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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?

drhab
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Idonotknow
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    write explicitly the definitions of limit superior and limit inferior and you are almost done. By last note that if $A\subset\Bbb R$ then $\sup A=-\inf(-A)$ –  Oct 15 '19 at 11:08
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    https://math.stackexchange.com/questions/392129/proof-that-inf-a-sup-a – Calvin Khor Oct 15 '19 at 11:08
  • but here we are dealing with functions and not sets @Masacroso –  Oct 15 '19 at 14:09
  • but here we are dealing with functions and not sets @CalvinKhor , so why there is an equality? is not it should be inequality? –  Oct 15 '19 at 14:10
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    @Smart note that $$\limsup\int_E f_n:=\lim_{n\to\infty}\sup_{k\ge n}\int_E f_k$$ and $$\sup_{k\ge n}\int_E f_k:=\sup\left{\int_E f_k:k\ge n\right}$$ –  Oct 15 '19 at 21:04

1 Answers1

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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 $a_n=\int_Ef_n$.)

David C. Ullrich
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  • More generally, if $f$ is an increasing function, then $\sup f(A) \le f(\sup A)$ and if $f$ is decreasing then $\sup f(A) \le f (\inf A)$. These are equalities if $f$ is continuous at $\sup A$ or $\inf A$ respectively. – Sera Gunn Oct 15 '19 at 15:04
  • Does the question in the title includes supremum over sets or functions? – Intuition Oct 15 '19 at 15:05
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    @Secretly Strictly speaking the answer is neither - the question asks about the sup and inf of sequences. If I have to say sets or functions, I'd say sets, because if $a_n$ is a sequence of reals and $A$ is the set ${a_n:n\in\Bbb N}$ then $\sup_na_n=\sup A$. – David C. Ullrich Oct 15 '19 at 15:13
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    @Secretly But to answer what you're probably really asking, no, the question has nothing whatever to do with the fact that $f_n$ is a function; the question is just about the sequence $a_n$, where it just happens that $a_n=\int_Ef_n$. – David C. Ullrich Oct 15 '19 at 15:15