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 $\mu$ if and only if:

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

$\Box$

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 $\map \nu A = 0$.

$\blacksquare$

Sources

• 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $4.2$: Absolute Continuity