# Singular Absolutely Continuous Measure is Zero

## Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a measure on $\struct {X, \Sigma}$.

Let $\nu$ be a signed measure or complex measure on $\struct {X, \Sigma}$ such that:

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

Then $\nu = 0$.

## Proof

If $\nu$ is a measure, we take $\cmod \nu = \nu$ for the sake of convenience.

Since $\nu$ is mutually singular with respect to $\mu$, there exists a $\mu$-null set $N \subseteq X$ such that:

- $\map {\cmod \nu} {N^c} = 0$

Let $A \in \Sigma$.

We have:

- $\map {\cmod \nu} A = \map {\cmod \nu} {A \cap N} + \map {\cmod \nu} {A \cap N^c}$

From Null Sets Closed under Subset, we have:

- $\map \mu {A \cap N} = 0$
- $\map {\cmod \nu} {A \cap N^c} = 0$

Since $\nu$ is absolutely continuous with respect to $\mu$, we have:

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

Hence we have $\map {\cmod \nu} A = 0$.

From Characterization of Null Sets of Variation of Signed Measure and Characterization of Null Sets of Variation of Complex Measure, we have $\map \nu A = 0$.

Since $A \in \Sigma$ was arbitrary we have $\nu = 0$.

$\blacksquare$