Write $[0,1]^\mathbb{N}$ as a disjoint union of two measurable subsets such that non of which contain a cylinder.

Question

Consider the space $X=[0,1]^\mathbb{N}$ endowed with the Borel $\sigma$-algebra (with respect to the product topology). I'd like to think of $X$ as the set of all sequences with values in $[0,1]$.

My Question: Is it possible to split $X$ into a disjoint union of two measurable subsets $X=\mathcal{A}_1\sqcup\mathcal{A}_2$ such that no cylinder of $X$ lies in any of the $\mathcal{A}_i$?

By a cylinder I mean a set of all sequences whose values at some finitely many indices $n_1,...,n_k$ are fixed. (i.e. sets of the form $\{a_n\in X : a_{n_1}=b_1,...,a_{n_k}=b_k\}$ where $b_1,...,b_k\in[0,1]$ are given, for example the set of all the sequences which begin with $\frac{1}{2}$ is a cylinder).

Edit: It turns out I recieved a positive answer in which one of the sets $\mathcal{A}_1$ is of measure zero (with respect to the standard Borel measure on $X$). I wonder if one could generalize the given construction for the case where both $\mathcal{A}_1$ and $\mathcal{A}_2$ are of positive measure.

Answer 1

Let $$A_1 = \{x: x_n = 0\ \text{eventually}\}$$ and $A_2$ its complement.

EDIT: For an example with the additional requirement that both $A_i$ are of positive measure, let $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j \ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.

Answer 2

Let $a_m,b_m \in [0,1]^{\mathbb N}$ be such that $a_m(n)=\delta_{m,n}$ and $b_m(n)=1-\delta_{m,n}=1-a_m(n)$. Then $$ A_1 = \{x\in [0,1]^{\mathbb N}: x_0 \leq 1/2\} \cup \{b_m:m\in\mathbb N\} \setminus \{a_m:m\in\mathbb N\} $$ and $$ A_2 = \{x\in [0,1]^{\mathbb N}: x_0 > 1/2\} \cup \{a_m:m\in\mathbb N\} \setminus \{b_m:m\in\mathbb N\} $$ should do the job.

Moreover the measures are $|A_1|=|A_2|=1/2$.

Edit: I had previously read $\{0,1\}^{\mathbb N}$ instead of $[0,1]^{\mathbb N}$. Now it is fixed.