I am reading the following statement:
If $Y ∈ m{F}$ (F-measurable) and
$\mathbb{E}(X1A) = \mathbb{E}(Y1A)$ for all $A$ in a $π$-system that contains $Ω$ and generates $F$,
then $Y = \mathbb{E}(X|F)$ a.s..
Does this simply follow directly from the fact that $\mathbb{E}(X1A)= \int_{A}Xd\mathbb{P}$ and $\mathbb{E}(Y1A)= \int_{A}Yd\mathbb{P}$, which are equal by the definition of conditional expectation, or is there something more interesting going on here?
The point is that if we have to check that $Y$ is indeed $\mathbb E\left(X\mid F\right)$, we have to check the equality $\mathbb{E}(X1A) = \mathbb{E}(Y1A)$ for all $A\in F$, which can be hard. There are case where the previous equality is easier to check for the elements of a $\pi$-system generating $F$.
For example, let $X$ be a random variable and let $\mathcal F_1$, $\mathcal F_2$ be two $\sigma$-algebras such that $\mathcal F_2$ is independent of the $\sigma$-algebra generated by $X$ and $\mathcal F_1$. Then we can check that $$ \mathbb E\left(X\mid \sigma(\mathcal F_1,\mathcal F_2)\right)=\mathbb E\left(X\mid \mathcal F_1 \right) $$ by using the previous result with the $\pi$-system of elements of the for $F_1\cap F_2$, where $F_1\in\mathcal F_1$ and $F_2\in\mathcal F_2$.