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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 $\mu$.

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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 $x\in A$.

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  • $\begingroup$ Very slick, nice! $\endgroup$
    – Nate River
    Commented Feb 4, 2023 at 22:03
  • $\begingroup$ Hm, wait actually, how do you get the countable union? Since the set of $x \in A$ may be uncountable, you may have to take an uncountable union of sets of the form $A \cap [0, x]$. Am I missing something? $\endgroup$
    – Nate River
    Commented Feb 4, 2023 at 22:13
  • $\begingroup$ Let $\alpha=\sup A$. If $\alpha\in A$, then $A=A\cap [0,\alpha]$, otherwise $A=\cup (A\cap [0,\alpha_n])$, where $\alpha_n\in A$ approach $\alpha$ $\endgroup$
    – Fedor Petrov
    Commented Feb 4, 2023 at 22:15
  • $\begingroup$ Ah yes, I just thought of that as well. You don’t have to union over all $x \in A$. Nice. $\endgroup$
    – Nate River
    Commented Feb 4, 2023 at 22:15

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