Showing a Transformation increases measure (Ergodic Theory)

Hoi, ive been breaking my head on this fora few days.. Ive been trying to show that $T:[0,1)\to [0,1)$ given by

$$ T(x)= \begin{cases} 3x & \mbox{ if } x\in [0,1/3)\\ & \\ \frac{3x}{2}-\frac{1}{2} & \mbox{ if } x\in [1/3,1) \end{cases} $$ is Ergodic.

Let B be T-invariant,i.e. $T^{-1}B=B$. If $\mu(B)>0$, I want to show that $\mu(B)=1$.

Can we show that if $\mu(B)<1$ that $\mu(TB)>\mu(B)$,

is that obvious? So that it is clear that B can not T-invariant?

We know nothing of the set B, which may be little complicated set. Any idea how we can show ergodic property of this transformation. (it is measure preserving).

For each k there are partition of $[0,1)$ in intervals $I{'s}$ such that $T^k$ restricted to each $I$ is a bijection on $[0,1)$. In each $I$

$$T^k(x)=ax+c $$

where $a$ and $c$ depends on $k$ and $I$.

Hence for every pair of measurable sets $A$ and $B$ in $I$ we have

$$ \dfrac{m(T^k(A))}{m(T^k(B))}=\dfrac{a^km(A)}{a^km(B)}=\dfrac{m(A)}{m(B)} . $$

Let $B$ be an invariant measurable set with positive measure, then,

$$m(B)\geqslant \dfrac{m(T^k(I\cap B ))}{m(T^k(I))} =\dfrac{m(I\cap B)}{m(I)}.$$

Fix a Lebesgue point $p$ in $B$ (we can choose a such point because $B$ has positive measure). For each $k$ choose a interval $I_k$ how above containing $p$. Since the diameter of the $I_k$ converge to zero and $p$ is Lebesgue point, we conclude that

$$ m(B)\geqslant \lim_{k\to\infty} \dfrac{m(I_k\cap B)}{m(I_k)}=1.$$