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 $\mu$ :


 * $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.

Proof
We have that $\nu$ is absolutely continuous with respect to $\mu$ :


 * $\size \nu$ 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 $\mu$.

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 $\mu$.