From point-wise to essential supremum of a set of real-valued measurable functions For an uncountable collection of uncountable sets of real-valued random variables (i.e. measurable with respect to a $\sigma$-algebra) $\{S_i\}_{i\in I}$, with
$\inf \left(\bigcap_{i\in I}S_i \right)= \sup\{\inf S_i|{i\in I}\}$
I want to to show
$\mathrm{ess}\inf \left(\bigcap_{i\in I}S_i \right)= \mathrm{ess}\sup\{\mathrm{ess}\inf S_i|{i\in I}\}$
I tried something in analogy to the proof of existence of the essential supremum, but failed. 
It would be great to get some help on this. Does it hold and if so, how prove it and if not, why not?

This is a cross-posting from this question on math.stackexchange

Edit: I change the notation. No the all infima/suprema are understood to be pointwise suprema of a set of functions. And the essential suprema/infima of a set of functions should be read like on this page.
The indexed intersection is defined as usual: $\bigcap_{i\in I}S_i = \{x|\forall i\in I: x\in S_i\}$
 A: Okay. Give $[0,1]$ Lebesgue measure. For each $t \in [0,1]$ let $S_t = \{1_{\{t\}}\} \cup \{a\cdot 1_{[0,1]}: a \geq 1\}$.
Then $\bigcap S_t = \{a\cdot 1_{[0,1]}: a \geq 1\}$ and its inf is the function $1_{[0,1]}$. For each $t$ the inf of $S_t$ is the function $1_{\{t\}}$, and the sup of these is also the function $1_{[0,1]}$. So your premise holds.
The essential inf of $\bigcap S_t$ is also the function $1_{[0,1]}$ but the essential inf of each $S_t$ is the zero function and their essential sup is again the zero function. So the conclusion fails.
Is this really what you meant?
A: The question is not clear. Are you asking whether the complete distributive law holds in (probably the unit ball of) $L^\infty(X,\mu)$? The answer is no: complete distributivity is characteristic of atomic measure spaces. An easy way to see this is to use the fact that a complete lattice is completely distributive iff it has the property that for all $c$ and $d$ with $c \not\geq d$ there exist $c' \not\leq c$ and $d' \not\geq d$ such that every element of the lattice lies above $c'$ or below $d'$. I refer you to Theorem 6.42 of my book Lipschitz Algebras (second edition) for a proof. Let $A$ be a positive measure set that contains no atoms, find $B \subset A$ with $0 < \mu(B) < \mu(A)$, and take $c = \chi_B$ and $d = \chi_{A\setminus B}$.
