# Essential infimum and essential supremum

In mathematics, the concepts of essential infimum and essential supremum are related to the notions of infimum and supremum, 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.

While the exact definition is not immediately straightforward, intuitively the essential supremum of a function is the smallest value that is larger or equal than the function values everywhere when allowing for ignoring what the function does at a set of points of measure zero. For example, if one takes the function ${\displaystyle f(x)}$ that is equal to zero everywhere except at ${\displaystyle x=0}$ where ${\displaystyle f(0)=1}$, then the supremum of the function equals one. However, its essential supremum is zero because we are allowed to ignore what the function does at the single point where ${\displaystyle f}$ is peculiar. The essential infimum is defined in a similar way.

## Definition

As is often the case in measure-theoretic questions, the definition of essential supremum and infimum does not start by asking what a function f does at points x (i.e., the image of f), but rather by asking for the set of points x where f equals a specific value y (i.e., the preimage of y under f).

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

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

is empty. Let

${\displaystyle U_{f}=\{a\in \mathbb {R} :f^{-1}(a,\infty )=\varnothing \}\,}$

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

${\displaystyle \sup f=\inf U_{f}\,}$

if the set of upper bounds ${\displaystyle U_{f}}$  is nonempty, and ${\displaystyle \sup f=+\infty }$  otherwise.

Alternatively, if for some ${\displaystyle a\in \mathbb {R} }$  we have ${\displaystyle f(x)\leq a}$  for all ${\displaystyle x\in X}$  then ${\displaystyle \sup f\leq a}$ , and ${\displaystyle \sup f=\inf\{a\in \mathbb {R} :f(x)\leq a{\text{ for all }}x\in X\}}$  (with this infimum being taken to be ${\displaystyle +\infty }$  if the set is empty).

Now assume in addition that ${\displaystyle (X,\Sigma ;\mu )}$  is a measure space and, for simplicity, assume that the function ${\displaystyle f}$  is measurable. A number ${\displaystyle a}$  is called an essential upper bound of f if the measurable set ${\displaystyle f^{-1}(a,\infty )}$  is a set of measure zero,[a] i.e., if ${\displaystyle f(x)\leq a}$  for almost all ${\displaystyle x}$  in ${\displaystyle X}$ . Let

${\displaystyle U_{f}^{\operatorname {ess} }=\{a\in \mathbb {R} :\mu (f^{-1}(a,\infty ))=0\}\,}$

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

${\displaystyle \operatorname {ess} \sup f=\inf U_{f}^{\mathrm {ess} }\,}$

if ${\displaystyle U_{f}^{\operatorname {ess} }\neq \varnothing }$ , and ${\displaystyle \operatorname {ess} \sup f=+\infty }$  otherwise.

Alternatively, if for some ${\displaystyle a\in \mathbb {R} }$  we have ${\displaystyle f(x)\leq a}$  for almost all ${\displaystyle x\in X}$  then ${\displaystyle \operatorname {ess} \sup f\leq a}$ , and ${\displaystyle \operatorname {ess} \sup f=\inf\{a\in \mathbb {R} :f(x)\leq a{\text{ for almost all }}x\in X\}}$  (with this infimum being taken to be ${\displaystyle +\infty }$  if the set is empty).

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

${\displaystyle \operatorname {ess} \inf f=\sup\{b\in \mathbb {R} :\mu (\{x:f(x)

if the set of essential lower bounds is nonempty, and as ${\displaystyle -\infty }$  otherwise; again there is an alternative expression as ${\displaystyle \operatorname {ess} \inf f=\sup\{a\in \mathbb {R} :f(x)\geq a{\text{ for almost all }}x\in X\}}$  (with this being ${\displaystyle -\infty }$  if the set is empty).

## Examples

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

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

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

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

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 ${\displaystyle +\infty }$ , and its essential infimum is ${\displaystyle -\infty }$ .

Lastly, consider the function

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

Then for any ${\displaystyle \textstyle a\in \mathbb {R} }$ , we have ${\displaystyle \textstyle \mu (\{x\in \mathbb {R} :1/x>a\})\geq {\tfrac {1}{|a|}}}$  and so ${\displaystyle \textstyle U_{f}^{\text{ess}}=\varnothing }$  and ${\displaystyle \operatorname {ess} \sup f=+\infty }$ .

## Properties

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

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