Fatou's Lemma: Difference between revisions

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===Theorem===
==Statement==
Suppose <math>\{f_n\}</math> is a sequence of non-negative measurable functions, <math> f_n: X \to [0,+\infty]</math>.  
Suppose <math>\{f_n\}</math> is a sequence of non-negative measurable functions, <math> f_n: X \to [0,+\infty]</math>.  
Then:
Then:
<math> \int \liminf_{n\rightarrow +\infty} f_n \leq \liminf_{n\rightarrow +\infty}\int f_n </math>. <ref name="Folland">Gerald B. Folland, ''Real Analysis: Modern Techniques and Their Applications, second edition'', §2.2 </ref>
<math> \int \liminf_{n\rightarrow +\infty} f_n \leq \liminf_{n\rightarrow +\infty}\int f_n </math>. <ref name="Folland">Gerald B. Folland, ''Real Analysis: Modern Techniques and Their Applications, second edition'', §2.2 </ref>
==Proof<ref>Craig, Katy. ''MATH 201A Lecture 14''. UC Santa Barbara, Fall 2020.</ref>==
Define <math> g_n := \inf_{k\geq n} f_k </math> for all <math> n \in \mathbb{N} </math>.
By definition, <math> \liminf_{n\rightarrow +\infty} f_n= \lim_{n\rightarrow +\infty} (\inf_{k\geq n}f_k)=\lim_{n\rightarrow +\infty} g_n</math> and <math> g_n\leq g_{n+1}, \forall n \in \mathbb{N} </math>, so by Monotone Convergence Theorem,
<math> \lim_{n\rightarrow +\infty} \int g_n=\int \lim_{n\rightarrow +\infty} g_n = \int \liminf_{n\rightarrow +\infty} f_n</math>.
Furthermore, by definition we have <math> g_n\leq f_n,  \forall n \in \mathbb{N}</math>, implying that <math> \int g_n\leq \int f_n </math>.
Since <math> \lim_{n\rightarrow +\infty} \int g_n </math> exists, taking <math> \liminf_{n\rightarrow +\infty} </math> of both sides yields:
<math> \int \liminf_{n\rightarrow +\infty} f_n=\lim_{n\rightarrow +\infty} \int g_n = \liminf_{n\rightarrow +\infty} \int g_n \leq \liminf_{n\rightarrow +\infty} \int f_n</math>.
==References==

Latest revision as of 02:33, 12 December 2020

Statement

Suppose is a sequence of non-negative measurable functions, . Then:

. [1]

Proof[2]

Define for all .

By definition, and , so by Monotone Convergence Theorem,

.

Furthermore, by definition we have , implying that .

Since exists, taking of both sides yields:

.

References

  1. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second edition, §2.2
  2. Craig, Katy. MATH 201A Lecture 14. UC Santa Barbara, Fall 2020.