Prove $\mathbb{E}(X_{1}+X_{2}|Y)=\mathbb{E}(X_{1}|Y)+\mathbb{E}(X_{2}|Y)$ I would like to prove
$$\mathbb{E}(X_{1}+X_{2}|Y)=\mathbb{E}(X_{1}|Y)+\mathbb{E}(X_{2}|Y)$$
This question is given in the text book as an exercise. It looks trivial but took me hours to think of a proof. I don't know what to do after expand it using definition of expectation...

Use the definition of conditional expectation. in others word, you have to prove that $E(X_1|Y)+E(X_2|Y)$ is $\sigma(Y)$-mesurable and for any $Z$ $\sigma(Y)$-mesurable, $E(Z(X_1+X_2))=E(Z(E(X_1|Y)+E(X_2|Y)))$.
First, $E(X_1|Y)+E(X_2|Y)$ is $\sigma(Y)$-mesurable (by definition of conditional expectation). And for any $Z$ $\sigma(Y)$-mesurable, we have :
$E(Z(X_1+X_2))=E(ZX_1)+E(ZX_2)= E(E(ZX_1|Y))+E(E(ZX_2|Y))$.
Since Z is $\sigma(Y)$-mesurable, we have:
$E(E(ZX_1|Y))+E(E(ZX_2|Y))=E(ZE(X_1|Y))+E(ZE(X_2|Y))$
so $ E(Z(X_1+X_2))=E(Z(E(X_1|Y)+E(X_2|Y)))$.
