Ergodic Measure Preserving Transformations on Compact Metric Spaces

Question

Peter Walters. An Introduction to Ergodic Theory. Chapter $6$. Page $152 - 153$. Theorem $6.10$. The fourth statement $(iv)$.

If $\mu_1<<m$ then $\mu_1\circ T^{-1}<<m\circ T^{-1}$, why ?

If $\mu_2$ and $m$ are mutually singular then $\mu_2\circ T^{-1}$ and $m\circ T^{-1}$ are mutually singular, why ?

Answer 1

We just apply the property of $T$ to be measure preserving over and over:
If $\mu$ and $m$ are mututally singular, we have $\mathcal{A} , \mathcal{B} \subset X$ measurable, such that for all $A \in \mathcal{A} , B \in \mathcal{B}$ $$\mu(B)=0 \text{ and } m(A)=0.$$Therefore, for every $A \in \mathcal{A}$ we have $$\big( m\circ T^{-1} \big)(A) = m \big( T^{-1} (A) \big) = m(A) = 0$$and likewise for $\mu$.

If $\mu \ll m$ then $m(A)=0$ implies $\mu(A)=0$. Therefore take $A \subset X$ measurable with $\big( m \circ T^{-1} \big)(A)=0$, then we have $$m(A) = m\big(T^{-1} (A) \big) = \big( m \circ T^{-1} \big) (A) =0 ,$$ therefore $\mu(A)=0$. Hence $\mu \big( T^{-1} (A) \big)=0$.