Convergence of Constant Measure Sets in Finite Measure Spaces?

[SpaceTag]

This question came up recently, and I'm wondering whether or not it's true:

Let (X,A,m) be a finite measure space. Let E_1,E_2,... be a sequence of measurable subsets of X of constant positive measure (i.e., there exists c>0 such that m(E_i) = c for all i). Then there exists a subsequence of the sequence E_1,E_2,... whose intersection has positive measure.

Any ideas?

 

[Billy Bob]

Got it! It's false.

I have an example in X=[0,1] (with ordinary Lebesgue measure) of a sequence of sets E_n all of measure 1/2, with the property that the intersection of any subsequence has measure zero.

Now that you know which way to go (F vs. T), do you want me to spoil your fun, or do you want to keep looking? :)

 

[SpaceTag]

Spoil my fun, please! This has been annoying me for too long, haha.

 

[Billy Bob]

$$E_n = \{ x\in [0,1] : \sin(2^n\pi x)\ge0\}$$

Equivalently, $$E_n$$ is the set of real numbers in [0,1] that have a binary expansion with 0 in the nth position.

$$E_n$$ is the disjoint union of $$2^{n-1}$$ intervals, each of length $$1/2^n$$

Use the binary expansion idea to prove the intersection of any subsequence has measure zero.

 

[SpaceTag]

Brilliant! Thanks a lot.