essential supremum

Essential supremum of a function

Let $(\mathrm{\Omega },ℱ,\mu )$ be a measure space and let $f$ be a Borel measurable function from $\mathrm{\Omega }$ to the extended real numbers $\overline{ℝ}$. The essential supremum of $f$ is the smallest number $a\in \overline{ℝ}$ 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 $\mathrm{ess}supf$ as follows. Let $a\in ℝ$, and define

$$M_a=\{x:f(x)>a\},$$

the subset of $X$ where $f(x)$ is greater than $a$. Then let

$$A_0=\{a\in ℝ:\mu (M_a)=0\},$$

the set of real numbers for which $M_a$ has measure zero. The essential supremum of $f$ is

$$\mathrm{ess}supf:=inf{A}_0.$$

The supremum is taken in the set of extended real numbers so, $\mathrm{ess}supf=\mathrm{\infty }$ if $A_0=\mathrm{\varnothing }$ and $\mathrm{ess}supf=-\mathrm{\infty }$ if $A_0=ℝ$.

Essential supremum of a collection of functions

Let $(\mathrm{\Omega },ℱ,\mu )$ be a measure space, and $𝒮$ be a collection of measurable functions $f:\mathrm{\Omega }\to \overline{ℝ}$. The Borel $\sigma$-algebra on $\overline{ℝ}$ is used.

If $𝒮$ is countable then we can define the pointwise supremum of the functions in $𝒮$, which will itself be measurable. However, if $𝒮$ 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 $𝒮$, written as $\mathrm{ess}sup𝒮$, if it exists, is a measurable function $f:\mathrm{\Omega }\to \overline{ℝ}$ satisfying the following.

• $f\ge g$, $\mu$-almost everywhere (http://planetmath.org/AlmostSurely), for any $g\in 𝒮$.

• if $g:\mathrm{\Omega }\to \overline{ℝ}$ is measurable and $g\ge h$ ($\mu$-a.e.) for every $h\in 𝒮$, then $g\ge f$ ($\mu$-a.e.).

Similarly, the essential infimum, $\mathrm{ess}inf𝒮$ is defined by replacing the inequalities ‘$\ge$’ by ‘$\le$’ in the above definition.

Note that if $f$ is the essential supremum and $g:\mathrm{\Omega }\to \overline{ℝ}$ 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 (http://planetmath.org/ExistenceOfTheEssentialSupremum). Furthermore, they are always equal to the supremum or infimum of some countable subset of $𝒮$.