Equivalent definition of essential supremum

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?

• What have you tried? May 31, 2022 at 9:06
• 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 May 31, 2022 at 9:06
• I think, for equivalence, you need to take $\sup|f|$ rather than $\sup f$ in $(1)$ May 31, 2022 at 9:08
• @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. May 31, 2022 at 9:11
• @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...) May 31, 2022 at 10:11

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, $$\inf A\leq \inf B$$.