# Absolute Continuity of Complex 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 complex measure on $\struct {X, \Sigma}$.

Let $\tuple {\nu_1, \nu_2, \nu_3, \nu_4}$ be the Jordan decomposition of $\nu$.

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

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

## Proof

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

### Sufficient Condition

Suppose that:

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

Then:

- whenever $A \in \Sigma$ has $\map \mu A = 0$, we have $\map {\nu_1} A = 0$, $\map {\nu_2} A = 0$, $\map {\nu_3} A = 0$ and $\map {\nu_4} A = 0$.

From Bound for Variation of Complex Measure in terms of Jordan Decomposition, we have:

- $\map {\cmod \nu} A \le \map {\nu_1} A + \map {\nu_2} A + \map {\nu_3} A + \map {\nu_4} A$

giving:

- $\map {\cmod \nu} A \le 0$

so:

- $\map {\cmod \nu} A = 0$

We therefore have:

- whenever $A \in \Sigma$ has $\map \mu A = 0$, we have $\map {\cmod \nu} A = 0$.

So:

- $\cmod \nu$ is absolutely continuous with respect to $\mu$.

$\Box$

### Necessary Condition

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

Then:

- $\cmod \nu$ is absolutely continuous with respect to $\mu$.

That is:

- whenever $A \in \Sigma$ has $\map \mu A = 0$, we have $\map {\cmod \nu} A = 0$.

From Characterization of Null Sets of Variation of Complex Measure, we have:

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

Let $\nu_R$ be the real part of $\nu$.

Let $\nu_I$ be the imaginary part of $\nu$.

From the definition of the Jordan decomposition of a complex measure, $\tuple {\nu_1, \nu_2}$ is the Jordan decomposition of $\nu_R$.

Let $\tuple {P_R, N_R}$ be a Hahn decomposition of $\nu_R$.

Then by the definition of Jordan decomposition and Uniqueness of Jordan Decomposition, we have:

- $\map {\nu_1} A = \map {\nu_R} {A \cap P_R}$

and:

- $\map {\nu_2} A = -\map {\nu_R} {A \cap N_R}$

for each $A \in \Sigma$.

From Intersection is Subset, we have:

- $A \cap P_R \subseteq A$

so:

- $\map \nu {A \cap P_R} = 0$

That is:

- $\map {\nu_R} {A \cap P_R} + i \map {\nu_I} {A \cap P_R} = 0$

giving:

- $\map {\nu_1} A + i \map {\nu_I} {A \cap P_R} = 0$

Since:

- $\map {\nu_1} A$ and $\map {\nu_I} {A \cap P_R}$ are real

we have, comparing real and imaginary parts:

- $\map {\nu_1} A = 0$

We handle $\nu_2$ similarly.

From Intersection is Subset, we have:

- $A \cap N_R \subseteq A$

so:

- $\map \nu {A \cap N_R} = 0$

giving:

- $\map {\nu_2} A + i \map {\nu_I} {A \cap N_R} = 0$

Since:

- $\map {\nu_2} A$ and $\map {\nu_I} {A \cap N_R}$ are real

we have, comparing real and imaginary parts:

- $\map {\nu_2} A = 0$

So:

- if $A \in \Sigma$ has $\map \mu A = 0$ then $\map {\nu_1} A = 0$ and $\map {\nu_2} A = 0$.

It remains to show that $\map {\nu_3} A = 0$ and $\map {\nu_4} A = 0$.

From the definition of the Jordan decomposition of a complex measure, $\tuple {\nu_3, \nu_4}$ is the Jordan decomposition of $\nu_I$.

Let $\tuple {P_I, N_I}$ be a Hahn decomposition of $\nu_I$.

Then by the definition of Jordan decomposition and Uniqueness of Jordan Decomposition, we have:

- $\map {\nu_3} A = \map {\nu_I} {A \cap P_I}$

and:

- $\map {\nu_4} A = -\map {\nu_I} {A \cap N_I}$

From Intersection is Subset, we have:

- $A \cap P_I \subseteq A$

so:

- $\map \nu {A \cap P_I} = 0$

giving:

- $\map {\nu_R} {A \cap P_I} + i \map {\nu_3} A = 0$

Since:

- $\map {\nu_R} {A \cap P_I}$ and $\map {\nu_3} A$ are real

we have, comparing real and imaginary parts:

- $\map {\nu_3} A = 0$

From Intersection is Subset, we have:

- $A \cap N_I \subseteq A$

so:

- $\map \nu {A \cap N_I} = 0$

giving:

- $\map {\nu_R} {A \cap N_I} + i \map {\nu_4} A = 0$

Since:

- $\map {\nu_R} {A \cap N_I}$ and $\map {\nu_4} A$ are real

we have, comparing real and imaginary parts:

- $\map {\nu_4} A = 0$

So:

- if $A \in \Sigma$ has $\map \mu A = 0$ then $\map {\nu_1} A = \map {\nu_2} A = \map {\nu_3} A = \map {\nu_4} A = 0$.

So:

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

$\blacksquare$

## Sources

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