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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 $\int gd\mu$.

Raheem Najib
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  • It is similar to this question http://math.stackexchange.com/questions/76478/exercise-on-convergence-in-measure-folland-real-analysis –  Dec 05 '15 at 14:21
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    Do you know that a subsequence of a sequence converging in measure is a.e. convergent? – PhoemueX Dec 05 '15 at 14:21
  • @PhoemueX, Yes I know that! I prove my above problem. But why $\int gd\mu$ exists? – Raheem Najib Dec 05 '15 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? – Raheem Najib Dec 05 '15 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. –  Dec 05 '15 at 14:51
  • @maxerize, I mean why it's integral is finite. – Raheem Najib Dec 05 '15 at 15:24
  • @RaheemNajib there is no reason for it to be, in general. if $f$ is infinite on a set of positive measure, $g$ must also be. since it is also positive, the integral will be infinite. –  Dec 05 '15 at 20:41

1 Answers1

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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:

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

Kadmos
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