Absolute Continuity of Signed Measure in terms of Jordan Decomposition
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$.
Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if:
- $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
Proof
We have that $\nu$ is absolutely continuous with respect to $\mu$ if and only if:
- $\size \nu$ is absolutely continuous with respect to $\mu$
where $\size \nu$ is the variation of $\nu$.
From the definition of variation, we have:
- $\size \nu = \nu^+ + \nu^-$
Suppose that $\nu$ is absolutely continuous with respect to $\mu$.
Then:
- $\nu^+ + \nu^-$ absolutely continuous with respect to $\mu$.
That is:
- for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$.
Since $\nu^+ \ge 0$ and $\nu^-$, we have that:
- $\map {\nu^+} A + \map {\nu^-} A = 0$ implies that $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
- for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$
so:
- $\nu^+$ is absolutely continuous with respect to $\mu$.
We also have:
- for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^-} A = 0$
so:
- $\nu^-$ is absolutely continuous with respect to $\mu$.
So:
- if $\nu$ is absolutely continuous with respect to $\mu$ then $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
$\Box$
Suppose that:
- $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
Then:
- for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
- for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$
giving:
- $\nu^+ + \nu^-$ is absolutely continuous with respect to $\mu$.
That is:
- $\size \nu$ is absolutely continuous with respect to $\mu$.
So:
- if $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$ then $\nu$ is absolutely continuous with respect to $\mu$.
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $4.2$: Absolute Continuity