# Integral with respect to Kernel Transformation of Measure

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## 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

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## Sources

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