TITLE: Intersection of an uncountable number of sets.
QUESTION [7 upvotes]: Let $\mathcal{I}$ be an uncountable set. Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space, and $E_i, i\in \mathcal{I}$ be a measurable set such that $\mathbb{P}(E_i)=1$. What can we say about $\mathbb{P}(\cap_{i\in \mathcal{I}} E_i)?$. 
I know that an uncountable intersection of measurable sets is not necessarily measurable. But are there in general conditions that allow such infinite intersections to be measurable? 
Apologies if the question is too elementary or unclear.

REPLY [3 votes]: There is a quite general case when such an uncountable intersection is measurable:  Let $I$  be the set of real numbers, and assume that not only every $E_i$ is measurable, but that these sets are Borel, and that moreover the set 
$$E:= \{ (x,i):  x \in E_i \}$$
is a Borel set.  Then the intersection of all $E_i$ is co-analytic (also called $\bf \Pi^1_1$) and hence measurable by a classical theorem. 
("Borel" can be replaced  by "co-analytic", but not by "measurable".) 
It has already been pointed out that if $I$ is has sufficiently small cardinality then the intersection still has  full measure, where "sufficiently small" depends on the underlying set theory.  
But there is an interesting case where $I$ has size continuum.  Assume that $I= \omega^\omega $ is the set of all functions from the natural numbers to the natural numbers. Assume (as above) that $E $ is a Borel or $\Pi^1_1$ set. Assume also that the map $i \mapsto E_i$ is decreasing in the following sense: 
Whenever $i,j\in \omega^\omega$ and $i(n)\le j(n)$ for all $n$, then $E_i \supseteq E_j$.
Then the intersection of all $E_i$  is not only measurable but has full measure.