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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: $\bigcap_{i\in I}S_i = \{x|\forall i\in I: x\in S_i\}$

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up vote 2 down vote accepted

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?

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Nice, yes that's what he's asking (I think). – Rabee Tourky Feb 21 '13 at 18:33
thanks for the (embarassingly easy) example. What I am actually after: I want to find out the conditions for the second equation. I thought, solving the question for the first equation would help. Here, a sufficient condition seems to be that the sets are upper sets. But I have no clue how to show the same for "almost surely upper sets". The greatest obstacle is, that I do not know how to handle the set of measurable functions ordered by $\le$ (a.s.), because it fails to be the simple product order. What properties does it share with the product order? (e.g. complete distributivity: No, ...) – Johannes Feb 22 '13 at 11:35
If the $S_i$ are upper sets, then the equation you are asking for IS complete distributivity. – Nik Weaver Feb 22 '13 at 13:16

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 5.3.5 of my book Lipschitz Algebras 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 $d = \chi_{A\setminus B}$.

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Thanks for your answer. Are the equations in my questions equivalent to the complete distributive law? I ultimately want to know, under what conditions for $S_i$ the equation for the ess inf/sup holds. After I found a condition for the inf/sup equation I am now trying to transfer it to the ess inf/sup and do not know, if this makes sense at all. – Johannes Feb 16 '13 at 10:27
What order relation are you talking about? Product order for real-valued functions? – Johannes Feb 16 '13 at 11:54
How can a $c′$ exists, such that every element of the lattice lies above it, if also $c′ \nleq c$? Doesn't "above" mean $c′\le c$? – Johannes 0 secs ago – Johannes Feb 16 '13 at 12:20
@Johannes: (1) it's hard for me to tell exactly what the equations in your question are supposed to mean. I am guessing. (2) $f \leq g$ if $f(x) \leq g(x)$ except on a set of measure zero. (3) every element of the lattice either lies above $c'$ or lies below $d'$. – Nik Weaver Feb 16 '13 at 18:53
ok. I tried to make the notation more clear. Is it understandable now? – Johannes Feb 21 '13 at 11:38

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