Motivation for definition of essentially bounded

Question

A function $f:(X,S,\mu)$ is said to be essentially bounded if $\Vert f\Vert_\infty$ is finite where, $$\Vert f\Vert_\infty:=\inf\{M:\mu(x:|f(x)|>M)=0\}$$

I would like to know why $\Vert f\Vert_\infty$ is defined the way it is?

My understanding is the following:

To say a function is essentially bounded means to say it is almost equal to a bounded function. So the set where the function is unbounded is a set of measure zero. The goal now would ideally be to find a bound for the function at the remaining points. So if $A$ is the set where $f$ is unbounded then ideally, $$\Vert f\Vert_\infty=\sup\{|f(x)|: x\in A\}=\inf\{M:|f(x)|\leq M, \forall x\in A\}.$$ To me this seems like a more natural way to define $\Vert f\Vert_\infty$. Is this equivalent to the actual definition?

Answer 1

It's not clear to me how you are defining $A$.

The definition of essentially bounded I prefer requires that $f$ is equal to some bounded function off a set of measure zero. This happens to be equivalent to finiteness of $||f||_{\infty}.$ This is because $||f||_{\infty}$ is finite if and only if $$\{M:\mu(x:|f(x)|>M)=0\}$$ is nonempty, if and only if there is some $M$ such that $$\mu(x:|f(x)|>M)=0,$$ if and only if $f$ is equal to some bounded function almost everywhere.

Answer 2

In estimating integrals, we often want to make the estimate $|f(x)| \leq M$ for a.e. $x$. The inf of constants $M$ that work is called $\|f\|_{\infty}$. You can show that this inf is actually attained.