2

It is well known that a non-Lebesgue measurable, and hence non Borel measurable subset of $[0,1]$ exists.

However, if I consider the set $\Omega=\{0,1\}^{\mathbb N}$ in the infinite independent coin tossing example, with the $\sigma$-algebra generated by cylinder sets and the product measure $\mu^{\mathbb N}$ where $\mu(\{0\})=p,\mu(\{1\})=1-p$ for $p\in(0,1)$, do there exist non-measurable sets?

Any references/answers are welcome.

One thing I thought is that $|\{0,1\}^{\mathbb N}|=|\mathbb R|$ so these two spaces are bijective, so if there is some non-measurable set in $\mathbb R$ then I may have a non-measurable set in $\{0,1\}^{\mathbb N}$. This may be completely wrong, though.

Martin Sleziak
  • 53,687
Landon Carter
  • 12,906
  • 4
  • 32
  • 79
  • 1
    This is not my field, but I'm pretty sure that the mapping $\Omega \to [0,1]$ which takes $\alpha$ to $(1-p)\sum_{i = 0}^\infty \alpha_i/p^i$ preserves everything about these spaces "up to a set of measure 0". The inverse image of a non-measurable set in $[0,1]$ is then one in $\Omega$. – Mees de Vries Sep 13 '16 at 14:59
  • http://math.stackexchange.com/questions/1504887/why-is-it-that-mathscrf-ne-2-omega – BCLC Sep 27 '16 at 14:32

2 Answers2

1

In this link you will find a proof that if $\mathcal{B}$ is a countably generated $\sigma$-algebra (that is the $\sigma$-algebra generated by countably many elements), then either $\Omega$ is finite or $|\Omega|=2^{\aleph_0}$.

Cardinality of Borel sigma algebra

Notice that the product measure $\mu^{\mathbb{N}}$ is generated by the sets $X=\prod_{n\in\mathbb{N}}A_n$ where $A_n\subseteq \{0,1\}$ and $A_n=\{0,1\}$ for all but finitely many $n\in\mathbb{N}$. These are countably many elements.

So, the product sigma-algebra $\mathcal{B}$ on $\Omega$ has $2^{\aleph_0}$ sets, but there are $2^{|\Omega|}=2^{|2^\mathbb{N}|}=2^{2^{\aleph_0}}$ subsets of $\Omega$, so there has to be subsets of $\Omega$ that are not $\mu^{\mathbb{N}}$-measurable.

Community
  • 1
Darío G
  • 4,878
0

When $p=1/2$ you can adapt the standard Vitali argument for Lebesgue non-measurable sets to show that there are non-measurable sets in $\{0,1\}^\mathbb{N}$. Where in the original argument we use translation by rationals, in this case we use flips ($0 \to 1$, $1 \to 0$) at some finite set of indices. With $p=1/2$ these flips are measure preserving and the Vitali argument goes through. I wrote up the details as a blog post: https://www.jyotirmoy.net/posts/2020-06-20-nonmeasurable-fair-coin.html

With $p=1$ or $p=0$ we can make all sets measurable.

I don't know the answer for other values of $p$.

Jyotirmoy Bhattacharya
  • 5,518