Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$

Question

The original question is the number 10.6 of this pdf:

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be a bounded Borel-measurable function, and let $X$, and $Y$ be independent random variables. Define the function $g:\mathbb{R}\rightarrow \mathbb{R}$ by $g(y) = E[f(X,y)]$. Show that the function g is Borel-measurable, and that $E[f(X,Y)|Y=y] = g(y)$.

This is what I tried:

$$g(y) = \int_{\Omega}f(X,Y)1_{\{Y = y\}}dP = \int_{\{Y = y\}}f(X,Y)dP$$

I know that $\{Y = y\} \in \sigma(Y)$ and it is a set where $Y$ is constant, so it follows:

$$g(y) = \int_{\{Y = y\}}f(X,Y)dP = E[f(X,Y)|Y=y]$$

Proving that $g$ is Borel-measurable: I will aproximate $f(X,y)$ by simple Borel-medible functions. The sets $A1, A2, A3 ... An$ are a partition of $\{Y=y\}$: $$f(X,y) = \lim_{n\rightarrow\infty}\sum_{i}^{n} f_i(X,y)1_{Ai}$$ $$g(y) = \int_{\{Y=y\}}[\lim_{n\rightarrow\infty}\sum_{i}^{n} f_i(X,y)(1_{Ai})]dP = \lim_{n\rightarrow\infty}\sum_{i}^{n} f_i(X,y)P(Ai)$$ The Lebesgue integral is a lineal operation on Borel measurable functions, so it follows that $g$ is Borel measurable.

I don't know if I'm right, and I don't know why it is necesary the independence of X and Y.

Answer 1

You need only consider function $f$ of the form \begin{align*} f(x, y) = \pmb{1}_{(x, y) \in A_1 \times A_2}, \end{align*} where $A_1$ and $A_2$ are Borel sets. The general case can be proved by approximation of linear combinations of such functions.

By definition, $E(f(X, Y) \mid Y=y) = h(y)$ is a Borel measurable function such that, for any Borel set $A$, \begin{align*} \int_{Y \in A} f(X, Y) dP = \int_A E(f(X, Y) \mid Y=y) P_Y(dy), \end{align*} where $P_Y(dy)$ is the Lebesgue-Stieltjes measure generated by the distribution function of $Y$, that is $P_Y(A) = P(Y \in A)$. Then \begin{align*} \int_A h(y) P_Y(dy) &=\int_{Y \in A} f(X, Y) dP\\ &= E\big(\pmb{1}_{X \in A_1} \pmb{1}_{Y \in A \cap A_2} \big)\\ &= E\left( \pmb{1}_{X \in A_1}\right)E\left( \pmb{1}_{Y \in A \cap A_1}\right)\\ &= P(X \in A_1)P(Y \in A \cap A_1). \end{align*} On the other hand, \begin{align*} g(y) &= E(f(X, y)) \\ &=\pmb{1}_{y \in A_2} P(X \in A_1). \end{align*} Then, \begin{align*} \int_A g(y) P_Y(dy) &=\int_A \pmb{1}_{y \in A_2} P(X \in A_1) P_Y(dy)\\ &= P(X \in A_1) \int_A \pmb{1}_{y \in A_2}P_Y(dy)\\ &= P(X \in A_1)P(Y \in A \cap A_1). \end{align*} Therefore, $h(y)=g(y)$, that is, \begin{align*} E(f(X, Y) \mid Y=y) = E(f(X, y)). \end{align*}