Example of non-measurable sets on $\mathbb R^{\mathbb N}$ For Kolmogorov extension theorem, a probability measure $\mu$ is constructed on $\mathbb R^{\mathbb N}$ in the following way:
Firstly, we define $\mu$ on collection of cylindrical sets $\mathscr A_0$, by $\mu(A \times \mathbb R^{\mathbb N})=\mu_n(A)$, where $\mu_n$ are probability measures on $(\mathbb R^n, \mathcal B_n)$, $\mu_n$ are consistent, and $A \in \mathcal B_n$. $\mathscr A_0$ is an algebra and $\mu$ is finitely additive on $\mathscr A_0$.
Then, we prove by contradiction that $\mu$ is also sigma-additive on $\mathscr A_0$. And eventually we extend $\mu$ to $\sigma(\mathscr A_0)$ by Caratheodory theorem.
The above proof is understandable, but I feel I gain very little intuition on $\sigma(\mathscr A_0)$. So my question is that could we have an example of non-measurable set in $\mathbb R^{\mathbb N}$ for this $\mu$?
 A: We can give an example of a non-measurable set, but maybe you find this a boring example.
Consider a set $C\subset\mathbb R$ that is not measurable (w.r.t. the Borel $\sigma$-algebra $\mathcal B_1$).
Then the set $C\times \mathbb R^{\mathbb N}\subset \mathbb R^{\mathbb N}$ is not measurable w.r.t. the $\sigma$-algebra $\sigma(\mathscr A_0)$.
sketch of proof:
Consider the function $f:\mathbb R\to\mathbb R^{\mathbb N}$ given by $x \mapsto (x,0,0,\dots)$.
Next, we will show that $f$ is measurable.
In order to show that $f$ is measurable it suffices to check if all preimages of the generating sets 
$A\times\mathbb{R}^{\mathbb N}\in \mathscr A_0$ (with $A\in\mathcal B_n$) 
are measurable (this is a nice property of $\sigma$-algebras that are generated by sets).
Let $A\times\mathbb{R}^{\mathbb N}\in \mathscr A_0$.
Then we have
$$
f^{-1}(A\times\mathbb{R}^{\mathbb N})
= \{x\in\mathbb R: (x,0,\ldots,0)\in A\subset \mathbb R^n\}
= g_n^{-1}(A),
$$
where $g_n:\mathbb R\to\mathbb R^n,\;x\mapsto (x,0,\ldots,0)$ is a continuous (and therefore measurable) function.
It follows that $f^{-1}(A\times\mathbb{R}^{\mathbb N})$ is measurable.
Therefore $f$ is measurable.
Then we note that 
$$
f^{-1}(C\times \mathbb R^{\mathbb N}) = C.
$$
So if $C\times \mathbb R^{\mathbb N}$ was measurable, then $C$ would be measurable, which is a contradiction.
