## 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.

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