$\aleph_1$ almost sure events that almost never all hold This recent question sparked my curiosity. Is there a family of events $(E_k)_{k \in I}$ such that$\def\pp{\mathbb{P}}$ $\pp(E_k) = 1$ for any $k \in I$ but $\pp( \bigcap_{k \in I} E_k ) = 0$? Clearly any such family must be uncountable, and my intuition tells me that it exists, so I proceeded to try constructing one.
Let $X \sim U([0,1])$, and let $E_r = ( X \ne r )$ for each $r \in [0,1]$. Then $\pp(E_r) = 1$ for any $r \in [0,1]$, but $\pp( \bigcap_{r \in [0,1]} E_r ) = 0$.
So now my question is, is there a family of size $\aleph_1$ (that ZFC proves has the above property)?
At first I meant my question to be general, where one is allowed to choose any maximal probability measure (non-negative, total measure 1, countable additivity, cannot be extended) based on which all the events are defined. But Eric's comment suggests that the question may be non-trivial even for the Lebesgue measure on a Euclidean space. So I'm now interested in both variants. =)
 A: Let me just address the question on $\mathbb{R}$.
One way to ask this question is:

Is there a family of $\aleph_1$-many subsets of $\mathbb{R}$, each of which is Lebesgue null, whose union is all of $\mathbb{R}$?

(I'm interpreting the sets in this phrasing as the complements of the events in the question, to match with language used elsewhere.)
It turns out (IIRC) that this is independent of ZFC!
If $\vert\mathbb{R}\vert=\aleph_1$ (the Continuum Hypothesis), then the answer is clearly "yes": just take each set to be a singleton!
If $\vert\mathbb{R}\vert>\aleph_1$, the answer is not obviously no; and indeed if I remember correctly, it is consistent that $\vert\mathbb{R}\vert=\aleph_2$ and $\mathbb{R}$ is the union of $\aleph_1$-many null sets.
A related question is:

What is the smallest number of null sets, whose union is non-null?

(Here "null" refers to the usual Lebesgue measure on $\mathbb{R}$.) This number, $\mathfrak{m}$, is a cardinal characteristic of the continuum (see https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#non.28N.29 and http://www.math.lsa.umich.edu/~ablass/hbk.pdf); it is provably uncountable, and provably at most $2^{\aleph_0}$, but consistently we can have $$\aleph_1<\mathfrak{m}<2^{\aleph_0}.$$
