Let $X: \Omega \to \mathbb{R}$ be a random variable with $\mathbb{E}[|X|]< \infty$, and suppose that $Y: \Omega \to \mathbb{R}^d$ is a discrete random vector.
Define $$f(y) = \begin{cases}\frac{1}{P(Y=y)} \int_{\{Y=y\}}XdP \quad P(Y=y) > 0\\ 0 \quad P(Y=y) =0\end{cases}$$
I'm trying to show that $f$ is a version of $E[X|Y=y]$. For this, I'm trying to prove that for $B$ a borel part of $\mathbb{R}^d$:
$$E[XI_{\{Y \in B\}}] = \int_B f(y) P_Y(dy)$$
I tried to work out both sides. Define $S:= \{y \in \mathbb{R}^d: P(Y=y) > 0\}$. This is by assumption at most countable, thus a Borel subset.
$$E[XI_{\{Y \in B\}}] = \int_{\{Y \in B\}} X dP$$
$$\int_B f(y) P_Y(dy) = \int_{B \cap S} \left(\frac{1}{P(Y=y)} \int_{\{Y=y\}}XdP\right)P_Y(dy)$$
Now, I'm stuck as to how to continue. I tried things like Fubini and applying the formula
$$\int_{A} gdP_Y = \int_{\{Y \in A\}} g \circ Y dP$$
but I could not find anything useful.
Any help will be appreciated!
Let $A:=\left\{ y\mid P\left(Y=y\right)>0\right\} $.
Then $A$ is countable, so that:
$$\mathbb{E}X\mathbf{1}_{Y\in B}=\mathbb{E}X\mathbf{1}_{Y\in A\cap B}=\mathbb{E}\sum_{y\in A\cap B}X\mathbf{1}_{Y=y}=\sum_{y\in A\cap B}\mathbb{E}X\mathbf{1}_{Y=y}=$$$$\sum_{y\in A\cap B}f\left(y\right)P\left(Y=y\right)=\sum_{y\in A}f\left(y\right)\mathbf{1}_{y\in B}P\left(Y=y\right)=\mathbb{E}f\left(Y\right)\mathbf{1}_{Y\in B}$$This proves that $\mathbb E[X\mid Y]=f(Y)$.