Problem similar to folland chapter 2 problem 51.
The problem is easy if $f$ and $g$ are characteristic functions of measurable sets. Then we use the fact that we can approximate pointwise a measurable functions by linear combinations of characteristic functions of measurable sets.
We assume first that $f$ and $g$ are non-negative. We use the definition of the product measure when $f=\chi_A$ and $g=\chi_B$. Then we extend to simple functions and use monotone convergence theorem.
I think I solved the problem. If anyone has comments, I would greatly appreciate them.
$\text{a)}$ Define $\phi:X\times Y\to\Bbb R\times\Bbb R$ by $\phi(x,y)=\left(f(x),g(y)\right)$ and define $\psi:\Bbb R\times\Bbb R\to\Bbb R$ by $\psi(m,n)=mn$. Notice that $h$ is the composition of these two functions, i.e., $$h(x,y)=f(x)g(y)=\left(\psi\circ\phi\right)(x,y).$$ Since a continuous function of a measurable function is measurable, it suffices to show that $\phi(x,y)$ is a $(\mathcal M\otimes\mathcal N)$-measurable function from $X\times Y$ into $\Bbb R\times \Bbb R$. Let $R$ be an open rectangle in $\Bbb R\times\Bbb R$ such that $R=A\times B$ for some open sets $A$ and $B$ in $\Bbb R$. Then $$\begin{align} \phi^{-1}(R)=\phi^{-1}(A\times B)&=\{(x,y):f(x)\in A,g(y)\in B\}\\\,\\ &=\{(x,y):f(x)\in A\}\cap\{(x,y):g(y)\in B\}\\\,\\ &=\left(f^{-1}(A)\times Y\right)\cap\left(X\times g^{-1}(B)\right)\\\,\\ &=f^{-1}(A)\times g^{-1}(B). \end{align}$$Since $f^{-1}(A)\in\mathcal M$ and $g^{-1}(B)\in\mathcal N$, and $f$ and $g$ are $\mathcal M$ and $\mathcal N$ measurable, respectively. Thus, $\phi^{-1}(R)\in\mathcal M\otimes\mathcal N$, and thus, $h(x,y)$ is $(\mathcal M\otimes\mathcal N)$-measurable. $\,\blacksquare$
$\text{b)}$ Let us suppose $f$ and $g$ are non-negative measurable functions. Let $\{\phi_n\}$ and $\{\psi_n\}$ be sequences of increasing functions such that $\phi_n\to f$ and $\psi_n\to g$ converge pointwise on $X$ and $Y$, respectively. Then $\{\phi_n\psi_n\}$ is an increasing sequence such that $\phi_n\psi_n\to h$. By the monotone convergence theorem, we have $$\begin{align} \int\,fgd(\mu\times v)=\lim_{n\to\infty}\int\,\phi_n\psi_nd(\mu\times v)&=\lim_{n\to\infty}\left(\int\phi_nd\mu\int\psi_ndv\right)\\ &=\left(\lim_{n\to\infty}\int\phi_nd\mu\right)\left(\lim_{n\to\infty}\int\psi_ndv\right)\\ &=\int\,fd\mu\int\,gdv. \end{align}$$$\quad$Since $|fg|=|f||g|$ and $f\in L^{-1}(\mu)$ and $g\in L^{-1}(v)$, then $fg=h\in L^{-1}(\mu\times v)$. Also, since $f$ and $g$ are real valued, $$\begin{align} \int\,fgd(\mu\times v)&=\int(fg)^+d(\mu\times v)-\int(fg)^-d(\mu\times v)\\ &=\int f^+g^+d(\mu\times v)+\int f^{-}g^-d(\mu\times v)-\int f^+g^-d(\mu\times v)-\int f^-g^+d(\mu\times v)\\ &=\left(\int f^+d\mu-\int f^-d\mu\right)f g^+ dv-\left(\int ^+d\mu-\int f^-d\mu\right)\int g^-dv\\ &=\int fd\mu\int gdv. \qquad\blacksquare \end{align}$$