Invariant Measure of Image under Bijection

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

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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