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



Sources