ThmDex – An index of mathematical definitions, results, and conjectures.
Measure of set in backward orbit under measure-preserving endomorphism

Let $M = (X, \mathcal{F}, \mu, T)$ be a D2827: Measure-preserving system such that
 (i) $E \in \mathcal{F}$ is a D1109: Measurable set in $M$
Then $$\forall \, n \in \mathbb{N} : \mu(T^{-n} E) = \mu(E)$$

Let $M = (X, \mathcal{F}, \mu, T)$ be a D2827: Measure-preserving system such that
 (i) $E \in \mathcal{F}$ is a D1109: Measurable set in $M$
Then $$\mu(E) = \mu(T^{-1} E) = \mu(T^{-2} E) = \mu(T^{-3} E) = \cdots$$
Proofs
Proof 0
Let $M = (X, \mathcal{F}, \mu, T)$ be a D2827: Measure-preserving system such that
 (i) $E \in \mathcal{F}$ is a D1109: Measurable set in $M$
Fix a natural number $n \in \mathbb{N}$. We define $T^{-0} E : = E$ and the equality $T^{-1} E = E$ is included in the definition of a measure-preserving endomorphism, so it is left to establish the claim for $n > 1$. Since we define $$T^{-n} : = \underbrace{T^{-1} T^{-1} \cdots T^{-1}}_{n \text{ times}}$$ then, applying R1973: Map composition is associative, we have the recurrence relation $$\begin{split} T^{-n} E = T^{-n + 1 - 1} E = T^{- n + 1} T^{-1} E = T^{- n + 1} E \end{split}$$ By a repeated application of this recurrence relation, we have the equalities $$\begin{split} T^{-n} E & = T^{- n + 1} E \\ T^{-n + 1} E & = T^{- n + 2} E \\ T^{-n + 2} E & = T^{- n + 3} E \\ & \; \; \vdots \\ T^{-3} E & = T^{-2} E \\ T^{-2} E & = T^{-1} E \\ T^{-1} E & = E \\ \end{split}$$ By transitivity of equality, the claim follows. $\square$