Take $f: \mathbb{R}\to \mathbb{R}$ be continuous, then w.r.t. Lebesgue measure $L$, we have $$\operatorname{ess sup} |f|=\sup |f|$$ w.r.t. $L$.
I have been provided a proof in a Measure theory course I studied many years ago.
Hence $M\le N$.
I understood first part of the proof, but in the proof of the 2nd part, I do not understand why $2\epsilon$ is used. (Why not just $\epsilon?$)
$\endgroup$1 Answer
$\begingroup$The first $\epsilon$ comes from the fact that you don't necessarily know there is an $x$ such that $|f(x)|=M$, you only know that $|f(x)|$ gets arbitrarily close to $M$, so you can say $|f(x)|>M-\epsilon$. The second $\epsilon$ comes from the definition of continuity at $x$: you know that in some neighborhood of $x$, $f$ is within $\epsilon$ of the value $f(x)$. Since $|f(x)|>M-\epsilon$, this means $$|f(y)|>|f(x)|-\epsilon>M-\epsilon-\epsilon=M-2\epsilon$$ for all $y$ in some neighborhood of $x$.
$\endgroup$ 0
