Given measure space $(S, \mathcal{S}, \mu)$, and measurable function $\phi: S \to S$. $\phi$ is measure-preserving if $\forall A \in \mathcal{S}, \mu(A) = \mu(\phi^{-1}(A))$. My confusion is that why we do not define measure-preserving as $\forall A \in \mathcal{S}, \mu(\phi(A)) = \mu(A)$? It seems more natural to me and I have not found any inconsistency with this definition.



Understanding measure-preserving transformation
As pointed out by user LSpice, your definition would be different from the accepted one. 


However, it is not a well-constructed definition at all. Indeed, it is possible to have a situation when a function $\phi\colon S \to S$ is $\mathcal{S}$-measurable and $A \in \mathcal{S}$, but $\phi(A)\notin\mathcal{S}$ and hence $\mu(\phi(A))$ has no meaning. For a simplest example, suppose that $S=\{1,2\}$, $\mathcal S=\{\emptyset,S\}$, $\phi(1)=\phi(2)=1$, and $A=S$. Then $\phi$ is $\mathcal{S}$-measurable and $A \in \mathcal{S}$, but $\phi(A)=\{1\}\notin\mathcal{S}$. 


