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.
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1$\begingroup$ Suppose that $S = \{a, b\}$, with $\mu(a) = 0$ and $\mu(b) = 1$. Then the constant function at $b$ is measure-preserving, but doesn't satisfy your definition. $\endgroup$– LSpiceCommented Mar 9, 2018 at 19:45
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2$\begingroup$ For one thing, $\mu(\phi(A))$ isn't even defined in general. $\endgroup$– Johannes HahnCommented Mar 9, 2018 at 21:19
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$\begingroup$ See also this Math.Stackexchange thread: math.stackexchange.com/questions/1768257/… $\endgroup$– D. ThomineCommented Mar 9, 2018 at 21:31
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1$\begingroup$ When $\phi$ is bijective, your definition is fine. But if $\phi$ is not injective it is not what you want. A nice measure-preserving map is $x \mapsto (2x \;\mathrm{ mod }\; 1)$ on $[0,1)$ with Lebesgue measure. This satisfies the correct definition, but not yours. $\endgroup$– Gerald EdgarCommented Mar 9, 2018 at 22:27
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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 $\phi(A)=\{1\}\notin\mathcal{S}$.
answered Mar 9, 2018 at 20:18
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2$\begingroup$ This defect is easy to correct, though. In localizable measurable spaces (which include all σ-finite measurable spaces) one can easily define the image of a measurable set, which will be an equivalence class of measurable sets up to a set of measure 0. $\endgroup$ Commented Mar 11, 2018 at 0:57