Why is $\limsup \int_E f_n = -\liminf \left(-\int_E f_n \right) $?

Question

Why is $\limsup \int_E f_n = -\liminf \left(-\int_E f_n \right) $?

Could anyone give me a detailed proof of this?

I know that lim sup is the largest accumulation accumulation points of a sequence and lim inf is the smallest accumulation points of a sequence, but then what? how can I use this in proving the above statement?

Answer 1

This has nothing to do with integration, it's a basic fact about sequences of real numbers. For $A\subset\Bbb R$ write $$-A=\{-a:a\in A\}.$$

Trivial Exercise. $\alpha$ is an upper bound for $A$ is an only if $-\alpha$ is a lower bound for $-A$.

Now, since $\sup$ is simply the least upper bound and $\inf$ is the greatest lower bound,

Easy Exercise. $\sup A=-\inf-A$.

And then using the definitions and the Easy Exercise:

Exercise. If $(a_n)$ is any sequence of reals then $\limsup a_n=-\liminf -a_n$.

Proof: $$\begin{align}-\liminf-a_n&=-\lim_{n\to\infty}\inf_{j>n}-a_k=-\lim_n-\sup_{k>n}a_k=\lim_n\sup_{k>n}a_k=\limsup a_n.\end{align}$$

(In case the relevance of all this is not clear: Let $a_n=\int_Ef_n$.)