If $Y=\mathbb{E}[X\mid\mathcal{B}]$, then $Y(\omega)=\frac{1}{P(A_i)}\int_{A_i}XdP$

Question

Let $(\Omega, \mathcal{A}, P)$ be a probability space, and $\mathcal{B}\subset \mathcal{A}$ be a sub-$\sigma$-algebra generated by the partition $\{A_1, \ldots, A_n\}$ of $\Omega$, where $P(A_i)>0$, $i=1,\ldots, n$. I want to prove that if \begin{align} 1)&\int_BYdP= \int_BXdP\quad\forall B\in \mathcal{B} \\ 2)&\: Y \:\mbox{is} \: \mathcal{B}-\mbox{measurable} \end{align} Then $\forall \omega \in A_i$: $$Y(\omega)=\frac{1}{P(A_i)}\int_{A_i}XdP$$ I know that $Y=\mathbb{E}[X\mid\mathcal{B}]$, and I also know that $$\mathbb{E}[X\mid A_i]=\frac{1}{P(A_i)}\int_{A_i}XdP$$ Then all I need to know it that if $\omega \in A_i$, then $\mathbb{E}[X\mid\mathcal{B}](\omega)=\mathbb{E}[X\mid A_i]$ (this would be well-defined since $\omega$ can only be in one $A_i$). But I am not sure how can I conclude this. Can someone help me?

Answer 1

Hint: Convice yourself that $E[X|]\mathcal{B}]$ must be a linear combination of the functions $\mathbb{1}_{A_1},\ldots,\mathbb{1}_{A_n}$. That is $$ E[X|\mathcal{B}]=a_1\mathbb{1}_{A_1}+\ldots+a_n\mathbb{1}_{A_n}$$ The rest, is like in geometry or linear algebra, find the Fourier coefficients $c_j$ noticing that $E[\mathbb{1}_{A_j}\mathbb{1}_{A_i}]=0$ for $j\neq i$ (since the $A_j$'s form a partition, $\mathbb{1}_{A_j}\mathbb{1}_{A_i}=\mathbb{1}_{A_j\cap A_i}=0$ if $j\neq i$).

For example: $$E[X\mathbb{1}_{A_1}]=E\big[E[X|\mathcal{B}]\mathbb{1}_{A_1}\big]=\sum^n_{j=1}c_jE[\mathbb{1}_{A_j}\mathbb{1}_{A_1}]=c_1 P[A_1]$$ so $c_1=\frac{1}{P[A_1]}E[X\mathbb{1}_{A_1}]$

I hope this helps you finish the problem. Let me know if you still have some questions.