Suppose that $X$ is an integrable random variable, and let $\mathcal{G} \subseteq \mathcal{F}$ be a $\sigma$-algebra. Show that $E(E(X | \mathcal{G})) = E(X)$.
(At this point in the notes, we have proven the existence and uniqueness of conditional expectations)
The notes say that it follows directly from the proof of the extisnce and uniqueness of conditional expectations theorem, but I can only get to the following:
We know $E(X 1_A) = E(Y 1_A)$, where $A \in \mathcal{G}$, and we let $Y = E(X | \mathcal{G})$ because we know from the theorem on conditional expectations that such a random variable $Y$ exists.
However, we want to show $EX = EY$. I was thinking of summing over all $A \in \mathcal{G}$, but then that seems problematic because then you might be overcounting regions of overlap between different elements of $\mathcal{G}$.
Hint:
Observe that the whole space $\Omega$ is an element on $\mathcal G$.
What happens if we take $A=\Omega$?