Permutation of binary expansion on (0,1)

Question

Suppose $n_1, n_2, n_3...$ is a fixed permutation of positive intergers.

For any $x\in(0,1)$, the binary expansion of $x$ is $0.x_1x_2x_3....$. Define a mapping $M$ such that $Mx = 0.x_{n_1}x_{n_2}....$.

Can we show that such a mapping is Borel measurable?'

I started with toy examples (1/2,1/4) and tried to see the inverse image of this interval. However, then I get lost since I would like to show that the inverse image is in fact an open interval. To me it seems more like a collection of isolated point then open intervals/sets.

Thanks

Answer 1

Given a number $u=0.u_1u_2u_3,\ldots,$ it is enough to show $\{x:Mx> u\}$ is a borel set. Let $\pi$ be the given permutation determining $M$. Then \begin{align} \{x:Mx\ge u\} &= \cup_{j=1}^\infty(\cap_{k=1}^{j-1}\{x : x_{\pi^{-1}(k)}\ge u_k\}\cap\{x : x_{\pi^{-1}(j)}>u_j\}). \end{align} The terms of the form $\{x : x_{\pi^{-1}(k)}\ge u_k\}$ are measurable, being a union of finite intervals, and so by the usual closure properties $\{x:Mx\ge u\}$ is measurable.

Answer 2

$x \to x_{n_k}$ is measurable for each k so $x \to \{x_{n_k}\}$ is measurable into $\{0,1\}^{\infty}$ with the product topology. Combine this with the measurable function $\{x_n\} \in \{0,1\}^{\infty} \to .x_1x_2.....$. The latter is in fact continuous.