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[parent] proof of Fatou's lemma (Proof)

Let $f(x)=\liminf_{n\to\infty} f_n(x)$ and let $g_n(x)=\inf_{k\ge n} f_k(x)$ so that we have $$ f(x) = \sup_n g_n(x). $$

As $g_n$ is an increasing sequence of measurable nonnegative functions we can apply the monotone convergence Theorem to obtain $$ \int_X f\, d\mu = \lim_{n\to\infty} \int_X g_n\, d\mu. $$ On the other hand, being $g_n\le f_n$ we conclude by observing $$ \lim_{n\to\infty} \int_X g_n\, d\mu = \liminf_{n\to\infty}\int_X g_n\, d\mu \le \liminf_{n\to\infty}\int_X f_n\, d\mu. $$




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Cross-references: monotone convergence theorem, functions, measurable, sequence, increasing

This is version 1 of proof of Fatou's lemma, born on 2003-03-07.
Object id is 4076, canonical name is ProofOfFatousLemma.
Accessed 9107 times total.

Classification:
AMS MSC28A20 (Measure and integration :: Classical measure theory :: Measurable and nonmeasurable functions, sequences of measurable functions, modes of convergence)

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Discussion
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Circular? by IgorZ on 2007-10-10 16:15:38
This proof certainly works, but using the Dominated Convergence Theorem seems fundamentally circular, since proofs of that theorem (at least any I've seen) typically use either Fatou's Lemma or the Fatou-Lebesgue Theorem, which also rests on Fatou's Lemma.

Seems a more direct measure theoretic proof is in order.
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question about this proof of fatou lemma by castilla on 2006-02-13 19:09:34
Can someone explain why the g_n are measurable functions? Each g_n is the infimum of a set of measurable functions, but I dont knew that such thing is enough... thanks.
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