Undergrad Showing equivalence of two definitions of essential supremum

[psie]

TL;DR

I've encountered the following two definitions of essential supremum and was wondering if someone could check if my "proof" of equivalence of the two definitions is correct.

Assume $f: X\to\mathbb R$ to be a measurable function on a measure space $(X,\mathcal A,\mu)$. The first definition is $\operatorname*{ess\,sup}\limits_X f=\inf A$, where $$A=\{a\in\mathbb R: \mu\{x\in X:f(x)>a\}=0\}$$ and the second is $\operatorname*{ess\,sup}\limits_X f=\inf B$ where $$B= \left\{\sup_X g:g=f\ \text{pointwise a.e.}\right\}.$$.

First, it is easily seen that if $h=f$ a.e., then $\mu\{x\in X: f(x) > \sup_X h\} = 0$, so $B\subset A$, which shows that $\inf B\geq \inf A$. Now, I can't show the other inclusion of the sets and I suspect these sets are not necessarily equal, but what I can show is that if $a\in A$, then $h:=\min\{f,a\}=f$ a.e. and $\sup_X h\leq a$ (*). So for $a\in A$, we can find a $b\in B$ such that $b\leq a$. Thus $\inf A\geq \inf B$.

Any thoughts on this?

(*) $h:=\min\{f,a\}$ means $h(x):=\min\{f(x),a\}$.

 

[Office_Shredder]

I agree with your proof.


I think the sets are equal because if you pick some a such that f is always smaller than a (which is the problem region for you attempt at a bijection), you can just arbitrarily define like, g(x)=f(x) if x is not zero, g(0)=a