For $f\in L^\infty[a,b]$, show that $$\|f\|_\infty = \min \big\{M : m\{x \in [a, b] : |f(x)|>M\} = 0\big\}\;,$$ and if, furthermore, $f$ is continuous on $[a, b]$, that $\|f\|_\infty = \|f\|_{\max}$.
I am having trouble getting started, but particularly, decomposing $$\|f\|_\infty = \min \big\{M : m\{x \in [a, b] : |f(x)|>M\} = 0\big\}\;.$$ It equals the minimum $M$ that $|f(x)|$ is always greater than such that the measure over $[a, b]$ equals $0$, correct? How do I start this? Is there another definition of $\|f\|_\infty$ that I should be working with as well?
This is not a complete answer, or even a complete hint; it’s an attempt to make the question a little clearer by disentangling some of the notation, together with a large hint for the second part.
For each $M\in\Bbb R$ let $A_M=\{x\in[a,b]:f(x)>M\}$; clearly $m(A_M)\ge m(A_N)$ if $M\le N$. That is, the sets $A_M$ get smaller as $M$ gets bigger. In particular, if $m(A_M)=0$ and $M\le N$, we must have $m(A_N)=0$ as well. Let $Z=\{M\in\Bbb R:m(A_M)=0\}$; these are the cutoffs $M$ with the property that $f(x)\le M$ almost everywhere, and we’ve just seen that if $N\ge M\in Z$, then $N\in Z$.
The claim that you’re asked to prove is that $\|f\|_\infty=\min Z$ and further that if $f$ is continuous, then $\|f\|_{\max}=\min Z$ as well. The second part is rather easy: if $f$ is continuous, there is an $x_0\in[a,b]$ such that $f(x_0)=\max_{x\in[a,b]}f(x)$, and it’s not hard to use the continuity of $f$ to show that if $M<f(x_0)$, then $M\notin Z$.