Characterization of Absolute Continuity of Complex Measure

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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 complex 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 $\tuple {\nu_1, \nu_2, \nu_3, \nu_4}$ 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 Complex 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 Complex Measure in terms of Jordan Decomposition, we have:

$\nu_1$, $\nu_2$, $\nu_3$ and $\nu_4$ are absolutely continuous with respect to $\mu$.

So:

for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu_1} A = \map {\nu_2} A = \map {\nu_3} A = \map {\nu_4} A = 0$

From the definition of the Jordan decomposition, this implies:

$\map \nu A = \map {\nu_1} A - \map {\nu_2} A + i \paren {\map {\nu_3} A - \map {\nu_4} A} = 0$

So:

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

$\blacksquare$


Sources

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