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Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f$ be a Borel measurable function from $\Omega$ to the extended real numbers $\mathbb{\bar R}$ The essential supremum of $f$ is the smallest number $a\in\mathbb{\bar R}$ for which $f$ only exceeds $a$ on a set of measure zero. This allows us to generalize the maximum of a function in a useful way.
More formally, we define $\mathrm{ess } \sup f$ as follows. Let $a \in \reals$ and define
\begin{equation*} M_{a} = \set{x: f(x) > a}, \end{equation*}the subset of $X$ where $f(x)$ is greater than $a$ Then let \begin{equation*} A_{0} = \set{a \in \reals: \mu(M_a) = 0}, \end{equation*}the set of real numbers for which $M_a$ has measure zero. The essential supremum of $f$ is \begin{equation*} \mathrm{ess } \sup f \defined \inf A_0. \end{equation*}The supremum is taken in the set of extended real numbers so, $\mathrm{ess}\sup f=\infty$ if $A_0=\emptyset$ and $\mathrm{ess}\sup f = -\infty$ if $A_0=\mathbb{R}$
Let $(\Omega,\mathcal{F},\mu)$ be a measure space, and $\mathcal{S}$ be a collection of measurable functions $f\colon\Omega\rightarrow\mathbb{\bar R}$ The Borel $\sigma$ algebra on $\mathbb{\bar R}$ is used.
If $\mathcal{S}$ is countable then we can define the pointwise supremum of the functions in $\mathcal{S}$ which will itself be measurable. However, if $\mathcal{S}$ is uncountable then this is often not useful, and does not even have to be measurable. Instead, the essential supremum can be used.
The essential supremum of $\mathcal{S}$ written as $\esssup\,\mathcal{S}$ if it exists, is a measurable function $f\colon\Omega\rightarrow\mathbb{\bar R}$ satisfying the following.
- $f\ge g$ $\mu$ almost everywhere, for any $g\in\mathcal{S}$
- if $g\colon\Omega\rightarrow\mathbb{\bar R}$ is measurable and $g\ge h$ ($\mu$ a.e.) for every $h\in\mathcal{S}$ then $g\ge f$ ($\mu$ a.e.).
Similarly, the essential infimum, $\essinf\mathcal{S}$ is defined by replacing the inequalities `$\ge$ by `$\le$ in the above definition.
Note that if $f$ is the essential supremum and $g\colon\Omega\rightarrow\mathbb{\bar R}$ is equal to $f$ $\mu$ almost everywhere, then $g$ is also an essential supremum. Conversely, if $f,g$ are both essential supremums then, from the above definition, $f\le g$ and $g\le f$ so $f=g$ ($\mu$ a.e.). So, the essential supremum (and the essential infimum), if it exists, is only defined almost everywhere.
It can be shown that, for a $\sigma$ finite measure $\mu$ the essential supremum and essential infimum always exist. Furthermore, they are always equal to the supremum or infimum of some countable subset of $\mathcal{S}$
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