Characterization of Absolute Continuity of Signed Measure

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 $\struct {X, \Sigma}$.

Then $\nu$ is absolutely continuous with respect to $\mu$ :


 * for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.

Proof
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$.

Let $\size \nu$ be the variation of $\nu$.

Sufficient Condition
Suppose that:


 * for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.

We aim to show that:


 * for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$

which will give:


 * $\size \nu$ is absolutely continuous with respect to $\mu$

from which we will obtain:


 * $\nu$ is absolutely continuous with respect to $\mu$.

Suppose that $A \in \Sigma$ has $\map \mu A = 0$.

From Null Sets Closed under Subset, we have:


 * $\map \mu B = 0$ for each $\Sigma$-measurable $B \subseteq A$.

Using the assumption on each such $B$, we have:


 * for each $\Sigma$-measurable $B \subseteq A$ we have $\map \nu B = 0$.

From Characterization of Null Sets of Variation of Signed Measure, this implies that:


 * $\map {\size \nu} A = 0$

So:


 * for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\size \nu} A = 0$

Necessary Condition
Suppose that $\nu$ is absolutely continuous with respect to $\mu$.

Then from Absolute Continuity of Signed Measure in terms of Jordan Decomposition, we have:


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

So:


 * for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = \map {\nu^-} A = 0$

From the definition of the Jordan decomposition, this implies:


 * $\map \nu A = \map {\nu^+} A - \map {\nu^-} A = 0$

So:


 * for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map \nu A = 0$.