# The Unapologetic Mathematician

## Fatou’s Lemma

Today we prove Fatou’s Lemma, which is a precursor to the Fatou-Lebesgue theorem, and an important result in its own right.

If $\{f_n\}$ is a sequence of non-negative integrable functions then the function defined pointwise as

$\displaystyle f_*(x)=\liminf\limits_{n\to\infty} f_n(x)$

is also integrable, and we have the inequality

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

In fact, the lemma is often stated for a sequence of measurable functions and concludes that $f_*$ is measurable (along with the inequality), but we already know that the limit inferior of a sequence of measurable functions is measurable, and so the integrable case is the most interesting part for us.

So, we define the functions

$\displaystyle g_n(x)=\inf\limits_{i\geq n}f_i(x)$

so that each $g_n$ is integrable, each $g_n\leq f_n$ and the sequence $\{g_n\}$ is pointwise increasing. Monotonicity tells us that for each $n$ we have

$\displaystyle\int g_n\,d\mu\leq\int f_n\,d\mu$

and it follows that

$\displaystyle\lim\limits_{n\to\infty}\int g_n\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu<\infty$

We also know that

$\displaystyle\lim\limits_{n\to\infty}g_n(x)=\liminf\limits_{n\to\infty}f_n(x)=f_*(x)$

which means we can bring the monotone convergence theorem to bear. This tells us that

$\displaystyle\int f_*\,d\mu=\lim\limits_{n\to\infty}\int g_n\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu$

as asserted.

If it happened that $f_*$ were not integrable, then some of the $f_n$ would have to be only measurable — not integrable — themselves. And it couldn’t just be a finite number of them, or we could just drop them from the sequence. No, there would have to be an infinite subsequence of non-integrable $f_n$, which would mean an infinite subsequence of their integrals would diverge to $\infty$. Thus when we take the limit inferior of the integrals we get $\infty$, as we do for the integral of $f$ itself, and the inequality still holds.

June 16, 2010 - Posted by | Analysis, Measure Theory

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4. Would you mind explaining why each g_n is integrable? Many thanks!

Comment by rich | December 7, 2012 | Reply

5. Sorry, rich, I don’t have a solid answer off the top of my head (I was never really an analyst). My intuition is that the infimum of any (countable?) collection of integrable functions is integrable.

Comment by John Armstrong | December 9, 2012 | Reply

• Thanks for the reply! I think your intuition is correct – I’ll poke around in my copy of Apostol to see if I can find a substantiating theorem.

Comment by rich | December 10, 2012 | Reply