Explain why $\int_\Omega 1_B(X+Y)X \, dP = \iint_{\mathbb R^2} 1_B(x+y)x\,d\mu_{X,Y}(x,y)$, where $\mu_{X,Y}$ denotes the distribution of $(X,Y)$ 
Let $X,Y$ be independent r.v.s on $(\Omega, \mathscr{F}, P)$ having
   the same distribution i.e. the measures $P\circ X^{-1}= P\circ Y^{-1} :
 \mathscr{B}(\mathbb{R}) \rightarrow [0,1]$ are equal. Also, $E(|X|), E(|Y|)<\infty$.  Then
   $E[X|X+Y]=E[Y|X+Y]$.

In the proof given here,$\mu_X$, $\mu_Y$ and $\mu_{X,Y}$ denote the distribution measures. I do not understand why 

$$\int 1_B(X+Y)X \, dP = \int 1_B(x+y)x\,d\mu_{XY}(x,y)$$

A more general question would be, 

if $f= \hat
{f} (X,Y):\Omega \rightarrow \mathbb{R}$ where $\hat{f} :\mathbb{R}^2 \rightarrow \mathbb{R}$ is Borel measurable, then does it hold that,$$\int \hat{f}(X,Y) \, dP = \int \hat{f} \, d\mu_{X,Y} $$

 A: Let $(\Omega,\mathcal A,\mathsf P)$ be a probability space.
Every  random variable vector $X\to\mathbb R^n$ induces a probability measure $\mathsf P_X$ on $(\mathbb R^n,\mathcal B^n)$.
This by $B\mapsto\mathsf P(X\in B)$, or equivalently $\mathsf P_X(B)=\mathsf P(X\in B)$.
We could also write this equality as:$$\int 1_B(x)\mathsf P_X(dx)=\int 1_B(X(\omega))\mathsf P(d\omega)\tag1$$
In $(1)$ we use indicator function $1_B$ but it can be extended to all Borel measurable functions:$$\int f(x)\mathsf P_X(dx)=\int f(X(\omega))\mathsf P(d\omega)$$
A: Consider the function $(X,Y) : \Omega\to\mathbb R^2$. This defines the pushforward measure $\mu_{X,Y} = P\circ (X,Y)^{-1}$ on $\mathbb R^2$. It is well known that you integrate a measurable function $f : \mathbb R^2\to\mathbb R$ over $A\subset\mathbb R^2$ with respect to that measure as follows:
$$
\int_A f\,d\mu_{X,Y} = \int_{(X,Y)^{-1}(A)}f\circ (X,Y)\,dP.
$$
Here, $A = \mathbb R^2$ and $f(x,y) = 1_B(x+y)\cdot x$. So, $(X,Y)^{-1}(A)= \Omega$ and $f\circ (X,Y) = 1_B(X+Y)X$.
