# Strict inequality in Reverse Fatou lemma: $\varlimsup \int f_n\le \int \varlimsup f_n$

Let $\{f_n\}$ be a sequence of nonnegative functions dominated by some function $$g \in L^1.$$ Then, the reverse Fatou lemma says $$\limsup \int f_n \le \int \limsup f_n.$$

Is it possible to give an example where the inequality is strict?

I tried functions like $$f_n=\chi_{(n,n+1)},$$ but the dominating function is not integrble.

$$f_{2n}=\chi_{[0,1]}\qquad f_{2n+1}=\chi_{[1,2]}\qquad g=\chi_{[0,2]}$$
• Can you explain how this gives the answer? Don’t we end up with $\int$ lim sup $f_n(x) = 0$? – TuringTester69 Nov 4 '18 at 22:10
• @JaneDoe No we do not end up with what you say. What is $\limsup f_n$ according to you? – Did Nov 5 '18 at 7:18
• Sorry, maybe I misunderstood. Are we taking the integral over $\mathbb{R}$ or over $[0,2]$? If we are taking it over the reals, I think $\limsup f_n$ should be 0 because $f_n (x) = 0$ $\forall x > 2$. However, if it is over $[0,2]$ then I think $\limsup f_n$ should be 1. – TuringTester69 Nov 6 '18 at 1:11
• @JaneDoe You seem to have in mind $$\limsup_{x\to\infty}f_n(x)$$ for some unspecified value of $n$, that is, a limsup which is irrelevant, instead of, for each given $x$, $$h(x)=\limsup_{n\to\infty}f_n(x)$$ which defines the function of interest $$h=\limsup_{n\to\infty}f_n$$ – Did Nov 6 '18 at 16:03
What about $f_n(x)=x(-1)^n \chi_{[-1,1]}(x)+\chi_{[-1,1]}, ~g=2\chi_{[-1,1]}$?
• nonnegative functions. – Did Apr 29 '13 at 16:58
• sorry, but adding $1$ should fix that – Julian Apr 29 '13 at 17:01