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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 26J / c / Solution

The <doubling map> preserves normalized <Lebesgue measure> on the unit circle. Each point has exactly two preimages, $x/2$ and $(x+1)/2$ modulo one, and each inverse branch scales lengths by $1/2$.

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 Incoming links: Lebesgue invariance of the doubling map