Fatou's Lemma for Integrals/Positive Measurable Functions

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.



Let $\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit inferior of the $f_n$.


Then:

$\ds \int \liminf_{n \mathop \to \infty} f_n \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$

where:

the integral sign denotes $\mu$-integration
the right hand side limit inferior is taken in the extended real numbers $\overline \R$.


Proof

For each $n \in \N$, define $g_n : X \to \overline \R$ by:

$\ds g_n = \inf_{k \mathop \ge n} f_k$

That is:

$\map {g_n} x = \inf \set {\map {f_k} x : k \ge n}$

for each $x \in X$.

For each $n \in \N$, we have that:

$\set {\map {f_k} x : k \ge n + 1} \subseteq \set {\map {f_k} x : k \ge n}$

From Infimum of Subset:

$\inf \set {\map {f_k} x : k \ge n + 1} \ge \inf \set {\map {f_k} x : k \ge n}$

That is:

$\map {g_{n + 1} } x \ge \map {g_n} x$

for each $n \in \N$.

From Pointwise Infimum of Measurable Functions is Measurable, we also have:

$g_n$ is $\Sigma$-measurable for each $n \in \N$.

We have:

$\ds \lim_{n \mathop \to \infty} g_n = \lim_{n \mathop \to \infty} \paren {\inf_{k \mathop \ge n} f_k}$

By the definition of limit inferior:

$\ds \lim_{n \mathop \to \infty} g_n = \liminf_{n \mathop \to \infty} f_n$

So:

$\sequence {g_n}_{n \mathop \in \N}$ is an increasing sequence of $\Sigma$-measurable functions convergent to $\ds \liminf_{n \mathop \to \infty} f_n$.

We are therefore able to apply the monotone convergence theorem.

We then have::

\(\ds \int \liminf_{n \mathop \to \infty} f_n \rd \mu\) \(=\) \(\ds \int \lim_{n \mathop \to \infty} \paren {\inf_{k \mathop \ge n} f_k} \rd \mu\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \int \paren {\inf_{k \mathop \ge n} f_k} \rd \mu\) applying the monotone convergence theorem to $\sequence {g_n}_{n \mathop \in \N}$

Clearly we have:

$\ds \inf_{k \mathop \ge n} f_k \le f_n$

for any $n \in \N$.

So, by Integral of Integrable Function is Monotone:

$\ds \int \paren {\inf_{k \mathop \ge n} f_k} \rd \mu \le \int f_n \rd \mu$

and so:

$\ds \int \paren {\inf_{k \mathop \ge n} f_k} \rd \mu \le \inf_{k \mathop \ge n} \int f_k \rd \mu$

Then:

\(\ds \lim_{n \mathop \to \infty} \int \paren {\inf_{k \mathop \ge n} f_k} \rd \mu\) \(\le\) \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{k \mathop \ge n} \int f_k \rd \mu}\)
\(\ds \) \(=\) \(\ds \liminf_{n \mathop \to \infty} \int f_n \rd \mu\)

$\blacksquare$


Also see


Source of Name

This entry was named for Pierre Joseph Louis Fatou.


Sources