Let $(\Omega,\mathcal A, \mathbb P)$ be a probability space, $\mathcal B$ a sub-$\sigma$-algeba of $\mathcal A$, $B \in \mathcal B$ with $\mathbb P(B) > 0$ and $A \in \mathcal A$ with $\mathbb P(A | \mathcal B) > \varepsilon$ almost surely. The task is to show that $\mathbb P(A|B) > \varepsilon$.
I saw the following solution: $\mathbb P(A \cap B) = \int_{B} \mathbf{1}_{A}d\mathbb P = \int_{B} \mathbb P(A | \mathcal B) d\mathbb P > \varepsilon \mathbb P(B)$.
The first and the last equality is clear to me. But I don't see why $\mathbf{1}_{A} = \mathbb P(A | \mathcal B)$ holds there because $\mathbb P(A | \mathcal B) := \mathbb E[\mathbf{1}_{A} | \mathcal B]$ and to my knowledge $A \cap B$ is in general not in $\mathcal B$.
No, $\mathbf{1}_{A} = \mathbb P(A | \mathcal B)$ does not hold.
But $\mathbf{1}_{A}$ and $\mathbb P(A | \mathcal B)$ do have the same integral over $B$, by definition of the conditional expectation.
Use the definition $\mathbb P(A | \mathcal B) = \mathbb E[\mathbf1_A | \mathcal B]$.