A transformation $T$ being $\mu$-invariant is by definition a transformation satisfying $$\mu(T^{-1} E) = \mu(E)$$ for all measurable sets $E$. I was wondering what are sufficient conditions for being able to conclude that $$\mu(T E)=\mu(E).$$ From an answer to my earlier question (see #3), it seems that $T$ being bijective and $T^{-1}$ being measure preserving are sufficient conditions?
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Sufficient condition
As you said: if the inverse $S=T^{-1}$ exists and $S$ is measure preserving, then $$\mu(TE) = \mu(S^{-1} E )= \mu(E)$$ so $T$ is forward-measure-preserving.
Necessary conditions
For $T$ to be forward-measure-preserving, it must be
- "measure-wise injective", meaning that for every two measurable sets $A,B$, $$\mu(A\cap B)=0\implies \mu(TA\cap TB)=0$$
- "measure-wise surjective", meaning that the complement of the range of $T$ has zero measure.
The surjectivity part is clear since the inverse image of the complement of the range is empty.
For injectivity, observe that $$\begin{split} \mu(A\cap B)&=\mu(A)+\mu(B)-\mu(A\cup B) \\ & = \mu(TA)+\mu(TB)-\mu(T(A\cup B)) \\ & = \mu(TA)+\mu(TB)-\mu(T A\cup TB) \\ & = \mu(T A\cap TB) \end{split}$$
Conclusion
Since we don't really care about sets of measure zero, there is not much gap between the sufficient and necessary conditions. Which makes the condition of being forward-measure-preserving pretty boring: we just have the usual measure-preserving property, stated for the inverse.