Computing $\mathbb{E}[h(Y)|X]$ given the distribution of $Y|X=x$ Say I know the random variables $X$ and $Y|X=x$ to have the pdfs $f_X(x)$ and $f_{Y|X=x}(y) = g(y,x)$ respectively. I would like to show that the following (perhaps under some assumptions regarding g) is true:
$$\mathbb{E}[h(Y)|X] = \int_{-\infty}^{\infty}h(y)g(y,X)dy$$ 
As far as I know this has to be done by showing that for all $A \in \sigma(X)$ we have that:
$$\int_A\bigg[\int_{-\infty}^{\infty}h(y)g(y,X)dy\bigg]d\mathbb{P} = \int_Ah(Y)d\mathbb{P}$$ 
But I'm not sure how to proceed from here. The forumla isn't mentioned in the literature I'm studying but it intuitively matches the way I imagine probabilities should work, so my guess is that it's true. Any help appreciated!
 A: Remember that the typical element $A\in\sigma(X)$ has the form $X^{-1}(B)$, where $B$ is a Borel subset of $\Bbb R$. Thus $1_A=1_B(X)$, and your left-hand integral can be written
$$
\int_\Omega 1_B(X)\int_{-\infty}^\infty h(y)g(y,X)\,dy\,d\Bbb P=\int_{-\infty}^\infty \int_{-\infty}^\infty 1_B(x)h(y)g(y,x)f_X(x)\,dy\,dx.
$$
Now use the connection between $g$ and the joint density of $X$ and $Y$, switch the order of integration, and integrate.
A: To elaborate on @John Dawkin's answer, we have
\begin{align}
\int_A\int_\mathbb R h(y)g(y,X)\ \mathsf dx \ \mathsf d\mathbb P &= \int_\Omega \mathsf 1_B(X)\int_{\mathbb R} h(y)g(y,X)\ \mathsf dy\ \mathsf d\mathbb P\\
&= \int_{\mathbb R}\int_{\mathbb R} h(y)f_{X,Y}(x,y)\mathsf 1_B(x)\ \mathsf dy\ \mathsf dx\\
&= \int_{\mathbb R} \int_{\mathbb R} h(y)f_{Y\mid X=x}(x)f_Y(y)\mathsf 1_B(x)\mathsf dy\ \mathsf dx\\
&= \int_{\mathbb R}f_{Y\mid X=x}(x) \mathsf 1_B(X)\int_{\mathbb R}h(y)f_Y(y)\ \mathsf dy\\
&= \int_{\mathbb R} f_{Y\mid X=x}(x) \int_{\Omega} h(Y)\mathsf 1_A\mathsf d\mathbb P\ \mathsf dx\\
&= \int_{\mathbb R}f_{Y\mid X=x}(x)\mathsf dx \int_A h(Y)\ \mathsf d \mathbb P\\
&= \int_A h(Y)\ \mathsf d \mathbb P,
\end{align}
so that $$\mathbb E[h(Y)\mid X] = \int_{\mathbb R}h(y)g(y,X)\ \mathsf dy.$$
