The Unapologetic Mathematician

Mathematics for the interested outsider

The Fatou-Lebesgue Theorem

Now we turn to the Fatou-Lebesgue theorem. Let \{f_n\} be a sequence of integrable functions (this time we do not assume they are non-negative) and g be some other function which dominates this sequence in absolute value. That is, we have \lvert f_n(x)\rvert\leq g(x) a.e. for all n. We define the functions

\displaystyle\begin{aligned}f_*(x)&=\liminf\limits_{n\to\infty} f_n(x)\\f^*(x)&=\limsup\limits_{n\to\infty} f_n(x)\end{aligned}

These two functions are integrable, and we have the sequence of inequalities

\displaystyle\int f_*\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu\leq\limsup\limits_{n\to\infty}\int f_n\,d\mu\leq\int f^*\,d\mu

Again, this is often stated for a sequence of measurable functions, but the dominated convergence theorem allows us to immediately move to the integrable case. In fact, if the sequence \{f_n\} converges pointwise a.e., then f_*=f^* a.e. and the inequality collapses and gives us exactly the dominated convergence theorem back again.

Since g dominates the sequence \{f_n\}, the sequence \{g+f_n\} will be non-negative. Fatou’s lemma then tells us that

\displaystyle\begin{aligned}\int g\,d\mu+\int f_*\,d\mu&=\int g+f_*\,d\mu\\&=\int\liminf\limits_{n\to\infty}(g+f_n)\,d\mu\\&\leq\liminf\limits_{n\to\infty}\int g+f_n\,d\mu\\&=\int g\,d\mu+\liminf\limits_{n\to\infty}\int f_n\,d\mu\end{aligned}

Cancelling the integral of g we find the first asserted inequality. The second one is true by the definition of limits inferior and superior. The third one is essentially the same as the first, only using the non-negative sequence \{g-f_n\}.

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June 17, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. [...] Then you can prove the monotone convergence theorem, followed by Fatou’s lemma, and then the Fatou-Lebesgue theorem, which leads to dominated convergence theorem, and we’re pretty much back where we [...]

    Pingback by An Alternate Approach to Integration « The Unapologetic Mathematician | June 18, 2010 | Reply


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