Question

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 $[0,1]^\omega$.

Answer 1

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 $$\mathcal{F}\otimes\left(\bigcap_{k=1}^\infty\mathcal{G}_k\right) = \bigcap_{k=1}^\infty\left(\mathcal{F}\otimes\mathcal{G}_k\right)$$