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?


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 $N_* = N_{**}$.

  • $\begingroup$ Thanks a lot for your reply! I will accept this answer as the best. Yeah in the case when $\mu X=0$ they are not the same but why in books authors does noe inlude this case? They just write measure space and do not mention that measure of whole space should be positive. $\endgroup$ – ZFR Jul 19 '20 at 21:02
  • $\begingroup$ @ZFR: I don't know. I prefer to avoid such ambiguities. $\endgroup$ – copper.hat Jul 19 '20 at 21:06
  • $\begingroup$ So you mean we are considering measure space with $\mu(X)>0$? $\endgroup$ – ZFR Jul 19 '20 at 21:10
  • $\begingroup$ I am a bit confused $\endgroup$ – ZFR Jul 19 '20 at 21:10
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    $\begingroup$ I would consider it strange to have the essential supremum of the function $f=0$ to be $-\infty$. $\endgroup$ – copper.hat Jul 19 '20 at 21:23

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.
  • $\begingroup$ I have not read your answer in detail but are you considering the case $\mu(X)>0$? If yes then I know that they are the same. But what if $\mu(X)=0$? $\endgroup$ – ZFR Jul 19 '20 at 19:05
  • $\begingroup$ @ZFR Then your measure space is not very interesting. $\endgroup$ – Reveillark Jul 19 '20 at 19:10
  • $\begingroup$ I don't think so. Because I reviewed many books in measure theory including Stein, Royden and others and most of them when they considering $L_{\infty}$ space they do not exclude the case when $\mu(X)=0$. $\endgroup$ – ZFR Jul 19 '20 at 19:13
  • $\begingroup$ @ZFR Those books are operating under the tacit assumption that $\mu$ is not identically $0$. The property in question is false if the space has measure $0$. But then again, so is every theorem which has the words "almost everywhere" in its statement. $\endgroup$ – Reveillark Jul 19 '20 at 19:23
  • $\begingroup$ Thanks! Two more questions: 1) "The property in question is false if the space has measure 0." Which property are you talking about? 2) "But then again, so is every theorem which has the words "almost everywhere" in its statement." Could you explain this, please? $\endgroup$ – ZFR Jul 19 '20 at 19:36

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