Essential supremum

Remarks

Example: on $\R$ with Lebesgue measure, let $f = \mathbf{1}_{\Q}$, the indicator function of the rationals. The set $\set{f > 0} = \Q$ is countable, so it has measure $0$, and $\operatorname{ess\,sup} f \le 0$. For $c < 0$ the set $\set{f > c}$ is all of $\R$, so $\operatorname{ess\,sup} f = 0$, while $\sup f = 1$.

Non-example: the number $1/2$ is not an essential upper bound of $g(x) = x$ on $[0, 1]$ with Lebesgue measure. The set $\set{g > 1/2} = (1/2, 1]$ has measure $1/2 \ne 0$. In fact $\operatorname{ess\,sup} g = 1$.

The essential supremum is itself an essential upper bound, since $\set{f > c}$ is a countable union of the sets $\set{f > c + 1/n}$. The essential infimum is $\operatorname{ess\,inf} f = -\operatorname{ess\,sup}(-f)$. The quantity $\operatorname{ess\,sup} \abs{f}$ is the norm $\norm{f}_\infty$ of the space L-infinity.