supremum inequality 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|
If by contradiction $N>M$ then $$\mu(|f|