I think this might just follow from the definition of $L^\infty$ norm: Let $\Omega$ be a measure space and $u$ be measurable. Suppose that for all $C > a$, we have shown the set $A := \{ x \in \Omega: |u(x)| > C \}$ is a null set. Is it true that $\| u \|_{L^\infty} \leq a$? In my opinion, if we wish to show that $\| u \|_{L^\infty} \leq a$, we must show that for all $C$ such that $|u(x)| \leq C$ for all $x \in \Omega - A$, where $A$ is a null set, we have $C \leq a$. How is this implied from the hypothesis given above?
Update: Just realized this when I was typing this question up: If we know for all $C > a$, we have $|u(x)| \leq C$ for all $x \in\Omega - A_C$ i.e. almost everywhere, then as $$\{ C : C > a \} \subseteq \{ C: |u(x)| \leq C \text{ almost everywhere} \},$$ we have $$ \| u \|_{L^\infty} = \inf\{ C: |u(x)| \leq C \text{ almost everywhere} \} \leq \inf \{ C : C > a \} = a. $$ Is this reasoning correct?