Question involving $\mathbb{E}\left(X \mid A_i \cap B\right)=\mathbb{E}\left(X \mid A_1 \cap B\right)$

Question

Let $\mathcal{A}=\left\{A_1, \ldots, A_k\right\}$ be a partition and $B$ be an event with $\mathbb{P}(B)>0$. Prove that if $\mathbb{E}\left(X \mid A_i \cap B\right)=\mathbb{E}\left(X \mid A_1 \cap B\right)$ for all $i$, then $\mathbb{E}\left(X \mid A_i \cap B\right)=\mathbb{E}(X \mid B)$. In this case we say that $X$ is conditionally independent of $\mathcal{A}$ given $B$.

Stumped on this question, any help would be appreciated.

Answer 1

$$\frac {\int_{A_i\cap B} XdP} {P(A_i\cap B)}=\frac {\int_{A_1\cap B} XdP} {P(A_1\cap B)}.$$ So $$\int_{A_i\cap B} XdP=\frac {P(A_i\cap B)}{P(A_1\cap B)}\int_{A_1\cap B} XdP.$$ Sum over $i$ to get $$\int_{ B} XdP=\frac {P( B)}{P(A_1\cap B)}\int_{A_1\cap B} XdP.$$ Write this as $$\frac {\int_{B} XdP} {P(B)}=\frac {\int_{A_1\cap B} XdP} {P(A_1\cap B)}.$$ This means $E(X|B)=E(X|A_1 \cap B)$.

So $E(X \mid A_i \cap B)=E(X \mid A_1 \cap B)=E(X|B)$.

Answer 2

\begin{gather*} E[X \mid B]= \sum E[X \mid A_i\cap B]\frac{P(A_i\cap B)}{P(B)}=\sum E[X \mid A_1\cap B]\frac{P(A_i\cap B)}{P(B)}=\\ =\frac{E[X \mid A_1\cap B]}{P(B)}\sum P(A_i\cap B)= E[X \mid A_1\cap B] \end{gather*}