I'm trying to prove this property that I read on my lectures notes but wasn't proven.
Given X real valued random variable over ($\Omega,\mathcal F,P $). $\mathcal G \subset \mathcal F$ sub $\sigma$-algebra and V $\mathcal G$-measurable and bounded $$ E[VX] = E[VE[X|\mathcal G]] $$
Notice that I use this property later on to prove the Pull-out property
I think the safest way would be to choose $V=\chi_G $ for some $G\in \mathcal G$ but I'm not able to prove it even in this way
SOLUTION:
Consider $V=\chi_G $ $$ E[VX]=\int_G X~dP = \int_G E[X|\mathcal G]~dP = \int_\Omega V E[X|\mathcal G]~dP =E[VE[X|\mathcal G] $$ Then this result also hold for linear combination of characteristic functions, i.e. simple functions. Using monothone convergence theorem and the fact that every measurable function can be seen as the pointwise limiti of a sequence of simple functions, we get the result for non-negative X. Thus considering the decomposition $X=X^+ - X^-$ we can extend the result to all integrable X
Does this work out in your opinion?
EDIT:
The definition of E(X∣$\mathcal G$) is:
E(X∣$\mathcal G$) is a $\mathcal G$-measurable random variable such that $\forall G \in \mathcal G, \int_G E(X|\mathcal G)~dP = \int_G X~dP $