Problem
Let $T:[0,1) \to [0,1)$ be defined by $$T(x)= \begin{cases} 2x &\text{if } 0 \leq x< \dfrac{1}{2} \\ 2x-1 & \text{if } \dfrac{1}{2} \leq x <1 \\ \end{cases}$$
Show that if $E \subset [0,1)$, then $T^{-1}(E)$ is measurable and $ |T^{-1}(E)|=|E|$
I am pretty lost on the problem but I'll try to explain what I thought of:
For each point $t$ of the interval $[0,1)$, there are exactly two points $x,y$ which are sent to $t$ under this function, one belonging to $[0,\dfrac{1}{2})$ and the other to $[\dfrac{1}{2},1)$. So if we consider a measurable set $E$ contained in $[0,1)$, we could separate its preimage as $T^{-1}(E)=S_1 \cup S_2$ where $S_1$ belongs to $[0,\dfrac{1}{2})$ and $S_2$ belongs to $[\dfrac{1}{2},1)$. If I could show that these two sets are measurable, then since $T(S_1)=T(S_2)=E$, then we would have $|T(S_i)|=2|S_i|$ and so $|E|=|S_1|+|S_2|=|S_1 \cup S_2|=|T^{-1}(E)|$.
Any help would be greatly appreciated. Thanks in advance.
You can do this in the following steps:
Can you first prove it for an open interval $E = (a,b)$? Here you can explicitly write down $T^{-1}(E)$ as a disjoint union of two intervals, both of which have length $(b-a)/2$.
Now suppose $E$ is any open set, then write it as a countable union of disjoint open sets, and appeal to the fact that $T^{-1}(E_1 \sqcup E_2) = T^{-1}(E_1)\sqcup T^{-1}(E_2)$. Thus, you have proved it for any open set $E$.
Now if $E_n \supset E_{n+1}$ is a decreasing sequence of open subsets of $[0,1)$, then prove that $$ |T^{-1}(\cap E_n)| = \lim |T^{-1}(E_n)| $$ (simply because the map $E \mapsto |T^{-1}(E)|$ defines a measure and $|T^{-1}(E_1)| < \infty$
If $m(E)$ is zero, conclude that $|m(T^{-1}(E)| = 0$ (use part (2))
If $E$ is any measurable set, then write it as $E = F\setminus Z$ where $F$ is a $G_{\delta}$ set and $m(Z) = 0$. Now appeal to (4) and (5).