Prove that $\int f(x)g(y) \,d(\mu \times \nu) = [\int f\, d\mu ][\int g \,d\nu]$ I got stuck on this problem to figure out how to calculate the integral on left-handed side, because we can't use Tonelli-Fubini theorem for this problem (lack of $\sigma$-finite condition). Hope someone can help me or give some hint to solve it. 
I'm thinking about building two simple function sequences $\{f_n\}$ and $\{g_n\}$ which converge to $f$ and $g$ correspondingly, but can't come to the conclusion. Thanks so much.


Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, v)$ be arbitrary measure spaces. If $f \in L^1(\mu)$ and $g \in L^1(v)$, prove that $h(x, y) = f(x)g(y) \in L^1(\mu \times v)$ and $\int h \, d(\mu \times v) = [\int f\,d\mu][\int g\,dv]$.


 A: You're on the right track with using simple functions. Let's just reduce to the case where $f\geq 0, g\geq 0$ (this is a typical argument, assuming it holds for this special case, proving it holds for $f=f^+-f^-$ and then $f=\text{Re} f+i\text{Im} f$).
Let $(f_n)_{n\in\mathbb{N}},(g_n)_{n\in\mathbb{N}}$ be a sequence of simple functions converging monotonically pointwise to $f,g$ respectively. Then define $F_n:X\times Y\to[0,\infty]$ by $F_n(x,y)=f_n(x)$ and similarly for $G_n(x,y)=g_n(y)$. It is very easy to show that this is $\mu\times \nu$ measurable. Moreover, $F_nG_n$ converges pointwise monotonically to $h$.
Now the proof: Prove that $$\int F_n G_n d(\mu\times\nu)=(\int F_nd\mu)\cdot(\int G_nd\nu)$$for fixed $n$. This is relatively simple:
Let $$F_n=\sum_{i=1}^k a_i\chi_{A_i\times Y}, G_n=\sum_{j=1}^m b_j\chi_{X\times B_j}$$where $f_n=\sum_i a_i\chi_{A_i},g_n=\sum_j b_j\chi_{B_j}$ and for $a_i,b_j\in[0,\infty]$, $A_i\in \mathcal{M},B_j\in\mathcal{N}$. Then it is easy to see that $$F_nG_n(x,y)=\sum_{i=1}^k\sum_{j=1}^m a_ib_j\chi_{A_i\times B_j}$$by mutliplying these functions. Then \begin{align*}\int F_nG_nd(\mu\times\nu)=\sum_{i=1}^k\sum_{j=1}^m a_ib_j\mu\times\nu(A_i\times B_j)=\sum_{i=1}^k\sum_{j=1}^m a_ib_j\mu(A_i)\nu(B_j)\\=(\sum_{i=1}^k a_i\mu(A_i))\cdot(\sum_{j=1}^m b_j\nu(B_j)=(\int F_nd\mu)\cdot (\int G_nd\nu)\end{align*}Then the result follows by the Monotone Convergence Theorem and the fact that the limit of the product is the product of the limits.
