Fatou’s Lemma for Convergence in Measure

Suppose f_k\to f in measure on a measurable set E such that f_k\geq 0 for all k, then \displaystyle\int_E f\,dx\leq\liminf_{k\to\infty}\int_E f_k\,dx.

The proof is short but slightly tricky:

Suppose to the contrary \int_E f\,dx>\liminf_{k\to\infty}\int_E f_k\,dx. Let \{f_{k_l}\} be a subsequence such that \displaystyle \lim_{l\to\infty}\int f_{k_l}=\liminf_{k\to\infty}\int_E f_k<\int_E f
(using the fact that for any sequence there is a subsequence converging to \liminf).

Since f_{k_l}\xrightarrow{m}f, there exists a further subsequence f_{k_{l_m}}\to f a.e. By Fatou’s Lemma, \displaystyle \int_E f\leq\liminf_{m\to\infty}\int_E f_{k_{l_m}}=\lim_{l\to\infty}\int f_{k_l}<\int_E f, a contradiction.

The last equation above uses the fact that if a sequence converges, all subsequences converge to the same limit.

Author: mathtuition88


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