Let $X$ and $Y$ be $1$ dimensional random variables such that $(X,Y)$ has density $g_{(X,Y)}$. Let $h(x):\mathbf{R}\to \mathbf{R}$ be a borel function such that $h(X)$ is integrable. Then the conditional expectation $E(X \mid Y)$ can be written as a function $\eta(Y)$, where $$\eta:y \mapsto \int\limits_{\mathbf{R}} h(x) g_{X \mid Y}(x, y)\ dx,$$ where $$g_{X \mid Y}(x, y) = \begin{cases} \dfrac{g_{(X, Y)}(x, y)}{g_Y(x)} &\text{ if } g_Y(x) > 0,\\ 0 &\text{ otherwise} \end{cases}$$ Why is $\eta(Y)$ $\sigma(Y)$-measurable? Why does $\int\limits_{\mathbf{R}} h(x) g_{X \mid Y}(x, y)\ dx$ even exist? Does it exist for every $y$ or just almost everywhere?

I'v read that $\sigma(Y)$-measurability follows from Fubini's theorem. I suppose it means that we use Fubini to a function $\phi:(x,y) \mapsto h(x) g_{X \mid Y}(x, y)$ and conclude that $\eta(y)$ is a borel function. But to use Fubini we need to know that $\phi$ is both borel and integrable (since $h(x)$ doesn't have to be positive). I see why it is borel, but not why it is integrable.

I'v got similar problem with the case of $X$,$Y$ being discrete. Denote $S_X$ set of atoms of $X$ and $P_X(x)=P(X=x)$. Why is $\sum_{x\in S_X} \frac{P_{(X,Y)}(x,Y)}{P_Y(Y)}$ (I guess we put $0$ when $P(Y=y)=0$) $\sigma(Y)$-measurable?

Any help would be appreciated. I'm just starting with conditional expectation.


First off there's a typo in your definition of the $g_{X \mid Y}(x, y)$. It should be $$g_{X \mid Y}(x, y) = \begin{cases} \dfrac{g_{(X, Y)}(x, y)}{g_Y(y)} &\text{ if } g_Y(y) > 0,\\ 0 &\text{ otherwise} \end{cases}$$ Let $g_X$ denote the marginal density of $X$: $g(x)=\int g(x,y) dy$. Since $$\int \int |h(x) g(x,y)| dx dy = \int |h(x)| \int g(x,y) dy dx = \int |h(x)|g_X(x) dx =E(|h(X)|)<\infty $$ the function $(x,y)\mapsto h(x) g(x,y)$ is integrable.


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