Do non-measurable sets exist for $\{0,1\}^{\mathbb N}$? 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.
 A: 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.
A: 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 $p$.
