Kernel Transformation of Measure is Measure

From ProofWiki
Jump to navigation Jump to search

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 $(1)$.

$\Box$

Proof of $(2)$

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

Then:

\(\ds \map {\paren {\mu N} } {\bigcup_{n \mathop = 1}^\infty E_n}\) \(=\) \(\ds \int \map N {x, \bigcup_{n \mathop = 1}^\infty E_n} \rd \map \mu x\)
\(\ds \) \(=\) \(\ds \int \paren {\sum_{n \mathop = 1}^\infty \map N {x, E_n} } \rd \map \mu x\) since for each $x \in X$, $E \mapsto \map N {x, E}$ is a measure, we use countable additivity in the second argument
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\int \map N {x, E_n} \rd \map \mu x}\) Integral of Series of Positive Measurable Functions
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\paren {\mu N} } {E_n}\)

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

$\Box$

Proof of $(3)$

We have:

\(\ds \map {\paren {\mu N} } \O\) \(=\) \(\ds \int \map N {x, \O} \rd \map \mu x\)
\(\ds \) \(=\) \(\ds \int 0 \rd \map \mu x\) Empty Set is Null Set
\(\ds \) \(=\) \(\ds 0\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral
\(\ds \) \(\in\) \(\ds \R\)

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

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

verifying $(3)$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Kernel_Transformation_of_Measure_is_Measure&oldid=564156"