Suppose we have a fixed measure space $(X,M,\mu)$
$||f||_\infty=\inf\{a\geq 0: \mu\big(\{x:|f(x)|> a\}\big)=0\}$
I been looking at this definition for quite a while, but I don't really understand what exactly it is. If I take a simple function $\displaystyle f=\sum^n_{k=1}a_k1_{E_k}$ where $\mu(E_k)>0$ for all $k$ then what exactly does $||f||_\infty$ give me? assuming $a_k>0$ for all $k$
If I call $\{a\geq 0: \mu\big(\{x:|f(x)|> a\}\big)=0\}=A$ then I want to consider $0\leq b\leq a_k$ for all $k$, in this case since every $E_k$ has positive measure, and for all $x$, $f(x)=a_j$ for some $1\leq j\leq k$,$b\notin A$. And for any $b^\prime \geq a_k$ for all $k$ , $b^\prime \in A$, since $\{x:|f(x)|>b^\prime\}=\emptyset$. And Suppose $a=\operatorname{max}\{a_j:1\leq j\leq k\}$, then $a\in A$ by similar argument, but $a_j\notin A$ for all $a_j\neq a$. So we have $||f||_\infty=a$? in other words $\max\{a_j\}$?
Can anyone check my understanding of the definition?
You need to understand what essentially bounded means. Formally, in a measure space $(X, \boldsymbol{\mathfrak A}, \mu), $ take an extended complex-valued $\boldsymbol{\mathfrak A}$-measurable function $f$ on $X.$ If there exists a null set $N$ such that $|f|\leq M, ~M\in [0, \infty), $ on $X\setminus N$ (which means $\mu\{x\in X:|f(x) > M\}=0$), then $M$ is an essential bound of $f.$
Now we can seek the infimum of the set of all such essential bounds for $f$ i.e. $\inf\{M\in [0, \infty) :\mu\{x\in X:|f(x) > M\}=0\}=:\Vert f\Vert_\infty.$ This is how essential supremum of $f$ is defined.
What does it mean when $\Vert f\Vert_\infty=\infty$? Based on the idea that the smaller the set, greater the infimum and subsequently the infimum of an empty set is $\infty, $ we can say that $\Vert f\Vert_\infty=\infty$ if the set of all essential bounds of $f$ on $X$ is empty.
Now do you have the insight to carry on your work with your example?