Given a sequence of compact subsets of the real line which converges in the Hausdorff metric, it is not guaranteed that the sequence of indicator functions of the given subsets converge in measure (with respect to the Lebesgue measure).
For instance, given a dense subset $\{a_n\}_{n=1}^{\infty}\subset[0,1]$, one can show that the sequence of compact subsets $A_n=\{a_1,...,a_n\}$ converges to $[0,1]$ in the Hausdorff distance. But of course, the sequence of indicators $\chi_{A_n}$ does not converge in measure to $\chi_{[0,1]}$.
My question is - are there any nice additional conditions one can impose under which this convergence does occur? Any relevent condition would be nice to know.
Thanks in advance.
When the sequence $A_n$ is decreasing, $A_1 \supseteq A_2 \supseteq A_3\supseteq\cdots$, then $A_n$ converges in Hausdorff metric to the intersection $\bigcap_{n=1}^\infty A_n$. In that case, we have $\mu(A_n) \to \mu(A)$ for any finite Borel measure, not merely for Lebesgue measure.
As the OP noted, this is not true in general for an increasing sequence.