Suppose in measure on a measurable set such that for all , then .
The proof is short but slightly tricky:
Suppose to the contrary . Let be a subsequence such that
(using the fact that for any sequence there is a subsequence converging to ).
Since , there exists a further subsequence a.e. By Fatou’s Lemma, a contradiction.
The last equation above uses the fact that if a sequence converges, all subsequences converge to the same limit.