About integral with regular conditional distribution I'm currently dealing with a problem from my assignment:
 The problem is:
X, Y are two random variables. and suppose Y $\mathcal{G}-measurable$. Let K(w, ·) be a regular conditional distribution for X given $\mathcal{G}$. Prove that, for bounded measurable $f$,
$\mathrm{E}(f(X,Y)|\mathcal{G})(w)=\int f(X,Y(w))K(w,dx)\ \  a.s.$
I want to use 4-step proof which starts from Indicator function $I$
But I'm not quite sure how to write down this, I guess it should be $I_A(X)I_B(Y)$ But I then got stuck, as I'm not quite sure about the how to do integral in right-hand side and thus need some help about this.
 A: Yes, in the first step, we assume that $f\colon (x,y)\mapsto f(x,y)$ has the form $f(x,y)=\mathbf{1}_A(x)\mathbf{1}_B(y)$ where $A$ and $B$ are two Borel subsets of the real line. Then 
$$
\mathbb E\left(f(X,Y)\mid\mathcal G\right)=\mathbb E\left(\mathbf{1}_A(X)\mathbf{1}_B(Y)\mid\mathcal G\right)=\mathbf{1}_B(Y)\mathbb E\left(\mathbf{1}_A(X)\mid\mathcal G\right)
$$ 
since the random variable $\mathbf{1}_B(Y)$ is $\mathcal G$-measurable. 
Now, replacing $f$ by its particular expression, the following equality hold for a fixed $\omega$:
$$
\int f(X,Y(\omega))K(\omega,dx)=\int \mathbf{1}_A(X)\mathbf{1}_B\left(Y(\omega)\right)K(\omega,dx).
$$
The term $\mathbf{1}_B\left(Y(w)\right)$ is not concerned by the integration with respect to the measure $F\mapsto K(\omega,F)$, and we deduce that 
$$
\int f(X,Y(\omega))K(\omega,dx)=\mathbf{1}_B\left(Y(\omega)\right)\int \mathbf{1}_A(X)K(\omega,dx).
$$
Then $\int \mathbf{1}_A(X)K(\omega,dx)$ can be written in a simpler way as $ K(\omega,A)$ and by definition of a regular condition distribution, this is 
$\mathbb E\left(\mathbf{1}_A(X)\mid\mathcal G\right)(\omega)$ hence 
$$
\int f(X,Y(\omega))K(\omega,dx)=\mathbf{1}_B\left(Y(\omega)\right)\mathbb E\left(\mathbf{1}_A(X)\mid\mathcal G\right)(\omega).
$$
