Existence of essential supremum for an uncountable sequence of measurable functions

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.