Here is the question:
Let $(X, \mathcal{S}, \mu)$ and $(Y, \mathcal{T}, \lambda)$ be $\sigma$-finite measure spaces. Suppose that $g: X \rightarrow \mathbb{R}$ is a $\mu-$integrable function and that $h: Y \rightarrow \mathbb{R}$ is a $\lambda-$integrable function. Define $f: X \times Y \rightarrow \mathbb{R}$ by $f(x,y) = g(x)h(y).$ Prove that $f$ is $\mu \times \lambda$ integrable and that $$\int_{X\times Y} f d(\mu \times \lambda) = (\int_{X}g d\mu) (\int_{Y} hd\lambda)$$
I found this question here Proving a formula for the product of two measurable functions but I am wondering if someone can give me more details about the first and last step?
Best Answer
First of all, we may assume that $f$ and $g$ are non-negative.
This is due to the fact that we can write $f$ and $g$ as $f=f_+-f_-$ and $g=g_+-g_-$, where $f_\pm=\max\{\pm f,0\}$ and $g_\pm=\max\{\pm g,0\}$. So if $f$ and $g$ are integrable, so are $f_\pm$ and $g_\pm$ and vice versa. Then $fg=f_+g_++f_-g_--f_+g_--f_-g_+$, where $|fg|\ge |f_+g_+|, |f_-g_-|, |f_+g_-|, |f_-g_+|$.
If $f,g\ge 0$, then there exist increasing sequences of simple measurable functions $f_n\to f$ and $g_n\to g$, with Monotone Convergence Theorem (MCT) providing that $\int_X f_n\,d\mu\to \int_X f\,d\mu$ and $\int_Y g_n\,d\lambda\to \int_Y g\,d\lambda$.
It is now straight-forward that $$ \Big(\int_X f_n\,d\mu\Big)\cdot\Big(\int_Y g_n\,d\lambda\Big)=\int_{X\times Y}f_ng_n\,d(\mu\times\lambda) $$ and, again applying MCT we get that $$ \Big(\int_X f_n\,d\mu\Big)\cdot\Big(\int_Y g_n\,d\lambda\Big)\to \Big(\int_X f\,d\mu\Big)\cdot\Big(\int_Y g\,d\lambda\Big) $$ while $$ \int_{X\times Y}f_ng_n\,d(\mu\times\lambda)\to \int_{X\times Y}fg\,d(\mu\times\lambda) $$