Put $S=[0,1)$ and $\theta(x)=ax(\modul1)$ for $a-1=1/a$, $a > 0$. Then Lebesgue measure is not invariant under $\theta$. Find a measure $\mu$ on $S$ invariant under $\theta$. Hint: Assume $\mu$ has constant density $m_1$ on $(0,1/a)$ and constant density $m_2$ on $(1/a,1)$.
The set $[\theta\leq t]$ is given by:
$$
[0,t/a]\cup[1/a,1/a+t/a]
$$
if $t\leq a-1=1/a$ and by
$$
[0,t/a]\cup[1/a,1]
$$
if $t\geq 1/a$. Thus its measure is
$$
ta^{-1}m_1+ta^{-1}m_2
\quad\mbox{and}\quad
ta^{-1}m_1+(1-a^{-1})m_2
$$
This gives us the equations:
$$
(a^{-1}-1)m_1+a^{-1}m_2=0
\quad\mbox{and}\quad
-a^{-1}m_1+m_2=0
$$
Which has a non trivial solution: $m_1=a$, $m_2=1$.