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 $\int gd\mu$.

  • $\begingroup$ It is similar to this question math.stackexchange.com/questions/76478/… $\endgroup$
    – user42268
    Dec 5, 2015 at 14:21
  • 1
    $\begingroup$ Do you know that a subsequence of a sequence converging in measure is a.e. convergent? $\endgroup$
    – PhoemueX
    Dec 5, 2015 at 14:21
  • $\begingroup$ @PhoemueX, Yes I know that! I prove my above problem. But why $\int gd\mu$ exists? $\endgroup$ Dec 5, 2015 at 14:25
  • $\begingroup$ @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? $\endgroup$ Dec 5, 2015 at 14:29
  • $\begingroup$ 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. $\endgroup$
    – user42268
    Dec 5, 2015 at 14:51

1 Answer 1


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:

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


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