Product measure and measurability

Let $(X,\mathcal{G})$ and $(Y,\mathcal{H})$ be measure spaces, and $f:X\times Y\rightarrow \mathbb{R}$ be measurable with respect to the product measure space $(X\times Y,\sigma(\mathcal{G}\times\mathcal{H}))$.

Show that $x\mapsto f(x,y)$ is a Borel measurable function.

Any help is greatly needed

• Could you show us what you have attempted so far? – Lost in a Maze May 5 '15 at 15:21
• One idea was to try bootstrapping as $1_{A\times B}=1_A.1_B$ but that seems not to work and too much brute force. Another idea is consider it as a composition of maps, so for a fixed $y$, $x\mapsto (x,y)\mapsto f(x,y)$, but I am not sure why $x\mapsto (x,y)$ is measurable, and it seems to be the same question I am trying to show. I am also confused as what is the pre-image relation for the map $x\mapsto f(x,y)$ as I don't think it is $f^(-1)$ as that would send it to $X\times Y$ and not just $X$ Are any of these right. Can you clear things up for me? Thanks – Edwyn May 5 '15 at 18:29
• @Edwyn: Yes, it is easy enough to show that the map $x \mapsto (x,y)$ is measurable for any fixed $y$, because each of the component functions are measurable. – Shalop May 6 '15 at 3:31

Fix $y_0\in Y$ and let $g(x)=f(x,y_0)$. To show that $g$ is measurable, it suffices to show that if $a<b$, then $g^{-1}(a,b)\in\mathcal G$. We have $$g^{-1}(a,b) = \{x\in X: a<g(x)<b\} = \{x\in X: a<f(x,y_0)<b\}.$$ Since $f$ is $\sigma(\mathcal G\times\mathcal H)$-measurable, we know that $$f^{-1}(a,b) = \{(x,y)\in X\times Y : a<f(x,y)<b\}\in\sigma(\mathcal G\times\mathcal H).$$ Recall that if $(W,\mathcal M)$ and $(Z,\mathcal N)$ are measure spaces and $E\in\sigma(\mathcal M\times\mathcal N)$, then for any $w\in W$, $$E_w = \{z\in Z: (w,z)\in E\}\in\mathcal N$$ and for any $z\in Z$, $$E^z = \{w\in W: (w,z)\in E\}\in\mathcal M.$$
It follows that $$g^{-1}(a,b) = f^{-1}(a,b)^{y_0} = \{x\in X:(x,y_0)\in f^{-1}(a,b)\}\in\mathcal G.$$
• Thanks, but I don't see why the last line is true. What is a generic element of $\sigma(\mathcal{G}\times\mathcal{H})$ I think that's were my confusion lies as $\mathcal{G}\times\mathcal{H}$ is not necessarily $\sigma(\mathcal{G}\times\mathcal{H})$. Thanks again and would really appreciate if you can explain further. – Edwyn May 5 '15 at 19:30
• Thanks again but I am a bit slow today so if you don't mind can you please show me why $E_w\in\mathcal{N}$ – Edwyn May 5 '15 at 19:49