Quasi-measure-preserving map

Remarks

Condition 2 says that the pushforward $T_* \mu$ is absolutely continuous with respect to $\nu$, that is, preimages of null sets are null sets. Every measure-preserving map satisfies it. Example: $T(x) = 2x$ on $\R$ with Lebesgue measure $\lambda$. For a Borel set $B$ one has $\lambda(T^{-1}(B)) = \lambda(B)/2$, so null sets pull back to null sets, but $T$ is not measure-preserving. Non-example: the constant map $T(x) = 0$ from $[0,1]$ to $\R$, both with Lebesgue measure, is measurable. But $\lambda(\set{0}) = 0$ while $\lambda(T^{-1}(\set{0})) = \lambda([0,1]) = 1$, so condition 2 fails.