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.

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