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Let $(X,\mathcal{A},\mu)$ be a measure space and let $f\colon X\to [-\infty,+\infty]$ be a measurable function. Denote with $\mathcal{N}_\mu$ the collection of $\mu$-null sets.

On same text the definition of essential supremum is

$$\operatorname{esssup}f:=\inf\left\{\sup_{x\in X\setminus N} f(x)\;\middle|\; N\in\mathcal{N}_\mu \right\}\tag 1$$

In other texts it is:

$$\operatorname{esssup}f:=\inf\left\{a\ge 0\;\middle|\;\mu\left(\{x\in X\;\middle|\; f(x)>a\}\right)=0 \right\}\tag2$$

Question Are $(1)$ and $(2)$ equivalent? Why?

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  • $\begingroup$ What have you tried? $\endgroup$ May 31, 2022 at 9:06
  • $\begingroup$ I’ve edited out the complement macro and replaced it with ^c as that is far more common on this site, never once have I seen \complement used, to improve readability. So too did I remove the large-text commands because - what’s the point? They also mess with readability especially for mobile users $\endgroup$
    – FShrike
    May 31, 2022 at 9:06
  • $\begingroup$ I think, for equivalence, you need to take $\sup|f|$ rather than $\sup f$ in $(1)$ $\endgroup$
    – FShrike
    May 31, 2022 at 9:08
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    $\begingroup$ @KaviRamaMurthyI haven't tried anything, because I don't know how to start not because I don't want to. The idea would be to show that one is less than the other and vice versa, but I don't know how. $\endgroup$
    – NatMath
    May 31, 2022 at 9:11
  • $\begingroup$ @FShrike Speaking of readability, the current version looks to me just like $$\sup_{x\in N};$$ the little $c$ in the $N^c$ is almost invisible. I honestly didn't realize that the c was supposed to be there until I looked at the OP's original (of course that made the problem wrong...) $\endgroup$ May 31, 2022 at 10:11

1 Answer 1

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I think you need $a\in \mathbb{R}$ and f(x) without the absolute value. Otherwise f konstant -1 is a couterexample. But if you correct this, the two definitions are equivalent:

Let $A:= \{\sup_{x\in X\setminus N}f(x)\mid N\in \mathcal{N}_{\mu}\}$ and $B:= \{a\in \mathbb{R}\mid \mu(\{x\in X\mid f(x)>a\})=0\}$. We show $\inf A=\inf B$:

"$\geq$": For $N\in \mathcal{N}_{\mu}$ we have $\{x\in X \mid f(x)>\sup_{x\in X\setminus N}f(x)\}\subseteq N$, and hence $\mu(\{x\in X \mid f(x)>\sup_{x\in X\setminus N}f(x)\})=0$. Therefore, $A\subseteq B$ and $\inf A\geq \inf B$.

"$\leq $": Let $a$ be given such that $\mu(\{x\in X\mid f(x)>a\})=0$ and define $N:= \{x\in X\mid f(x)>a\}$. Then $N\in \mathcal{N}_\mu$ and for $x\in X\setminus N$ we have $f(x)\leq a$, hence $\sup_{x\in X\setminus N}\leq a$. So for all $x\in B$ there is a $y\in A$ with $y\leq x$. Therefore, $\inf A\leq \inf B$.

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