Equivalent definitions of essential supremum?

Question

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 inequality?

Can anyone provide the rigorous proof, please?

Answer 1

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 $N_* = N_{**}$.

Answer 2

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.