Let $\mu$ be a Radon measure on $[0, 1]$, and $f: [0, 1] \to \mathbb R$ a Borel measurable function.


**Question:** Is it true that for $\mu$ almost every $x \in [0, 1]$, we have


$$f(x) \leq \mu\text{-esssup}\_{[0, x]} \, f?$$


Here the esssup is taken with respect to $\mu$.



An inequality involving the essential supremum
Yes. It suffices to prove that for every rationals $p<q$ the set $A$ of those $x$ for which simultaneosly $\mu\text{-esssup}\_{[0,x]} f<p$ and $q<f(x)$ satisfies $\mu(A)=0$. Note that if $x\in A$, then $\mu(A\cap [0,x])=0$, otherwise we would get $\mu\text{-esssup}\_{[0,x]} f\geqslant q$. It remains to note that $A$ is at most countable union of the sets of form $A\cap [0,x]$ with $x\in A$.


