Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to\CC\) be measurable. The essential supremum of \(\abs{f}\) is \[ \norm{f}_\infty \coloneqq \inf \theset{\alpha \geq 0 \suchthat \mu\qty{\theset{x\in X \suchthat \abs{f(x)} \geq \alpha}} = 0} \in [0,\infty], \] with \(\inf\emptyset=\infty\). The function \(f\) is essentially bounded if there exists a real number \(c\) such that \(\mu\qty{\theset{x\in X \suchthat \abs{f(x)} > c}} = 0\), equivalently \(\norm{f}_\infty < \infty\).
When \(\norm{f}_\infty<\infty\), the set \(\theset{\abs{f}>\norm{f}_\infty}=\bigcup_{k\geq 1}\theset{\abs{f}\geq\norm{f}_\infty+1/k}\) is a countable union of null sets, so \(\abs{f(x)} \leq \norm{f}_\infty\) for almost every \(x\), and \(\norm{f}_\infty\) is the smallest constant with this property.