11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user223391 Apr 5 '15 at 4:37
  • 2
    $\begingroup$ The best way to understand Fatou is probably by picture, rather than words: math.stackexchange.com/questions/242920/… $\endgroup$ – William Stagner Apr 5 '15 at 4:38
  • $\begingroup$ @avid19 : I've posted a less pessimistic view of the matter below. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 5 '15 at 4:49
15
$\begingroup$

$$ (0,1),\quad(1,0),\quad(0,1),\quad(1,0),\quad\ldots $$ 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.

$\endgroup$
  • $\begingroup$ +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 $\endgroup$ – William Stagner Apr 5 '15 at 4:53
2
$\begingroup$

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 $\ast$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.