Show function is measure preserving Given $\Theta:[0,1]\to [0,1]$ of the following form, if we express $x = (x_1,x_2,x_3,x_4,...)$ in binary where $x\in [0,1]$, then $$\Theta(x)=(x_2,x_1,x_4,x_3,...)$$
I.e. $\Theta$ transposes the binary digits of $x$. I want to show that $\Theta$ is measure preserving. My idea was to show that it preserves (Borel) measure on some simple generating set of the Borel algebra. I.e. intervals of form $(0,t)$. But in practice this seems extremely hard, and I am finding it very hard to visualize how $\Theta$ behaves on open intervals. 
 A: For a dyadic interval of the form
$$I_{a}=\left(\sum_{k=1}^n a_i2^{-i}, \sum_{k=1}^n a_i2^{-i} +2^{-n}\right)=\left\{x=(a_1,..a_n,x_{n+1},x_{n+2},...) : x_i \in \{0,1\} \}\right\},$$
where $a=(a_i)=(a_1,a_2...,a_n) \in \{0,1\}^n$, then 
$$\Theta^{-1}(I_a)=I_b,$$ 
for $b=(a_2,a_1,a_3,a_4,...,a_{n},a_{n-1})$, provided $n$ is even.
Since $\mathcal{Leb}(I_a)=2^{-n}=\mathcal{Leb}(I_b)$, this proves that $\Theta_*\mathcal{Leb}$ coincide with $\Theta$ on the set of dyadic intervals. 
Now notice that the Borel set on which these two measure coincide is a monotone class, and by the previous equality this monotone class contains the (not sigma !) algebra generated by dyadic intervals (this is just finite unions of dyadic intervals). By the monotone class theorem (see https://en.wikipedia.org/wiki/Monotone_class_theorem), they coincide on the smallest $\sigma$-algebra containing dyadic intervals, so the full Borel $\sigma$-algebra.
A: $\Theta_1=(x_2,x_1,x_4,x_3,x_5,\cdots ),\ \Theta_2 =
(\cdots,x_6,x_5,x_8,x_7,\cdots )$ so that $\Theta=\cdots \Theta_2\circ \Theta_1$ is measure preserving, since $\Theta_i$ is measure preserving.
