# Absolute Continuity of Signed Measure in terms of Jordan Decomposition

## 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 $\mu$.

$\blacksquare$