Let $f\in\mathcal L^\infty(X,\Sigma,\mu)$, where $(X,\Sigma,\mu)$ is a measure space, and $\mathcal L^\infty(X,\Sigma,\mu)$ is the set of essentially bounded functions. That is,
$\|f\|_\infty:=\inf\{c>0:\mu\{x\in X:|f(x)|>c\}=0\}<\infty$.
How can we prove that $|f(x)|\leqslant\|f\|_\infty$ almost everywhere?
It almost seems obvious, since this condition is equivalent to $\mu\{x\in X:|f(x)|>\|f\|_\infty\}=0,$ which appears in the definition of $\|\cdot\|_\infty$ with $c\to\|f\|_\infty$. However, $\|f\|_\infty$ is the infimum of the set satisfying this condition. And there is in general no need for an infimum of a set $A$ to satisfy the condition defining $A$. For example, the infimum of $\{x\in\mathbb R:x>0\}$ certainly does not satisfy $x>0.$
Best Answer
There exist $c_n$ decreasing to $\|f\|_{\infty}$ such that $\mu \{x: |f(x)| >c_n\}=0$ for all $n$. Let $E$ be the union of the sets $\{x: |f(x)| >c_n\}$. Then $\mu (E)=0$ and $x \notin E$ implies $|f(x)| \leq c_n$ for all $n$. Letting $n \to \infty$ we see that $|f(x)| \leq \|f\|_{\infty}$ whenever $x \notin E$.