Conditional expectation of X given X+Y I suppose this is a trivial question, but sometimes you´re just stuck.

Let $X,Y$ be integrable i.i.d. Random Variables. Then $E(X|\sigma(X+Y)) = E(Y|\sigma(X+Y))$.

To show this I obviously have to verify $\int XZ \; dP = \int YZ \; dP $ for any $\sigma(X+Y)$-measurable $Z$. But I just can´t formulate a rigorous general proof...
Hint: $X$ and $Y$ may be realized on a probability space $\Omega = \Omega_1 \times \Omega_2$.   Consider the symmetry $(x,y) \mapsto (y,x)$ of $\Omega$.
Hint: $X$ and $Y$ may be realized on a probability space $\Omega = \Omega_1 \times \Omega_2$.   Consider the symmetry $(x,y) \mapsto (y,x)$ of $\Omega$.