# Essential supremum and essential infimum

In mathematics, the concepts of essential supremum and essential infimum are related to the notions of supremum and infimum, but adapted to measure theory and functional analysis, where one often deals with statements that are not valid for all elements in a set, but rather almost everywhere, i.e., except on a set of measure zero.

## Definition

Let f : X → R be a real valued function defined on a set X. A real number a is called an upper bound for f if f(x) ≤ a for all x in X, i.e., if the set

$f^{-1}(a, \infty) = \{x\in X: f(x)>a\}$

is empty. Let

$U_f = \{a \in \mathbb{R}: f^{-1}(a, \infty) = \emptyset\} \,$

be the set of upper bounds of f. Then the supremum of f is defined by

$\sup f=\inf U_f \,$

if the set of upper bounds $U_f$ is nonempty, and  sup f = +∞ otherwise.

Now assume in addition that (X, Σ, μ) is a measure space and, for simplicity, assume that the function f is measurable. A number a is called an essential upper bound of f if the measurable set f−1(a, ∞) is a set of measure zero,[a] i.e., if f(x) ≤ a for almost all x in X. Let

$U^{\mathrm{ess}}_f = \{a \in \mathbb{R}: \mu(f^{-1}(a, \infty)) = 0\}\,$

be the set of essential upper bounds. Then the essential supremum is defined similarly as

$\mathrm{ess } \sup f=\inf U^{\mathrm{ess}}_f \,$

if $U^{\mathrm{ess}}_f \ne \emptyset$, and ess sup f = +∞ otherwise.

Exactly in the same way one defines the essential infimum as the supremum of the essential lower bounds, that is,

$\mathrm{ess } \inf f=\sup \{b \in \mathbb{R}: \mu(\{x: f(x) < b\}) = 0\}\,$

if the set of essential lower bounds is nonempty, and as −∞ otherwise.

## Examples

On the real line consider the Lebesgue measure and its corresponding σ-algebra Σ. Define a function f by the formula

$f(x)= \begin{cases} 5, & \text{if } x=1 \\ -4, & \text{if } x = -1 \\ 2, & \text{ otherwise. } \end{cases}$

The supremum of this function (largest value) is 5, and the infimum (smallest value) is −4. However, the function takes these values only on the sets {1} and {−1} respectively, which are of measure zero. Everywhere else, the function takes the value 2. Thus, the essential supremum and the essential infimum of this function are both 2.

As another example, consider the function

$f(x)= \begin{cases} x^3, & \text{if } x\in \mathbb Q \\ \arctan{x} ,& \text{if } x\in \mathbb R\backslash \mathbb Q \\ \end{cases}$

where Q denotes the rational numbers. This function is unbounded both from above and from below, so its supremum and infimum are ∞ and −∞ respectively. However, from the point of view of the Lebesgue measure, the set of rational numbers is of measure zero; thus, what really matters is what happens in the complement of this set, where the function is given as arctan x. It follows that the essential supremum is π/2 while the essential infimum is −π/2.

On the other hand, consider the function f(x) = x3 defined for all real x. Its essential supremum is +∞, and its essential infimum is −∞.

Lastly, consider the function

$f(x)= \begin{cases} 1/x, & \text{if } x \ne 0 \\ 0, & \text{if } x = 0. \\ \end{cases}$

Then for any $\textstyle a \in \mathbb R$, we have $\textstyle \mu(\{x \in \mathbb R : 1/x > a\}) \geq \tfrac{1}{|a|}$ and so $\textstyle U_f = \emptyset$ and ess sup f = +∞.

## Properties

• If $\mu(X)>0$ we have $\inf f \le \mathrm{ess } \inf f \le \mathrm{ess }\sup f \le \sup f$. If $X$ has measure zero $\mathrm{ess }\sup f=-\infty$ and $\mathrm{ess }\inf f=+\infty$.[1]
• $\mathrm{ess }\sup (fg) \le (\mathrm{ess }\sup f)(\mathrm{ess }\sup g)$ whenever both terms on the right are nonnegative.

1. ^ For non measurable functions the definition has to be modified by assuming that $f^{-1}(a, \infty)$ is contained in a set of measure zero. Alternatively one can assume that the measure is complete
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