Let $(X,\mathfrak M, \mu)$ be a measure space, i.e. $\mathfrak M$ is a $\sigma$-algebra over the set $X$ and $\mu: \mathfrak{M}\to [0,+\infty]$ is a measure. Let $f:X\to \overline{\mathbb R}:=[-\infty,+\infty]$ be a measurable function. Let $M$ be the essential sup of $f$ over $X$. I'd like to prove the following fact:
$\forall \epsilon >0 \exists A_\epsilon\in\mathfrak M $, with $\mu(A_\epsilon)>0$, such that $|f(x)|> M_0-\epsilon$ for all $x\in A_\epsilon$.
My idea is to proceed for absurdity. Let's suppose that the thesis is false. I tried to have a contradiction but, I was not successful. Can anyone please help me?
I think that $A_\epsilon=f^{-1}(]M-\epsilon,\infty[)$ works, the only thing that needs to be proven is that $\mu(A_\epsilon)>0$ but if for some $\epsilon$, we have $\mu(A_\epsilon)=0$, then $M-\epsilon$ is an "essential upper bound" on $f$ and so $M$ is not the supremum.