Equivalent definitions of essential supremum? I was studying about Lebesgue spaces and ran into the definition of essential supremum.
Actually I have seen two very similar definitions: Let $(X,\mathfrak{M},\mu)$ be a measure space and $f:X\to [-\infty,+\infty]$ be a measurable function. Then $$\text{ess sup}|f(x)|:=\inf\{c\in \mathbb{R}: \mu(\{x\in X:|f(x)|>C\})=0\}. \qquad(*)$$ Also you can find exactly the same definition where infimum is taken over $c>0$, i.e. $$\text{ess sup}|f(x)|:=\inf\{c>0: \mu(\{x\in X:|f(x)|>C\})=0\}. \qquad (**)$$
And I think that probably $(*)=(**)$.
It follows easily that $(**)\geq (*)$. But how to show the converse ineqaulity?
Can anyone provide the rigorous proof, please?
 A: If $c<0$, then
$$
\{x\in X:|f(x)|>c\}=X
$$
Since $\mu(X)>0$, the set set that appears in the RHS of ($*$) is contained in $[0,\infty)$.
Now we consider two cases:

*

*If $f=0$ a.e. then the sets in $(*)$ and $(**)$ are $[0,\infty)$ and $(0,\infty)$ respectively, so their infimums are both $0$.

*If $\neg(f=0$ a.e.), then $0$ does not belong to the set in $(*)$, and so the sets in ($*$) and $(**)$ are the same.

A: Let $\phi(c) = \mu \{ x | |f(x)| > c \}$. Note that $\phi$ is non increasing, so
$\phi(0) \ge \phi(c)$ for all $c \ge 0$.
Let $N_* = \inf_{c \in \mathbb{R}} \phi(c), N_{**} = \inf_{c> 0} \phi(c)$.
It is clear that $N_* \le N_{**}$.
If $\mu X = 0$ then $N_* = -\infty$ and $N_{**} = 0$, so they are not equivalent in general.
Suppose $\mu X >0$.
If $c < 0$ then $\phi(c) = \mu X > 0$, so
$\{c | \phi(c)=0 \} \subset [0, \infty)$
and so $N_* \ge 0$.
Suppose $\phi(0) = 0$, then $\phi(c) = 0$ for $c \ge 0$ and so
$0 = N_* = N_{**}$, otherwise $\phi(0) >0$ and
$\{c \in [0,\infty] | \phi(c) = 0 \} = \{c \in (0,\infty] | \phi(c) = 0 \} $ and so taking
$\inf$ we have
$N_* = N_{**}$.
