# Almost Sure

## 6 January 19

### Essential Suprema

Filed under: Probability Theory — George Lowther @ 2:00 PM
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Given a sequence ${X_1,X_2,\ldots}$ of real-valued random variables defined on a probability space ${(\Omega,\mathcal F,{\mathbb P})}$, it is a standard result that the supremum

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\},\smallskip\\ &\displaystyle X(\omega)=\sup_nX_n(\omega). \end{array}$

is measurable. To ensure that this is well-defined, we need to allow X to have values in ${{\mathbb R}\cup\{\infty\}}$, so that ${X(\omega)=\infty}$ whenever the sequence ${X_n(\omega)}$ is unbounded above. The proof of this fact is simple. We just need to show that ${X^{-1}((-\infty,a])}$ is in ${\mathcal F}$ for all ${a\in{\mathbb R}}$. Writing,

$\displaystyle X^{-1}((-\infty,a])=\bigcap_nX_n^{-1}((-\infty,a]),$

the properties that ${X_n}$ are measurable and that the sigma-algebra ${\mathcal F}$ is closed under countable intersections gives the result.

The measurability of the suprema of sequences of random variables is a vital property, used throughout probability theory. However, once we start looking at uncountable collections of random variables things get more complicated. Given a, possibly uncountable, collection of random variables ${\mathcal S}$, the supremum ${S=\sup\mathcal S}$ is,

 $\displaystyle S(\omega)=\sup\left\{X(\omega)\colon X\in\mathcal S\right\}.$ (1)

However, there are a couple of reasons why this is often not a useful construction:

• The supremum need not be measurable. For example, consider the probability space ${\Omega=[0,1]}$ with ${\mathcal F}$ the collection of Borel or Lebesgue subsets of ${\Omega}$, and ${{\mathbb P}}$ the standard Lebesgue measure. For any ${a\in[0,1]}$ define the random variable ${X_a(\omega)=1_{\{\omega=a\}}}$ and, for a subset A of ${[0,1]}$, consider the collection of random variables ${\mathcal S=\{X_a\colon a\in A\}}$. Its supremum is

$\displaystyle S(\omega)=1_{\{\omega\in A\}}$

which is not measurable if A is a non-measurable set (e.g., a Vitali set).

• Even if the supremum is measurable, it might not be a useful quantity. Letting ${X_a}$ be the random variables on ${(\Omega,\mathcal F,{\mathbb P})}$ constructed above, consider ${\mathcal S=\{X_a\colon a\in[0,1]\}}$. Its supremum is the constant function ${S=1}$. As every ${X\in\mathcal S}$ is almost surely equal to 0, it is almost surely bounded above by the constant function ${Y=0}$. So, the supremum ${S=1}$ is larger than we may expect, and is not what we want in many cases.

The essential supremum can be used to correct these deficiencies, and has been important in several places in my notes. See, for example, the proof of the debut theorem for right-continuous processes. So, I am posting this to use as a reference. Note that there is an alternative use of the term `essential supremum’ to refer to the smallest real number almost surely bounding a specified random variable, which is the one referred to by Wikipedia. This is different from the use here, where we look at a collection of random variables and the essential supremum is itself a random variable.

The essential supremum is really just the supremum taken within the equivalence classes of random variables under the almost sure ordering. Consider the equivalence relation ${X\cong Y}$ if and only if ${X=Y}$ almost surely. Writing ${[X]}$ for the equivalence class of X, we can consider the ordering given by ${[X]\le[Y]}$ if ${X\le Y}$ almost surely. Then, the equivalence class of the essential supremum of a collection ${\mathcal S}$ of random variables is the supremum of the equivalence classes of the elements of ${\mathcal S}$. In order to avoid issues with unbounded sets, we consider random variables taking values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$.

Definition 1 An essential supremum of a collection ${\mathcal S}$ of ${\bar{\mathbb R}}$-valued random variables,

$\displaystyle S = {\rm ess\,sup\,}\mathcal{S}$

is the least upper bound of ${\mathcal{S}}$, using the almost-sure ordering on random variables. That is, S is an ${\bar{\mathbb R}}$-valued random variable satisfying

• upper bound: ${S\ge X}$ almost surely, for all ${X\in\mathcal S}$.
• minimality: for all ${\bar{\mathbb R}}$-valued random variables Y satisfying ${Y\ge X}$ almost surely for all ${X\in\mathcal S}$, we have ${Y\ge S}$ almost surely.

It is straightforward to see that the essential supremum is unique up to almost sure equivalence, although showing that it always exists is a bit trickier.

Theorem 2 For any collection ${\mathcal S}$ of ${\bar{\mathbb R}}$-valued random variables, its essential supremum exists and is uniquely defined up to almost-sure equivalence.

Proof: Uniqueness follows from the definition. If S and T are both essential suprema, then they are upper bounds of ${\mathcal S}$ under the almost sure ordering. By the minimality property, both ${S\ge T}$ and ${T\ge S}$ almost surely, so ${S=T}$ almost surely.

