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