Kernel Transformation of Measure is Measure

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

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

Then $\mu N: X \to \overline \R$, the kernel transformation of $\mu$, is a measure.

Proof
From the definition of the kernel transformation of $\mu$, we have:


 * $\ds \map {\paren {\mu N} } E = \int \map N {x, E} \rd \map \mu x$

for each $E \in \Sigma$.

We verify each of the conditions for a measure in turn.

Proof of $(1)$
Let $E \in \Sigma$.

From the definition of a kernel, we have that:


 * $x \mapsto \map N {x, E}$ is a positive $\Sigma$-measurable function.

From the definition of the $\mu$-integral of a positive $\Sigma$-measurable function, we have:


 * $\ds \int \map N {x, E} \rd \map \mu x \ge 0$

So:


 * $\map {\paren {\mu N} } E \ge 0$

for each $E \in \Sigma$, verifying $(1)$.

Proof of $(2)$
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets.

Then:

So $\mu N$ is countably additive, and we have $(2)$.

Proof of $(3)$
We have:

So there exists $E \in \Sigma$ such that:


 * $\map {\paren {\mu N} } E \in \R$

verifying $(3)$.