I want to show that
where the $\mathrm{ess }\inf$ of a measurable real function $f$ is defined as $\sup\{k\in \mathbb{R}:\mu (\mid f\mid<k)=0\}$ and $\mathrm{ess }\sup (f)=\inf\{k\in \mathbb{R}:\mu(\mid f \mid >k)=0\}$ on a given measure space $(E,\mathcal{A},\mu)$. Then $\mathrm{ess }\inf (f)$ is the supremum of those $k>0$ such that $\mid f\mid\geq k$ almost everywhere (w.r.t. $\mu$). Here I found equivalent definitions. Using these definitions I concluded that $$\mathrm{ess }\sup(f)=\inf\{\sup(f(A)):A\in \mathcal{A} ,\mu(A^c)=0\}\subseteq \{\sup(f(A)):A\in \mathcal{A}\}$$ which implies $$\sup(\mathrm{ess }\sup(f))=\mathrm{ess }\sup(f)\leq \sup\{\sup(f(A)):A\in \mathcal{A}\}=\sup f(E)$$ since $\mathrm{ess }\sup(f)$ is singleton. Is it correct?
Thanks!
Unfortunately, I don't understand your proof. I can give a proof of my own.
We want to show that the $\text{esssup}(f):= \inf\{k\in\mathbb{R}: \mu(|f|>k)=0\}$ is always smaller than the actual supremum.
To show this let $\sup(f)=M$. This means that for all $x\in \mathbb{R}$ we have that $f(x)\leq M$.
In particular this means that $\mu(|f|>M) = \mu(\emptyset)=0$. Therefore $M\in\{k\in\mathbb{R}: \mu(|f|>k)=0\}$ and so $M\geq \inf\{k\in\mathbb{R}: \mu(|f|>k)=0$. This proves the claim.
To show that $\text{esssup}(f) \geq \text{essinf}(f)$:
Let $\text{esssup}(f)=M$, $\text{essinf}(f)=N$. This means that $\mu(|f|>M)=0$ and $\mu(|f|<N)=0$.
If by contradiction $N>M$ then $$\mu(|f|<N+1) = \mu(|f|<N) + \mu(N\leq |f|<N+1)\leq 0+\mu(M<|f|)= 0$$ Therefore $N+1\in \{k\in\mathbb{R}:\mu(|f|<k)=0\}$ which contradicts the fact that $N$ is the supremum of this set.