Interchanging two essential suprema of a random variable

Question

Setting

Let's start with the definition of the essential supremum of a random variable:

A $\mathcal G$-measurable random variable (rv) $Y$ is called essential suprema of a family of $\mathcal G$-measurable rv $(X_i)_{i\in I}$ (with values in $[-\infty,\infty ]$), if

$a)$ $Y\ge X_i$ a.s. for every $i \in I$

$b)$ $Y \le Z$ a.s. for every $\mathcal G$-measurable rv $Z$ with $Z \ge X_i$

We write $Y=: \text{ess sup}_{i \in I} X_i.$

Question

I would like to know if the following equality is true or not: Let $I$ and $J$ be too arbitrary index sets and $(X_i^j)_{i\in I, i \in J}$ a bounded family of rvs, then

$$\text{ess sup}_{i \in I} \text{ess sup}_{j\in J} X_i^j \overset{?}=\text{ess sup}_{j \in J} \text{ess sup}_{i \in I} X_i^j.$$

Any help is appreciated!

Answer 1

Let \begin{gather*} U_i=\operatorname*{ess\,sup}_{j\in J}X_i^{(j)},\quad \forall i\in I, \qquad V= \operatorname*{ess\,sup}_{i\in I}U_i,\\ A_i^{(j)}=\{U_i\ge X_i^{(j)}\}, \quad B_i=\{V\ge U_i\}, \end{gather*} then \begin{gather*} \mathsf{P}(A_i^{(j)})=1,\qquad \forall j\in J,\quad \forall i\in I, \tag{1}\\ \mathsf{P}(B_i)=1,\qquad \forall i\in I. \tag{2} \end{gather*} In view of $B_iA_i^{(j)}\subset\{V\ge X_i^{(j)}\} $ and (1),(2), \begin{equation*} \mathsf{P}(V\ge X_i^{(j)})\ge \mathsf{P}(B_iA_i^{(j)})=1. \qquad \forall i\in I, \quad \forall j\in J, \tag{3} \end{equation*} Furthermomre, from (3) and the definition of $ \operatorname{ess\,sup} $, \begin{equation*} \mathsf{P}(V\ge \operatorname*{ess\,sup}_{i\in I}X_i^{(j)})=1,\qquad \forall j\in J, \end{equation*} and \begin{equation*} \mathsf{P}(V\ge \operatorname*{ess\,sup}_{j\in J}\,\operatorname*{ess\,sup}_{i\in I}X_i^{(j)})=1. \end{equation*} Now, by the symmetry of $ i $ and $ j $ in above proof, \begin{equation*} \mathsf{P}(\operatorname*{ess\,sup}_{i\in I} \operatorname*{ess\,sup}_{j\in J} X_i^{(j)} = \operatorname*{ess\,sup}_{j\in J} \operatorname*{ess\,sup}_{i\in I} X_i^{(j)})=1. \end{equation*}