# Fatou's lemma. Case of convergence in measure

Fatou's lemma: Let $f_1, f_2, f_3, \cdots$ be a sequence of non-negative measurable functions on a measure space $(S,\Sigma,\mu)$. Define the function $f:S\to [0,\infty]$ a.e. pointwise limit by $$f(s)=\lim \inf\limits_{n\to \infty}f_n(s), \quad s\in S.$$ Then $f$ is measurable and $$\int \limits_{S}fd\mu \le \lim \inf\limits_{n\to \infty}\int \limits_{S}f_nd\mu. \qquad (1)$$

It's very famous and important claim.

Prove that if $\{f_n\}$ converges in measure to some function $g$ then LHS of $(1)$ can be changed to $\int gd\mu$.

• It is similar to this question math.stackexchange.com/questions/76478/…
– user42268
Commented Dec 5, 2015 at 14:21
• Do you know that a subsequence of a sequence converging in measure is a.e. convergent? Commented Dec 5, 2015 at 14:21
• @PhoemueX, Yes I know that! I prove my above problem. But why $\int gd\mu$ exists? Commented Dec 5, 2015 at 14:25
• @PhoemueX, if it's wrong then exists subsequence $\{n_k\}$ such that $\int f d\mu > \lim \int f_{n_k} d\mu$. Then from this subsequence we can extract subsequence which converges a.e. Applying Fatou's lemma we get contradiction. But why integral over $g$ exists? Commented Dec 5, 2015 at 14:29
• Why would it not exist? g is measurable and a.e. positive. if you meant why is the integral finite, it isn't in general.
– user42268
Commented Dec 5, 2015 at 14:51

The following is a standard result:

Lemma 1: If $$g_n\xrightarrow[]{\mathcal{M}} g$$ there is a subsequence that converges pointwise to $$g$$ almost everywhere.

We will also use the following result from analysis which was proven for $$\limsup$$ in a previous MSE post, but shouldn't be hard to adapt for $$\liminf$$:

Lemma 2: If $$x_n$$ is a sequence of real numbers there is a subsequence such that $$x_{n_k}\rightarrow \liminf x_n$$.

With these two facts we are almost done.

There is a subsequence $$\int f_{n_k}d\mu \rightarrow \liminf \int f_n d\mu$$ because of Lemma 2.

It also holds that if $$f_n\xrightarrow[]{\mathcal{M}} g$$, any subsequence, in particular $$f_{n_k}$$, satisfies $$f_{n_k} \xrightarrow[]{\mathcal{M}} g$$.

By our first lemma, there is a subsequence of this subsequence such that $$f_{n_{k_j}}\rightarrow g$$ pointwise almost everywhere. Because of our usual Fatou:

$$\int g d\mu =\int \liminf f_{n_{k_j}}d\mu\leq \liminf \int f_{n_{k_j}}=\lim \int f_{n_{k_j}}=\lim \int f_{n_{k}}=\liminf \int f_n d\mu$$