Convergence of conditional means

Question

Suppose that $X$ and $Y$ are random variables and $\{b_n\}$ is a sequence such that $b_n \rightarrow b$ as $n\rightarrow \infty$.

Under what conditions $E[X|Y\leq b_n] \rightarrow E[X|Y \leq b]$?

Answer 1

This holds if $X$ is integrable and $P(Y=b) =0$.

In this case we have $E[X \mid Y \le b_n] = \frac{E[X 1_{\{Y \le b_n\}}]}{P(Y \le b_n)}$. As $n \to \infty$ we have $1_{\{Y \le b_n\}} \to 1_{\{Y \le b\}}$ almost surely (exercise), so by dominated convergence (using $|X|$ as the dominating function) the numerator converges to $E[X 1_{\{Y \le b\}}]$. The denominator is $E[1_{\{Y \le b_n\}}]$ and by the same argument (using 1 as the dominating function) converges to $E[1_{\{Y \le b\}}] = P(Y \le b)$. So the limit is $\frac{E[X 1_{\{Y \le b\}}]}{P(Y \le b)} = E[X \mid Y \le b]$.

It's pretty easy to get a counterexample with $P(Y=b) > 0$.