n,m}$ is an orthonormal basis, suppose that $f\in L^2(\mathbb{R}^{p}\times\mathbb{R}^{q})$ satisfies $\int_{\mathbb{R}^{p+q}}\alpha_n(x)\beta_m(y)f(x,y)dxdy = 0$ for all $n,m$. It is shown that $f=0$ a.e..
Because $\alpha_n(x)\beta_m(y)f(x,y) \in L^1(\mathbb{R}^{p+q})$, Fubini's Theorem for complete $\sigma$-finite measures gives
$$
           0=\int_{R^{p+q}}\alpha_n(x)\beta_m(y)f(x,y)dm=\int_{R^q}\left(\int_{R^p}\alpha_n(x)f(x,y)dx\right)\beta_m(y)dy.
$$
The inner integral on the right is defined for almost all $y$ and the resulting function of $y$ is measurable. By completeness of $\{ \beta_m \}$ in $\mathbb{R}^q$, the inner integral must be $0$ for almost all $y$. This holds for every $n$; though the exceptional set of measure $0$ may be change with $n$, there exists a set $E\subset\mathbb{R}^q$ of measure $0$ such that
\begin{align}
       1. & \;\;\int_{\mathbb{R}^p}\alpha_x(x)f(x,y)dx=0,\;\;\; y \in \mathbb{R}^q\setminus E,\\
       2. & \;\;x\mapsto f(x,y) \mbox{ is measurable for fixed } y \in \mathbb{R}^q\setminus E.
\end{align}
The completeness of $\{\alpha_n\}$ in $L^2(\mathbb{R}^p)$ then implies that, for any fixed $y\in\mathbb{R}^q\setminus E$, the function $F_y(x)=f(x,y)$ is $0$ for almost all $x\in\mathbb{R}^{p}$, which is enough to conclude that
$$
   \|f\|^2=\int_{R^{p+q}}|f(x,y)|^2 dm = \int_{R^{q}}\left(\int_{R^{p}}|f(x,y)|^2dx\right)dy = 0.
$$
Therefore, $f=0$ a.e., which completes the proof that $\{\alpha_n(x)\beta_{m}(y)\}_{n,m}$ is a complete orthonormal basis of $L^{2}(\mathbb{R}^{p+q})$.
