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.

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    $\begingroup$ You realize of course that an arbitrary set can be written as an uncountable intersection $\endgroup$ – Piero D'Ancona May 7 '12 at 14:50
  • $\begingroup$ The sigma in sigma algebra is partly for countable, so countable intersection of sets from the algebra are also in the algebra. There are generalizations to arbitrary cardinalities, but then one does not get as nice an object as a sigma algebra (by nice I mean having measure theoretic consequences). Gerhard "Ask Me About System Design" Paseman, 2012.05.07 $\endgroup$ – Gerhard Paseman May 7 '12 at 15:20

If the uncountable number of sets that you're intersecting is small enough, you might be able to guarantee that the intersection is measurable (and in fact of measure 1) --- it depends on information about the set-theoretic universe that the usual axioms (ZFC) don't decide. Specifically, the additivity of measure is defined to be the smallest cardinal $\kappa$ such that some $\kappa$ sets of Lebesgue measure 0 have a union that is not of Lebesgue measure 0. Obviously, this cardinal is at least $\aleph_1$ and at most the cardinality $\mathfrak c$ of the continuum. So if the continuum hypothesis holds, there's nothing more to be said about it. But in the absence of the continuum hypothesis, the additivity of measure is as small as $\aleph_1$ in some models of set theory, as large as $\mathfrak c$ in other models, and in between in still other models. If you intersect strictly fewer than this many sets of measure 1, the intersection will have measure 1. Whether that fact covers any uncountable famlilies of sets is, as indicated above, not decided by ZFC.


Take $[0,1]$ with Lebesgue measure, let $X$ be any subset of $[0,1]$, and for each $x \in X$ let $E_x =[0,1] - \{x\}$. Then each $E_x$ has full measure, but their intersection is the complement of $X$, which could be anything.

However ... would this help? Say that $A$ almost contains $B$ if $B - A$ is null.

Theorem. Let $\{E_i: i \in I\}$ be any (possibly uncountable) family of measurable subsets of a probability space. Then there is a measurable subset $E$ of the space such that (1) $E$ is almost contained in each $E_i$ and (2) $E$ almost contains any measurable set $E'$ which is almost contained in each $E_i$. The set $E$ is unique up to modification on a null set.

In other words, the family of measurable sets modulo null sets is a complete lattice.

Actually, this is true for any $\sigma$-finite measure space. To prove it in the finite measure case, let $\alpha$ be the supremum of $\mu(E)$ with $E$ ranging over those measurable subsets which are almost contained in every $E_i$. Then choose a sequence of such sets $E_n$ with $\mu(E_n) \to \alpha$, and take their union.

  • $\begingroup$ I guess your theorem allows us to conclude that $\cap_{i \in \mathcal{I}} E_i=E$?, but doesn't provide us with the measure of set E? $\endgroup$ – gmravi2003 May 9 '12 at 2:20
  • $\begingroup$ Try reading it again. $\endgroup$ – Nik Weaver May 9 '12 at 16:32

The question is asked in a generality that makes it difficult to say any positive results. Part of the problem is that, even for Lebesgue measure on the interval ${}[0,1]$, the usual axioms of set theory do not suffice to give a full answer.

For example, if CH holds, then even if we restrict our attention to intersections indexed by $\omega_1$, the smallest uncountable cardinal, any set of reals can appear this way. In fact, we only need to consider sets $E_i$ of the form ${}[0,1]$, or ${}[0,1]\setminus\{x\}$. Even if CH fails, the same idea shows that we do not really want the intersections to have "too many indices", i.e., we want $|{\mathcal I}|<|{\mathbb R}|$.

Now, it is a consequence of the very useful Martin's axiom that every ${\mathcal I}$-intersection of measurable subsets of ${}[0,1]$ is measurable, as long as $|{\mathcal I}|<|{\mathbb R}|$. (On the other hand, it is consistent that CH fails and yet there are intersections of size $\omega_1$ that are not measurable, so one really needs to appeal to additional axioms to get such a general positive statement.)

The appropriate context for this problem seems to be the theory of cardinal invariants of the continuum, see Joel's very nice answer here; the relevant invariant being the additivity of Lebesgue measure. (That Martin's axiom gives the result above is because all (classical) cardinal invariants have size $|{\mathbb R}|$ under Martin's axiom.) Andreas Blass and Tomek Bartoszynski wrote excellent introductions to this topic for the Handbook of Set Theory and elsewhere, you may want to take a look at their papers. (And I see Andreas just posted an answer while I was writing this.)

  1. 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".)

  2. 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.

  3. 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.

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    $\begingroup$ Since Martin is too modest to advertise his own papers, I'll do it for him (and advertise Shoenfield absoluteness too). Result 3 is in "An application of Shoenfield's absoluteness theorem to the theory of uniform distribution" [Monatshefte für Mathematik 116 (1993) pp. 237-243]. $\endgroup$ – Andreas Blass Dec 9 '12 at 0:55
  • $\begingroup$ Thank you, Andreas! Modesty.. or sloppiness... whatever. I also forgot to mention that the classical result that all (co)analytic are measurable is due to Lusin 1917. (See the textbooks by Moschovakis, 2H8, or Kechris, 21.10) My paper can also be found here: arxiv.org/abs/math/9308201 $\endgroup$ – Goldstern Dec 9 '12 at 13:04

Not that I know of necessarily , other than calculating it . To see how this can be the case , consider the complement of $E_i$ . It has measure 0 .Now the union of the complements (or the complement of $\cap_{i\in \mathcal{I}} E_i$) can be a very general set considering we have uncountable $i$'s . In particular , for $[0,1]$ with the usual topology , it can be an any subset . To see this , let $S$ be a subset of $[0,1]$ , such that the complement of $S$ , $C_S$ is uncountable , and let $b :[0,1] \rightarrow C_S $ be a bijection . Define $E_i$ , $i \in [0,1] $ , to be $[0,1] - \{ b(i) \} $ . Now $\cap_{i\in [0,1] } E_i$ is $S$ . Notice that if $\Omega$ is countable then you can not obtain non-measurable sets by intersecting $E_i$ . The general condition for this to happen is that the union of an uncountable ($R$ , the cardinality of the continuum ) number of the subsets with measure $0$ has measure $0$ .If this doesn't happen , there will be $E_i$'s such that the intersection is a non-measurable set . In general , for a triplet $(\Omega, \mathcal{F},\mathbb{P})$ , if there exists an uncountable subset $J \subseteq F $ , such that the elements of $J$ are pairwise disjoint , $\cup J= \Omega $ and the probability space $(J,M(J) ,\mathbb{P})$ (where $M(J)$ is the set of measurable subsets of the $powerset$ of $J$ )is isomorphic to $([0,1],M([0,1]) , \mathbb{P} _{ [ 0,1 ] } )$ , then there exist $E_i$ for $(\Omega, \mathcal{F},\mathbb{P})$ such that the union is non-measurable . In general , the condition for non-masurable unions of such sets not to exist for $(\Omega, \mathcal{F},\mathbb{P})$ is that $P(E_s)$ = 1 , where $E_s$ is the set of elementary events with non-zero probability .


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