Conjecture:
Suppose the probability space $(\Omega, \mathcal F, P)$, where $X \in m\mathcal F$ is a Random Variable which is integrable. Let $\mathcal A \subseteq \mathcal F$, and denote $\sigma(\mathcal A)$ to be the $\sigma$-algebra generated by the collection of sets $\mathcal A$. Then to verify that $\mathbb E[X; A] = \mathbb E[\mathbb E[X | \sigma(\mathcal A)]; A]$ (the "orthogonality condition") for any $A \in \sigma(\mathcal A)$, it suffices to show that the afforementioned holds looking only at $A \in \mathcal A$.
I am studying probability theory, and I found myself wondering whether the above is true. I know it holds if $\mathcal A$ is a partition of $\Omega$, but does it hold in the general case as well for any family of sets? If so, does anybody have a good proof resource, link, or guidance to offer? I can't figure out how to prove it myself.
What I know is true is that if $\mathcal A$ is a countable partition where $P(A_n) > 0$ for all $A_n \in \mathcal A$, then it suffices to check only whether the orthogonality condition holds on $A_n \in \mathcal A$, rather than considering the full $\sigma(\mathcal A)$, but I can't quite work out the proof as to why, nor find a good resource. I am having trouble justifying When $A \in \sigma(\mathcal A)$ is countable. Particularly, I can express $A = \bigsqcup_{n}A_n$ as the disjoint union of elements of the partition (due to the fact that the $\sigma$-algebra generated by a partition is the set containing unions of elements of the partition), and I am able to use the integrability of $X$ combined with the DCT to break $\mathbb E[X; \sqcup_n A_n]$:
\begin{align*}
\mathbb E[X; \sqcup_n A_n] &= \sum_n \mathbb E[X; A_n]
\end{align*}
But for $\mathbb E[\mathbb E[X | \sigma(\mathcal A)]; A]$ I can't figure out how the DCT would apply as I don't know that it's integrable. If I were able to conclude it were integrable the conclusion is immediate.
Best Answer
If $\mathcal A$ is a countable partition then $\sigma (\mathcal A)$ is nothing but the collection of all posssible unions of sets from $\mathcal A$. Hence the result is true in this case.
As pointed out in my comment above the result is not true in general.