MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A\subset\{0,1\}^\omega$ be a measurable set (w.r.t. the usual borel sigma algebra) which does not depend on any even coordinate (that is, if $x\in A$ and $x$ and $y$ agree except on a finite number of even coordinates, then $y\in A$).

Is it true that $A$ belongs to the sigma-algebra generated by all the odd coordinates + the tail sigma algebra?

To clarify: the tail sigma algebra consists of all the events which do not depend on any coordinate.

It seems to me that this should be some easy/well known measure theory fact/counterexample, but perhaps I'm wrong? Suggestions on where to look for an answer would be welcome.

Note that it is well known the this statement is false if the ground space would be $[0,1]^\omega$.

share|cite|improve this question
Could you clarify which sigma-algebra you mean? For example, what exactly do you mean by the "tail sigman algebra"? Also, by "measurable in the sigma-algebra", do you just mean that it should be an element of that algebra? And when you say that A is measurable, do you mean with respect to the usual probability measure on Cantor space? – Joel David Hamkins Dec 8 '09 at 19:22
edited the question for clarifications. Regarding your last question: this question is about sigma algebras, not about measures, I want to know whether this set belong to that sigma-algebra. – Ori Gurel-Gurevich Dec 8 '09 at 19:50
Sorry, I'm still not clear on what you mean by the tail sigma algebra. Do you mean the sets that are closed under all finite modifications? – Joel David Hamkins Dec 12 '09 at 3:50
Yes, all borel sets invariant under changing a finite number of coordinates. Anyway, in the meantime I was given a counterexample by a colleague, which I (or he) will put here as an answer. – Ori Gurel-Gurevich Dec 12 '09 at 6:14
I'd be interested in seeing your counterexample, if you're going to post it sometime. – George Lowther Dec 15 '09 at 1:21

Write $\{0,1\}^\mathbb{N}$ as cartesian product $\{0,1\}^E \times \{0,1\}^O$, where $E$ is the evens and $O$ is the odds. Your usual sigma-algebra is the product sigma-algebra that goes with this product.

So we need something like this: Write $\otimes$ for product sigma-algebra. Then is $$\mathcal{F}\otimes\left(\bigcap_{k=1}^\infty\mathcal{G}\_k\right) = \bigcap_{k=1}^\infty\left(\mathcal{F}\otimes\mathcal{G}\_k\right)$$

share|cite|improve this answer
Unfortunately, it is not true that the this sigma algebra is generated by the odd coordinates + the even tail, you do need the regular tail. – Ori Gurel-Gurevich Dec 8 '09 at 16:08
Did you mean to say "even tail" in the original question? – Alon Amit Dec 8 '09 at 19:13
No, as I wrote in the comment above, this is not true. The event ''$x_{2n}=x_{2n+1}$ for all but finitely many $n$'' is not generated by the odd coordinates + even tail. – Ori Gurel-Gurevich Dec 8 '09 at 19:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.