Let $T:[0,1]\to[0,1]$ be such that $T(x)=3x\pmod{1}$ which is measurable with respect to the $\sigma$-algebra of Borel on $[0,1]$, which we denote by $\mathscr{B}_{[0,1]}$. Prove that Lebesgue measure on $[0,1]$ is T-invariant.
I know that if $[a,b]\in T(x)$ then as the measure is invariant $T^{-1}([a,b])=I_1\cup I_2\cup I_3$. However I do not know how to determine the intervals.
Question: Can someone explain me how should I work with the inverse of $T$?
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + \frac{1}{3}$, and there's $u_3 = u_1 + \frac{2}{3}$
More generally, $$ T^{-1} ([a, b]) = [\frac{a}{3}, \frac{b}{3}] \cup [\frac{a+1}{3}, \frac{b+1}{3}] \cup [\frac{a+2}{3}, \frac{b+2}{3}]. $$
With that in mind, perhaps you can make some progress.