Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) \in L^1(\Omega,\mathcal{A},P)$. Moreover let $$ g(x)=E[f(x,Y)] \text{ if } \vert E[f(x,Y)]\vert <\infty \text{ and } 0 \text{ otherwise} $$ Then show that $g$ is Borel on $\mathbb{R}$ and that $$ E[f(X,Y)\mid X]=g(X) $$
My attempt 1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)\mid X]=g(X)$ we need to show that g(X) is $\sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A \in \sigma(X)$ we have \begin{equation} \label{1} E[f(X,Y) 1_A]=E[g(X) 1_A] \end{equation} Now I can show that if $A$ is of the form $A=\{X=x\}$ then the equation above holds since $$ E[f(X,Y) 1_A]=E[f(X,Y) 1_{\{X=x\}}]=E[f(x,Y) 1_{\{X=x\}}]=E[f(x,Y)] E[1_{\{X=x\}}]=g(x)E[1_{\{X=x\}}]=E[g(x)1_{\{X=x\}}]=E[g(X)1_{\{X=x\}}] $$ where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $\sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
Second part something like:$$\mathbb Ef(X,Y)\mathbf1_{\{X\in B\}}=\int\int f(x,y)\mathbf1_B(x)\;dF_Y(y)dF_X(x)=\int\mathbf1_B(x)\int f(x,y)\;dF_Y(y)dF_X(x)=$$$$\int\mathbf1_B(x) g(x)dF_X(x)=\mathbb Eg(X)\mathbf1_{\{X\in B\}}$$where the first equality is based on independence.
What we use here too is that $\sigma(X)=\{\{X\in B\}\mid B\in\mathcal B(\mathbb R)\}$