# Integral with respect to Kernel Transformation of Measure

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## Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $N: X \times \Sigma \to \overline{\R}_{\ge0}$ be a kernel.

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

Then:

- $\displaystyle \int f \, \mathrm d\mu N = \int N f \, \mathrm d\mu$

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 $\mu \left({f}\right)$ in place of $\displaystyle \int f \, \mathrm d\mu$, the theorem statement can be conveniently expressed as:

- $\mu N \left({f}\right) = \mu \left({N f}\right)$

## Proof

## Sources

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