Let $(X\mathcal M,\mu)$ and $(Y,\mathcal N,v)$ be $\sigma$-finite measure spaces.
If $f:X\to\Bbb R$ is $\mathcal M$-measurable, $g:Y\to\Bbb R$ is $\mathcal N$-measurable, and $h(x,y)=f(x)g(y)$, prove that $h$ is $\mathcal M\otimes \mathcal N$-measurable.
If $f\in L^1(\mu)$ and $g\in L^1(v)$, prove that $h\in L^1(\mu\times v)$ and $$\int\, h\,d(\mu\times v)=\left[\int\,f\,\mathrm d\mu\right]\left[\int\,g\,\mathrm dv\right].$$
Could someone show me how to prove this. I have been staring at it, and cannot even think where to start. I am a bit desperate here. I will highly rate for your help. Thank you.
$\endgroup$3 Answers
$\begingroup$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, $$\begin{align} \int\,fgd(\mu\times v)&=\int(fg)^+d(\mu\times v)-\int(fg)^-d(\mu\times v)\\ &=\int f^+g^+d(\mu\times v)+\int f^{-}g^-d(\mu\times v)-\int f^+g^-d(\mu\times v)-\int f^-g^+d(\mu\times v)\\ &=\left(\int f^+d\mu-\int f^-d\mu\right)f g^+ dv-\left(\int ^+d\mu-\int f^-d\mu\right)\int g^-dv\\ &=\int fd\mu\int gdv. \qquad\blacksquare \end{align}$$
$\endgroup$ 7 $\begingroup$@Sarah, here is my proof..... which I think is a little pretty because I don't use sequences.
Let $F(x,y):= f(x) \forall y \in Y$, $G(x,y):=g(y) \forall x\in X$. Then $$F^{-1}([-\infty, a))=\{(x,y) \in X\times Y : F(x,y)\in [-\infty,a)\}$$ $$=\{(x,y) \in X\times Y : f(x)\in [-\infty,a), \forall y\in Y\}$$ $$=\{x \in X : f(x)\in [-\infty,a) \}\times Y\in \mathcal{M}\times\mathcal{N}$$ as $Y\in \mathcal{N}$ and $\{x \in X : f(x)\in [-\infty,a) \}\in \mathcal{M}$ as $f$ is $\mathcal{M}$-measurable; similarly, we can do this for $g$.
By Tonelli's theorem we have $|F(x,y)G(x,y)|=|f(x)g(y)|=|h|:X\times Y \to [0,\infty]$ that $|h|$ is integrable because it is measurable (product of measurable functions is measurable and absolute value of measurable function is measurable) and nonnegative. On the other hand, $|h|$ is integralbe if and only if $h$ is integrable (hence $h$ is integrable).
now by Fubini's theorem,because $h$ is integrable and $f$ and $g$ are measurable we have
$$\int_{X\times Y} h d(\mu \times \nu) = \int_X (\int_Y f(x) g(y) d\nu (y) ) d\mu (x) = (\int_X f d\mu )(\int_Y g d\nu )$$
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