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.
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.