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I read a proof of following property: suppose $X$ is a random variable on the probability space $(\Omega,F,P)$, $A$ and $B$ are sub $\sigma$-field of $F$, $B$ is independent of $\sigma(X,A)$, then $E(X|\sigma(A,B))=E(X|A)$.

Proof: denote $E(X|A)$ by $Y$, we want to show for all $C \subset \sigma(A,B)$,$E(X1_C)=E(Y1_C)$. For $C=A' \cap B'$ where $A' \in A,B' \in B$, it is trivial, we can show it also holds for the field generated by $\{C| C=A' \cap B'\}$. Because the monotone class generated by a field equals to the $\sigma$-field generated by the same field, we get the conclusion.

Question: WHY $E(X1_C)=E(Y1_C)$ holds on the field generated by $\{C| C=A' \cap B'\}$?

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