To prove existence, we reduce to the existence of suprema of bounded subsets of ${{\mathbb R}}$ by taking expectations of a bounded function of the random variables. Start by choosing a continuous, bounded and strictly increasing function ${f\colon\bar{\mathbb R}\rightarrow{\mathbb R}}$. For example, we can take

$\displaystyle f(x)=\begin{cases} x/(1+\lvert x\rvert),&{\rm\ for\ }x\in{\mathbb R},\\ 1,&{\rm\ for\ }x=\infty,\\ -1,&{\rm\ for\ }x=-\infty. \end{cases}$

Also, let ${\mathcal T}$ be the collection of maxima ${X_1\vee\cdots\vee X_n}$ of finite sequences of random variables in ${\mathcal S}$, together with the constant function ${X=-\infty}$. Clearly, ${\mathcal T}$ is closed under taking the maximum of pairs of random variables. We set,

$\displaystyle \alpha=\sup\{{\mathbb E}[f(X)]\colon X\in\mathcal T\}.$

As f is measurable and bounded, the expectations are well-defined. Then, as ${\mathcal T}$ is nonempty, it contains a sequence ${X_n}$ such that ${{\mathbb E}[f(X_n)]}$ tends to ${\alpha}$. Replacing ${X_n}$ by ${X_1\vee\cdots\vee X_n}$ if necessary, we may suppose that ${X_n}$ is an increasing sequence. We show that

$\displaystyle S=\sup_nX_n$

is an essential supremum of ${\mathcal S}$. As ${S\ge X_n}$ and ${f}$ is increasing, we have ${{\mathbb E}[f(S)]\ge\alpha}$.

First of all, for any ${X\in\mathcal S}$, the maxima ${X\vee X_n}$ are in ${\mathcal T}$. By monotone convergence,

 $\displaystyle {\mathbb E}[f(S\vee X)]=\lim_n{\mathbb E}[f(X_n\vee X)]\le\alpha\le{\mathbb E}[f(S)].$ (2)

If the event ${X > S}$ has positive probability then the nonnegative random variable ${f(S\vee X)-f(S)}$ is strictly positive with positive probability giving

$\displaystyle {\mathbb E}[f(S\vee X)] > {\mathbb E}[f(S)]$

contradicting (2). So, ${X\le S}$ almost surely.

Next, suppose that Y is an ${\bar{\mathbb R}}$-valued random variable satisfying ${Y\ge X}$ almost surely, for all X in ${\mathcal S}$. Then ${Y\ge X_n}$ almost surely and, taking the limit, ${Y\ge S}$ almost surely. ⬜

In the case of countable collections of random variables the essential supremum coincides, almost surely, with the pointwise supremum (1), as we would expect.

Lemma 3 If ${\mathcal{S}}$ is a countable collection of ${\bar{\mathbb R}}$-valued random variables then

$\displaystyle {\rm ess\,sup\,}\mathcal{S}=\sup\mathcal{S}$

almost surely.

Proof: Assuming that ${\mathcal S}$ is nonempty, we can write it as ${\{X_1,X_2,\ldots\}}$. As noted above, the supremum of a countable sequence of random variables is measurable, so

$\displaystyle S\equiv\sup\mathcal S=\sup_nX_n$

is measurable and clearly satisfies the upper bound property. Next, suppose that X is an upper bound of ${\mathcal S}$ in the almost sure ordering. Then, ${X\ge X_n}$ almost surely, for all n. Countable additivity of probability measures gives ${X\ge S}$ almost surely, so S satisfies the minimality property. ⬜

Finally, we note that the essential supremum of ${\mathcal S}$ can always be expressed as the supremum of some countable sequence chosen from the collection of random variables ${X_n\in\mathcal S}$.

Lemma 4 Let S be a nonempty collection of ${\bar{\mathbb R}}$-valued random variables. Then, there exists a sequence ${X_1,X_2,\ldots}$ in ${\mathcal S}$ with

$\displaystyle {\rm ess\,sup\,}\mathcal{S}=\sup_nX_n$

almost surely.

Proof: If ${\mathcal T}$ is the collection of maxima of finite sequences of random variables in ${\mathcal S}$, the proof of theorem 2 constructed a sequence ${X_n\in\mathcal T}$ with ${S=\sup_nX_n}$ an essential supremum of S. As ${X_n}$ is the supremum of a finite subset ${\mathcal S_n}$ of ${\mathcal S}$, we have

$\displaystyle S=\sup_n\sup\mathcal S_n=\sup\bigcup_n\mathcal S_n.$

Letting ${Y_1,Y_2,\ldots\in\mathcal S}$ be an enumeration of the countable set ${\bigcup_n\mathcal S_n}$, we have ${S=\sup_nY_n}$. ⬜

## 1 Comment »

1. […] debut of a set is almost surely equal to a time in . Let be the set of all with , and be the essential supremum of . By standard properties of the essential supremum, we can write for a sequence in . It […]

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