Fatou: Reverse?

Question

Attention

The usual problems are about absolute convergence: $$\int|g_n|\mathrm{d}\mu\quad(g_n=f_n,f-f_n,s_m-s_n,\ldots)$$

(There Fatou may help out!)

But as proceeding with Fatou one encounters that one can't distort to the limessuperior: $$\int\limsup_n|g_n|\mathrm{d}\mu\nleq\int\liminf_n|g_n|\mathrm{d}\mu\leq\liminf_n\int|g_n|\mathrm{d}\mu\leq\limsup\int|g_n|\mathrm{d}\mu$$

So the real question is about the analogue for limessuperior!!

(And not the analogue for negative functions...)

Problem

Given a measure space $\Omega$.

The lemma of Fatou states: $$f_n\geq0:\quad\int\liminf_nf_n\mathrm{d}\mu\leq\liminf_n\int f_n\mathrm{d}\mu$$ Does the reverse hold true: $$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu\leq\limsup_n\int f_n\mathrm{d}\mu$$ Certainly, for convergent examples this holds true: $$f_n\geq0:\quad\int\limsup_nf_n\mathrm{d}\mu=\int\lim_nf_n\mathrm{d}\mu\leq\lim_n\int f_n\mathrm{d}\mu=\limsup_n\int f_n\mathrm{d}\mu$$ So one needs to dig deeper to find an honest counterexample!!!!

(I intend to answer my own question!)

(I have to admit that my earlier answer was lame!)

Answer 1

Here's a simpler counterexample.

For $x \in [0,1]$, $f_{n}(x) = 1$ if $n$ is odd, and $0$ otherwise.

For $x \in (1,2]$, $f_{n}(x)=1$ if $n$ is even, and $0$ otherwise.

For any other $x$, $f_{n}(x) = 0$.

In this case $$ \int \limsup_{n} f_{n}(x) dx = 2,\\ \limsup_{n} \int f_{n}(x) dx = 1. $$

Answer 2

Attention

The problematic examples are the divergent ones: $\liminf\neq\limsup$

Counterexamples

On the one hand, it holds by the usual Fatou: $$f_n:=\chi_{(n,n+1]}:\quad0=\int\limsup_nf_n\mathrm{d}\mu\leq\limsup_n\int f_n\mathrm{d}\mu=1$$ On the other hand, it fails the reverse Fatou: $$f_{1\leq k\leq n}:=\chi_{(\frac{k}{n},\frac{k}{n}]}:\quad1=\int\limsup_nf_n\mathrm{d}\mu\nleq\limsup_n\int f_n\mathrm{d}\mu=0$$ (So while the former forces the direction of the inequality the latter shows it can't hold.)