Integral with respect to Kernel Transformation of Measure

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $N: X \times \Sigma \to \overline \R_{\ge 0}$ be a kernel.

Let $f: X \to \overline \R$ be a positive measurable function.

Then:

$\ds \int f \map \rd {\mu N} = \int N f \rd \mu$

where:

The integral sign denotes integration with respect to a measure

$\mu N$ is the transformation of $\mu$ by $N$

$N f$ is the transformation of $f$ by $N$

Writing $\map \mu f$ in place of $\ds \int f \rd \mu$, the theorem statement can be conveniently expressed as:

$\map {\mu N} f = \map \mu {N f}$

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

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 9$: Problem $11 \ \text{(iii)}$