Measure-preserving transformation

Remarks

Condition 2 says that the pushforward measure $T_* \mu$ equals $\nu$. When $X = Y$ and $\mu = \nu$, one says that $T$ preserves $\mu$, and $(X, \mathcal{A}, \mu, T)$ is a measure-preserving system. Example: the doubling map $T(x) = 2x \bmod 1$ on $[0,1)$ with Lebesgue measure $\lambda$. For $0 \le a < b \le 1$ one has $T^{-1}([a, b)) = [a/2, b/2) \cup [(a+1)/2, (b+1)/2)$, of measure $b - a$. The intervals form a $\pi$-system that generates the Borel sets, so condition 2 holds for all Borel sets. Non-example: $T(x) = x^2$ on $[0,1]$ with Lebesgue measure is measurable but $\lambda(T^{-1}([0, 1/4])) = \lambda([0, 1/2]) = 1/2 \ne 1/4$.