Let $X,Y$ be discrete random variables from $(\Omega,\mathcal{A},\mathbb{P})$ to $(\mathbb{R}^n,\mathcal{B}^n)$ and $(\mathbb{R}^m,\mathcal{B}^m)$ respectively.
Define $Q:\mathcal{B}^m\times \mathbb{R}^n \rightarrow [0,1]$ as follows:
$Q(B,x)=\mathbb{P}(Y^{-1}(B) \cap X^{-1}(x))/ \mathbb{P}(X^{-1}(x))$ if $\mathbb{P}(X^{-1}(x))\neq 0$
$Q(B,x)=P(B)$ otherwise ($P$ is a probability on $(\mathbb{R}^m,\mathcal{B}^m)$
Now for fixed $B\in\mathcal{B}^m$, is the function $x\mapsto Q(B,x)$ measurable? If yes, how can I prove it? (my background in measure theory is very limited)
Thank you!
Yes, the function $f(x)=Q(B,x)$ is measurable. There are only countably many values of $x$ such that $\mathbb{P}(X^{-1}(x))\neq 0$ (if there were uncountably many, then you could pick a countable collection such that the values $\mathbb{P}(X^{-1}(x))$ sum to more than $1$). So, this function $f$ is just defined one way on some countable subset $C\subset\mathbb{R}^n$ and is the constant function with value $P(B)$ on the rest of $\mathbb{R}^n$. This implies $f^{-1}(S)$ is either countable or cocountable for any set $S$ (since $f^{-1}(S)$ must either contain $\mathbb{R}^n\setminus C$ or be a subset of $C$). Since any countable or cocountable subset of $\mathbb{R}^n$ is measurable, this means $f$ is measurable.