Solution
= Solution
Take the unit circle $X=\mathbb R/\mathbb Z$ with its <Borel sigma-algebra> and normalized <Lebesgue measure>, and let
$$
T(x)=2x\pmod1.
$$
This is the <doubling map>. Its two inverse branches halve length, so <Lebesgue invariance of the doubling map> gives $\mu(T^{-1}A)=\mu(A)$. Moreover
$$
\boxed{T^{-1}\{x\}=\left\{\frac x2,\frac{x+1}2\right\}\quad\text{and hence}\quad|T^{-1}\{x\}|=2}
$$
for every point of the circle.