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I'm working on Fatou's Lemma, which says for a sequence of measurable, nonnegative functions, $\int \lim\inf f_i \leq \lim\inf \int f_i$. I'm having trouble understanding what it means to have a pointwise $\inf f_i$ and was wondering if somebody could help me.

So we're dealing with a sequence of functions that are measurable, so what I think of is some random function, like $y = constant$, and then variations of it, like $y_n = c+1/n, n\in\mathbb{N}$. So then,

  1. Is $\lim\inf y_n = c$?
  2. What is $\lim\sup y_n$, is it $y = c+1$?
  3. Where does the "pointwise" thing come in?

Now with the pointwise I'm thinking, Wikipedia says we have to consider each value $f(x)$ of some function $f$. So if I have $y_n = |\sin(x+n)|,n\in\mathbb{N}$, is $\lim\inf y_n = 0$? And $\lim\sup y_n = 1$? Or should it be different?

Thanks for any clarification!!

letslearnmath
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1 Answers1

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$\liminf_{i} f_i$ is a function which given a point $x\in X$ returns $\liminf_i f_i(x)$. Note that for a fixed $x$, $\{f_i(x)\}_{i=1}^\infty$ is just a sequence of real numbers, for which its $\liminf$ is defined as usual.

A different concept is $\inf f_i$. This is the function which, given $x$, returns the infimum of the set $\{f_i(x):i\ge 1\}$.

In your first example, $\liminf y_n=\inf y_n=c$, while $\limsup y_n=c$ and $\sup y_n=c+1$. For $\limsup$, you only care about the limiting behavior, while $\sup$ is sensitive to all values.

In your second example, it is true that $\liminf y_n=0$ and $\limsup y_n=1$, since no matter what the value of $x$ is, the sequence $\sin(x+n)$ is dense in $[0,1]$.

For a different example where the $\liminf$ is a nonconstant function, let $f_n(x)=\frac{|x|^n}{1+|x|^n}$. Then $$ \liminf f_n=\limsup f_n = \begin{cases}0 & |x|<1\\\frac12 & |x|=1 \\ 1 & |x|>1\end{cases} $$ This is because $|x|^n$ tends to either $0,1$ or $\infty$ depending on $|x|$.

Mike Earnest
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