# What does Fatou's Lemma really say?

Fatou's lemma says that

if $f_n:X \rightarrow [0,\infty]$ are measurable,then

$$\liminf_{n\rightarrow \infty}\left(\int_X f_n \,\mathrm{d} \mu\right) \geq \int_X \liminf_{n\rightarrow \infty} f_n \,\mathrm{d}\mu$$

I like to know what this Lemma really says. That is, how can I express in words (rather informally) what this lemma actually says?

• Honestly I don't really think there's much intuition. It's just a technical lemma used to prove other things, most importantly dominated convergence theorem. By itself, it's pretty boring. Although I'd love to be proven wrong.
– user223391
Apr 5, 2015 at 4:37
• The best way to understand Fatou is probably by picture, rather than words: math.stackexchange.com/questions/242920/… Apr 5, 2015 at 4:38
• @avid19 : I've posted a less pessimistic view of the matter below. ${}\qquad{}$ Apr 5, 2015 at 4:49

$$(0,1),\quad(1,0),\quad(0,1),\quad(1,0),\quad\ldots$$ The terms of a sequence are alternately $(0,1)$ and $(1,0)$. In either case, the sum of the two components of each pair is $1$, so the lim inf of the sum of the two is $1$. But the lim inf of the sequence of pairs is $(0,0)$, and the sum of the two components of that pair is $0$.

In other words, for every value of $x$, $f_n(x)$ may be small for infinitely many $n$, but the values of $n$ for which $f_n(a)$ is small are not the same ones for which $f_n(b)$ is small.

• +1--this really distills the statement down to the fact that $\liminf$ only "cares" about a sequence of functions pointwise, and in minimizing pointwise, one can lose mass Apr 5, 2015 at 4:53

If $E_i$ is measurable then $$\lim\ \inf E_i =\{x \mid x\in E_i\ \text{ except finitely many } i \}$$

Then we have $$\mu (\lim\ \inf E_i)\leq \lim\ \inf \mu(E_i)\ \ast$$

If $E_i$ is a measurable, then $f_i=\chi_{E_i}$ is measurable. So $$\lim\ \inf \mu(E_i) = \lim\ \inf \int_X f_i$$

And if $E:=\lim\ \inf E_i$ then $$\lim\ \inf f_i= \chi_E$$

Hence Fatou is a generalization of $\ast$.