Let $Z$ be a random variable independent of $(X,Y)$. Prove that $\mathbf{E}(Y|\sigma(X))=\mathbf{E}(Y|\sigma(X,Z))$
My attempt: It is obvious that $\int_A\mathbf{E}(Y|\sigma(X,Z))d\mathbf{P}=\int_A\mathbf{E}(Y|\sigma(X))d\mathbf{P}=\int_AYd\mathbf{P}$ for all $A\in \sigma(X)$. However, I'm stuck in proving that $\mathbf{E}(Y|\sigma(X,Z))$ is $\sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?
Thanks in advance!
Proving that RHS (or a version of it) is meaurable w.r.t. $\sigma (X)$ is not easy. Since LHS is measurable w.r.t. $\sigma (X,Z)$ it is enough to show that $\int_{X^{-1}(A)\cap Z^{-1}(B)} E(Y|X)dP=\int_{X^{-1}(A)\cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $\mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P\{Z^{-1}(B)\} \int_{X^{-1}(A)} E(Y|X)dP=P\{Z^{-1}(B)\}\int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.