A measure theory problem Consider Lebesgue measure $\lambda$ restricted to the class $\mathscr B$ of Borel sets in $(0,1]$. For a fixed permutation $n_1, n_2, \dots$ of the positive integers, if $x$ has dyadic expansion $.x_1, x_2, \dots,$ take $Tx = .x_{n_1}x_{n_2}\dots$. Show that $T$ is measurable $\mathscr{B / B}$. 
Hint Enough. Not clear how to handle permuation ?
 A: Fix any $k\in\mathbb N$ and define $$T_k(x)=\mathord.\,x_{n_1}\ldots\, x_{n_k}\quad\forall x\in(0,1].$$
I leave it to you to check that this function is measurable. Then, check that for each $x\in(0,1]$, $\lim_{k\to\infty} T_k(x)=T(x)$ pointwise and use the fact that the pointwise limit of measurable functions defines a measurable function.

Hint: $T_k$ is the sum of the functions $x\mapsto \mathord.\, x_{n_1}$, $x\mapsto\mathord.\,0x_{n_2}$, $x\mapsto\mathord. 00 x_{n_3}$, and so forth. The measurability of these functions may be easier to check—these are actually simple functions. For example, if $n_3=17$, say, then all you need to do is check the $17$th digit of $x$. If it is $1$, then the value of the function $x\mapsto \mathord.\,00x_{n_3}$ is $\mathord.\,001=1/8$, and if the $17$th digit is $0$, then the value of the function is $0$. The $17$th digit of $x$ defines a partition consisting of $2^{18}$ disjoint subintervals of $(0,1]$, and the function $x\mapsto \mathord.\,00x_{n_3}$ assumes the value of $1/8$ on half of these subintervals and $0$ on the other half of them.
A: Each $x\in (0,1)$ has a unique dyadic expansion $.x_1 x_2 \dots $ that doesn't end in all $1$'s. Using only these expansions, the map $f_k(.x_1x_2\dots ) = x_k$ is well defined on $(0,1).$
Now it's easy to see that if $F$ is the floor function and $g:(0,1) \to \mathbb {R}$ is measurable (measurable = Borel measurable), then $F\circ g$ is measurable. Thus $f_1(x) = F(2x)$ is measurable. Supposing $f_1,\dots ,f_n$ are measurable, we have
$$f_{n+1}(x) = F(2^{n+1}x - (2^nf_1(x) + \cdots +2f_n(x))).$$
Hence $f_{n+1}$ is measurable. By induction, all $f_n$ are measurable. Thus the map 
$$T(x) = \sum_{k=1}^\infty \frac{f_{n_k}(x)}{2^k}$$
is measurable.
