Invariant Measure of Image under Bijection
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\theta: X \to X$ be an $\Sigma / \Sigma$-measurable bijection.
Let $\mu$ be an invariant measure.
Then:
- $\forall A \subseteq X: \map \mu {\theta \sqbrk A} = \map \mu A$
Proof
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Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): invariant measure
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): invariant measure