We are told that $X$ is a uniformly integrable random variable on $(\Omega, \mathcal{A}, \mathbb{P})$. Let $A \in \mathcal{A}$, with $\mathbb{P}(A) \ne 0$.
Then,
$$ \mathbb{E}[X|A] \equiv \frac{1}{\mathbb{P}(A)} \int_{A} X(\omega)\ d\mathbb{P}(\omega) \equiv \frac{1}{\mathbb{P}(A)} \int_{\Omega} X(\omega)\ \mathbb{I}_{A}(\omega)\ d\mathbb{P}(\omega) \equiv \frac{\mathbb{E}[X \mathbb{I}_{A}]}{\mathbb{P}(A)}$$
Where $\mathbb{I}$ is the indicator function. We are asked to show as an exercise that $\mathbb{E}[\mathbb{I}_{A_{1}}|A_{2}] = \mathbb{P}(A_{1} | A_{2})$.
I can see how $\mathbb{E}[\mathbb{I}_{A_{1}}|A_{2}] = \mathbb{E}(A_{1} | A_{2})$ is true, and can show that, but not how the conditional expectation is equal to the conditional probability. We are told nothing on $A_{1}$ or $A_{2}$. I assume by default they must both be $\mathcal{A}$-measurable.
Does anyone have any ideas?
$$ \mathbb{E}[I_{A_1} |A_2] = P[A_1 |A_2]\times 1 + P[\bar A_1 |A_2]\times 0 = P[A_1 | A_2] $$