# 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