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Let $(\Omega, \mathscr{F}, P)$ be a probability space and $\{X_i, i \in I\}\subset \mathbb{L}^0(\mathscr{F})$, where $I$ can be uncountable.

I want to prove that there exists unique (in P-a.s. sense) essential supremum $X := esssup_{i\in I}^PX_i$ in the following sense:

  • $X \in \mathbb{L}^0(\mathscr{F})$
  • $X\geq X_i, P-a.s. , \forall i \in I$
  • If another $\tilde{X}$ satisfies the above two properties, then $\tilde{X}\geq X, P-a.s.$

I know the definition for essential supremum of a function, which is in fact a number. But here we have a sequence of functions, where the essential supremum is itself a function. How are these definitions connected to each other? Any suggestions how I can solve this problem?

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John Dawkin's proof is almost correct, but you need to use Zorn's Lemma to make some of the details work. Just like in his proof, we can assume the $X_i$ are bounded by replacing them with $\arctan X_i$.

Let $\mathcal P$ be the collection of random variables satisfying your first two properties. Namely, $\mathcal P=\{X\in \mathbb L_0(\mathcal F):X\ge X_i\,\,a.s.\forall i\in I\}$. This is a partially ordered set, where $Y\ge Z$ if $Y\ge Z$ almost surely.

We will use Zorn's Lemma to show that $\mathcal P$ has a minimal element. To do this, we must show that any chain $\mathcal C\subset\mathcal P$ has a lower bound in $\mathcal P$. Given such a chain, let $\alpha=\inf_{Y\in \mathcal C} EY$, and let $Y_{n}$ be a sequence where $EY_{n}\downarrow\alpha$. Letting $Y=\inf_n Y_{n}$, I claim that $Y$ is a lower bound for $\mathcal C$. Given any $Z\in \mathcal C$, there are two cases.

  • If $Z\ge Y_n$ for some $n$, then clearly $Z\ge \inf Y_n=Y.$

  • If $Z\le Y_n$ for all $n$, then $Z\le Y$ as well, so $\alpha\le EZ\le EY=\alpha$, so $EZ=\alpha$. By the bounded convergence theorem, $E[Y-Z]=\lim_nE[Y_n-Z]=\lim_n EY_n-EZ=\alpha-\alpha=0$, which means that $Z=Y$ a.s, so $Z\ge Y$.

We have shown $Z\ge Y$ for all $Z\in \mathcal C$, so every chain has a lowed bound.

Finally, using Zorn's Lemma, we can conclude that $\mathcal P$ has a minimal element, $X$, which clearly satisfies all three of your properties.

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Let $g(x):=\arctan(x)$ and consider the real number $\beta:=\sup_{i\in I}E[g(X_i)]$. Choose a sequence $\{i_n\}$ of elements of $I$ such $E[g(X_{i_n})]$ is monotone non-decreasing and $\beta=\lim_n E[g(X_{i_n})]$. The random variable $X:=\sup_n X_{i_n}$ has the properties of essential supremum that you seek. The first property should be clear. Also, $g(X)\ge g(X_{i_n})$ for each $n$, so $\beta\ge E[g(X)]\ge E[g(X_{i_n})]$ for each $n$, hence $E[g(X)]=\beta$.

Now let $i$ be any fixed element of $I$. Define $X':=\max(X,X_i)$. Then $$ \beta\ge E[g(X')]\ge E[g(X)]=\beta, $$ and so $E[g(X')]=E[g(X)]$. But $X'\ge X$, so $g(X')\ge g(X)$, so $g(X')=g(X)$ a.s, and finally $X'=X$ a.s. This final equality amounts to the statement that $X\ge X_i$ a.s. As $i\in I$ was arbitrary, this proves that $X$ has the second required property. The third property follows in a similar way.

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  • $\begingroup$ @ John, thanks for your answer. To make sure that I completely understood your solution, the point of using $g(X)$ was just to have an increasing function that makes sure the expected value is bounded, right? So, if you do not use such function, where does your solution fail? $\endgroup$ – Mehdi Jafarnia Jahromi Sep 16 '15 at 5:09
  • $\begingroup$ Yes, transforming by $g$ ensures that all the expectations are finite and that $\beta$ is a finite number. $\endgroup$ – John Dawkins Sep 16 '15 at 16:17
  • $\begingroup$ In your fourth line, you say $\beta\ge E[g(X)]\ge E[g(X_{i_n})]$. How do you know $\beta\ge E[g(X)]$? This seems equivalent to saying $\sup_i EY_i\ge E(\sup_i Y_i)$, which is not true in general. $\endgroup$ – Mike Earnest Sep 17 '15 at 15:31

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