Here is the question:
Where $\|f\|_{\infty} $ is defined to be the infimum of the essential upper bounds for $f.$ And a function $f$ is called essentially bounded provided there is some $M \geq 0,$ called an essential upper bound for $f,$ for which $|f(x)| \leq M $ for almost all $x \in E.$
But still, I do not know how to prove this $\|f\|_{\infty} = \|f\|_{max}$ where $\|f\|_{max} = \max_{x \in [a,b]} |f(x)|.$ Could anyone help me in doing so, please?
For any function $f : [a,b] \to \mathbb{R}$ for which $\|f\|_{max}$ is well-defined (e.g. if $f$ is continuous), we have $\|f\|_\infty \leq \|f\|_{max},$ which follows from $$\{x \in [a,b] \mid |f(x)| > \|f\|_{max}\} = \emptyset$$
For a continuous function $f : [a,b] \to \mathbb{R}$, we can let $\varepsilon > 0$ be arbitrary and then note that $$U = \{x \in [a,b] \mid |f(x)| > \|f\|_{max} - \varepsilon\}$$ is an open subset of $[a,b]$ (as a subspace of $\mathbb{R}$). Since $f$ is continuous, $$f(c) = \|f\|_{max} > \|f\|_{max}-\varepsilon$$ for some $c \in [a,b]$, $U \ni c$ is nonempty, so $m(U) \neq 0$. Therefore $\|f\|_\infty \geq \|f\|_{max}-\varepsilon$. Now taking $\varepsilon \to 0^+$ yields $\|f\|_\infty \geq \|f\|_{max}$.
Note: I have assumed $m$ denotes a measure that satisfies $m(V) > 0$ for any nonempty open set $V$. This is true, for instance, of the Lebesgue measure on $[a,b]$ and $a<b$.