Product of $L^2$ functions on different measure spaces $\mathbf{Problem}$: This problem comes from Folland's Real Analysis book, exercise 5.61. I know that in Orthonormal basis for product $L^2$ space this question was asked and answered, but my question is on the more basic part of the problem: given $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$, and given functions $f \in L^2(\mu)$ and $g \in L^2(\nu)$, can we show that $fg \in L^1(\mu \times \nu)$ (where we define $(fg)(x,y) = f(x)g(y)$)? Moreover, why are we justified to write $\langle f_n\times g_m, f_s\times g_t\rangle = \langle f_n,f_s\rangle\,\langle g_m,g_t\rangle$ for $f_i$ and $g_j$'s being elements of orthonormal bases in $L^2(\mu)$ and $L^2(\nu)$ respectively?
$\mathbf{Attempt}$: I was not able to show $fg$ being in $L^1$ for the whole space $X \times Y$, but was able to do it for the finite measure spaces that partition $X \times Y$. Assume that $X = \cup_i X_i$ and $Y = \cup_i Y_i$, where $\{X_i\}_i$ and $\{Y_i \}_i$ are finite measure spaces. Consider the functions $F(x,y) = f(x)1(y)$ and $G(x,y) = g(y)1(x)$ where $1(\cdot) = 1 \in \mathbb{R}$. Then $F^{-1}((a,\infty)) = f^{-1}((a,\infty)) \times Y$ for arbitrary $a \in \mathbb{R}$, similarly for $G$, so the two new functions are $\in L^+(X \times Y)$. Now consider:
\begin{align}
\int_{X_i \times Y_j}|f(x)g(y)| d(\mu \times \nu)(x,y) 
&\le \int_{X_i \times Y_j}|f(x)1(y) \cdot g(y)1(x)| d(\mu \times \nu)(x,y) \\
& \le (\int_{X_i \times Y_j}|f(x)1(y)|^2 d(\mu \times \nu)(x,y)^{1/2}) \times \\
& (\int_{X_i \times Y_j}|g(y)1(x)|^2 d(\mu \times \nu)(x,y)^{1/2}) \\
& = (\int_{X_i}|f(x)|^2 d \mu(x) \int_{Y_j} 1(y) d\nu(y))^{1/2} \times \\
& (\int_{Y_j}|g(y)|^2 d \nu(y) \int_{X_i} 1(x) d\mu(x))^{1/2} \lt \infty
\end{align}
Where the second inequality follows from Holder's inequality, and the last equality follows from Tonelli's theorem. So it seems that the most I can show is $fg \in L^1(X_i \times Y_j)$, and the proof of orthonormality of the product of the bases then follows from this observation. Firstly, is my reasoning correct? Secondly, is there a more general answer for my question? Any help is appreciated!
 A: Your estimates are correct but they merely reprove the fact that on finite measure spaces $L^2 \subset L^1$. 
For $f\in L^1(\mu)$, $g\in L^1(\nu)$ you always have $fg \in L^1(\mu\nu)$ which is simply a consequence of Fubini's theorem. That's all you can hope for in that generality since for the singleton $(Y,\nu)=(\{\mathrm{pt}\},\#)$ the "product" $fg$ is just a scaled $f$.
For $L^2$ you can also prove that the map $f\otimes g \mapsto fg$ defines an isometric isomorphism $L^2(\mu)\hat\otimes L^2(\nu)\rightarrow L^2(\mu\nu)$ (where $\hat \otimes$ is the tensor product of Hilbert spaces). To even phrase the according sattement for $p\neq 2$ you need more sophisticated versions of the tensor product, that I'm not very familar with but I would guess that with the right choice the statement stays true.
A: I don't think it follows that $fg \in L^1(\mu \times v)$. The reason you were able to prove it for finite measure spaces is that $L^2 \subseteq L^1$. Consider $f(x) = \frac{1}{x}\chi_{x > 1}, g(y) = \frac{1}{y}\chi_{y > 1}$ on $\mathbb{R}$. These are in $L^2$ but not $L^1$. On $\mathbb{R}\times\mathbb{R}$, $fg$ is given by $(fg)(x,y) = f(x)g(y) = \frac{1}{x}\frac{1}{y}\chi_{x,y > 1}$. This is definitely not in $L^1(\mathbb{R}\times\mathbb{R})$. Just integrate it over the set $\{(x,y) : x> 1, 2 < y < 3\}$ for example.
