18
$\begingroup$

I learnt Fatou's lemma a while ago. I am able to prove it and use it. I know examples showing that the inequality may be strict. But I don't really have an intuitive way to understand it. Any good thoughts?

$\endgroup$
6
  • $\begingroup$ I don't know how rigorous this is, or if it matches the intuition you're looking for, but if you think of the integral as an type of averaging, then it says the average of the lim inf is less than or equal to the lim inf of the average, which seems fairly intuitive. $\endgroup$ Aug 12, 2016 at 20:32
  • $\begingroup$ As I see it, on Fatou's lemma when you take liminf on one side of the inequality you may loose some information about the function, which may not be recoverable. Later, when you integrate first and then apply liminf, you loose less information about the integral. Something like, it is less violent to apply limit to an integral then to a function $\endgroup$ Aug 12, 2016 at 20:37
  • $\begingroup$ Except it only applies to non-negative functions, @SquirtleSquad, so I'm not sure how intuitive your explanation is. $\endgroup$ Aug 12, 2016 at 20:39
  • 2
    $\begingroup$ Related: math.stackexchange.com/questions/242920/… $\endgroup$ Aug 13, 2016 at 14:24
  • $\begingroup$ @ThomasAndrews There is a [en.wikipedia.org/wiki/… to functions $f_1,\dots\ge-g$ that have an integrable lower bound. I am not saying I support SquirtleSquad's comment, but your criticism seems unfounded, in view of this generalization. $\endgroup$ Mar 25, 2020 at 13:32

3 Answers 3

23
$\begingroup$

Since the Lebesgue integral for nonnegative functions is built up "from below" by taking suprema of "obvious" integrals, the monotone convergence theorem has always seemed to me to be the most natural of the big three (MCT, FL, LDCT). And FL is a direct corollary of the MCT: Start with the obvious, i.e.,

$$\int \inf \{f_n,f_{n+1}, \dots \} \le \int f_n.$$

From that we get

$$\lim_{n\to \infty} \int \inf \{f_n,f_{n+1}, \dots \} \le \liminf_{n\to \infty} \int f_n.$$

Really, that should be $\liminf$ on the left, but since the integrands increase, so do the integrals, so the limit exists and we're fine. Now by MCT, that limit can be moved through the integral sign, and then you have FL.

$\endgroup$
1
  • 1
    $\begingroup$ Nice answer! +1 $\endgroup$
    – Mark Viola
    Aug 13, 2016 at 3:37
15
$\begingroup$

Fatou's lemma tells you that in the limit "mass" can only be lost but not generated. Let's recall the satement. If $f_n,f\geq 0$ are measurable and $f_n\to f$ pointwise a.e., then we have $\int f \leq \liminf_{n\to\infty} \int f_n$.

A classical example is $f_n= n \chi_{[0,1/n]}$ where $\int f_n=1$ for all $n$, but in the limit the mass escapes to "vertical" infinity, so it is lost, and we have that $f_n\to 0=:f$ a.e., with $\int f=0$.

The other example where, mass escapes to "horizontal" infinity, is $f_n= \chi_{[n,n+1]}$. Again $f_n$ has mass $1$, but the limit has mass $0$.

If we shut down these escape possibilities, then mass is preserved, i.e. $\int f=\lim_{n\to\infty} \int f_n$. For example, one way to do that is to assume that $f_n$ are bounded and all of them are supported on a large interval $[-M,M]$. This follows from the Dominated Convergence Theorem which gives a fairly general criterion for convergence of the integral: if $|f_n|\leq g$ where $\int g<\infty$, then the mass is preserved under the limit.

$\endgroup$
0
$\begingroup$

Fatou's Lemma is a description of "semi-continuity" of the integral operator $\int_{\Omega}(\bullet)=E(\bullet)$.

Think of the the integral operator as a mapping from a space $F_{\Omega}$ consisting of measurable functions $f:\Omega\rightarrow\mathbb{R}$. One may wonder whether the pointwise topology in this space $F_{\Omega}$ will allow the integral operator to be continuous. i.e. $$lim_{n\rightarrow\infty}E(f_n)=E(lim_{n\rightarrow\infty}f_n)$$ For arbitrary $f_n\in F_{\Omega}$, the best possible answer is given by Fatou's Lemma, i.e. $$E(liminf_{n\rightarrow\infty}f_n)\leq liminf_{n\rightarrow\infty}E(f_n)\leq limsup_{n\rightarrow\infty}E(f_n)\leq E(limsup_{n\rightarrow\infty}f_n)$$ And for those $f_n$ such that $E(liminf_{n\rightarrow\infty}f_n)= E(limsup_{n\rightarrow\infty}f_n)=E(lim_{n\rightarrow\infty}f_n)$, $E(\bullet)$ possesses the continuity. Two sufficient conditions that $f_n$ belongs to this class are given by Dominate Convergence Theorem and Monotonic Convergence Theorem.

(1)Dominate Convergence Theorem actually pushes a step further, finding another $g_n\equiv |g|$ sequence $$E(-liminf_{n\rightarrow\infty}g_n)\leq E(liminf_{n\rightarrow\infty}f_n)\leq liminf_{n\rightarrow\infty}E(f_n)\leq limsup_{n\rightarrow\infty}E(f_n)\leq E(limsup_{n\rightarrow\infty}f_n)\leq E(limsup_{n\rightarrow\infty}g_n)$$ such that $E(g)<\infty$ holds.

(2)Monotonic Convergence Theorem boldly asserts that $E(liminf_{n\rightarrow\infty}f_n)= E(limsup_{n\rightarrow\infty}f_n)=E(lim_{n\rightarrow\infty}f_n)$ because monotonic convergence function sequence cannot have different upper and lower limits at any point and this suffices to guarantee their integrals to be the same.

And this intuition is the start of investigation of the lattice structure of $F_{\Omega}$.

$\endgroup$
1
  • $\begingroup$ Actually I think Daniell integral Theorem is the best starting point when taking a second course in real analysis, along with Pratt's Lemma, we can have a more intuitive viewpoint over the whole subject and concern more about the "big picture" instead of spending time remembering scattered results with counter-intuitive assumptions. $\endgroup$
    – Henry.L
    Dec 2, 2016 at 0:55

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.