Consider any measurable function $f\colon \Omega \to \mathbb R.$ It is well known that $\sup f \geqslant f(x),$ for every $x \in \Omega$ (the supremum must be an upper bound). I was wondering: is there an analogue of this inequality using $\text{ ess $\sup$ }$ instead of $\sup$ ?
Clearly, one can't guarantee that $\text{ ess } \sup f \geqslant f(x),$ for every $x \in \Omega.$ A counter-example follows: consider $\Omega = \Bbb R$ and define
$$ f(x) = \begin{cases} 1 \quad \text{ if } x \in \Bbb Z, \\[.2cm] 0 \quad \text{otherwise.} \end{cases} $$
We have that $\text{ess } \sup f = 0,$ but $f(x) = 1, $ for $x \in \Bbb Z.$ Therefore, the inequality can't hold.
So I started wondering: altought the inequality $\text{ ess } \sup f \geqslant f(x),$ for every $x \in \Omega$ doesn't hold, could it be true that
$$ \color{red}{\text{ ess } \sup f \geqslant f(x), \, \, \text{ for almost every } x \in \Omega ?} $$
I couldn't come up with a counter example for this inequality but I also can't show that it holds (note that the precedent example doesn't show this inequality is false since $\Bbb Z$ has measure zero). So, essentially, my goal with this question is to find out where the inequality posted in red is true or false.
Thanks for any help in advance.
UPDATE (formal definition of essential supremum)
The definition I am using for essential supremum is the following:
$$ \text{ ess } \sup f = \inf\{ a \in \Bbb R \, \colon \, |\{x \in \Omega \, \colon \, f(x) > a\}| = 0\}.$$
This can be rewritten as
$$ \inf\{a \in \Bbb R \, \colon \, f(x) \leqslant a, \text{ almost everywhere on } \Omega \}.$$
(note that I denoted $|\cdot|$ as the measure in cause).
Best Answer
Yes, it follows that $\text{ess sup} f \geq f(x)$ for almost every $x\in \Omega$.
Indeed, if by contradiction this is false, then there exists a set $A$ of positive measure so that $f(x) > \text{ess sup} f$ for all $x\in A$.
For every $n\in \mathbb{N}$, let $A_n = \{ x\in \Omega : f(x) > \text{ess sup f} + \frac{1}{n}\}$, and let $B_n:= A_n \backslash \bigcup_{i=1}^{n-1} A_i$ (exercise: prove that these are measurable sets).
Now $\bigcup_n B_n = \bigcup_n A_n = A$, because if $f(x)>\text{ess sup} f$, then for some $n$ sufficiently large $f(x)> \text{ess sup} f + \frac{1}{n}$.
Since $B_n$ are disjoint, $\sum_n \mu(B_n) = \mu(A)>0$ and so there exists some $n_0$ so that $\mu(B_{n_0})>0$.
We want to show that this is a contradiction. Since $\text{ess sup} f + \frac{1}{n_0} > \text{ess sup} f$ and by the definition of infimum, we can find some $a\in \mathbb{R}$, $\text{ess sup} f + \frac{1}{n_0}\geq a \geq \text{ess sup} f$ so that $\mu(\{x\in \Omega : f(x)>a\})=0$. Since $B_{n_0}\subseteq \{x\in \Omega : f(x)>a\}$, we deduce that $\mu(B_{n_0})=0$. This is a contradiction.