Characterization of Absolutely Continuous Measures

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

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

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

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


 * for each $\epsilon > 0$ there exists $\delta > 0$ such that for each $A \in \Sigma$ with $\map \mu A < \delta$, we have $\map \nu A < \epsilon$

Necessary Condition
We prove the contrapositive, then the result follows from Rule of Transposition.

Suppose that:


 * for some $\epsilon > 0$, there exists no $\delta > 0$ such that for each $A \in \Sigma$ with $\map \mu A < \delta$, we have $\map \nu A < \epsilon$.

That is:


 * for some $\epsilon > 0$, for all $\delta > 0$ there exists $A \in \Sigma$ with $\map \mu A < \delta$ and $\map \nu A \ge \epsilon$.

Fix one such $\epsilon$.

For each $k$, pick $A_k \in \Sigma$ such that:


 * $\map \mu {A_k} < 2^{-k}$

and:


 * $\map \nu A \ge \epsilon$

We have:

From Intersection is Subset, we have:


 * $\ds \bigcap_{n \mathop = 1}^\infty \bigcup_{k \mathop = n}^\infty A_k \subseteq \bigcup_{k \mathop = m - 1}^\infty A_k$

for each $m$.

So, from Measure is Monotone, we have:


 * $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \bigcup_{k \mathop = n}^\infty A_k} \le \paren {\frac 1 2}^{m - 1}$

for each $m$.

Taking $m \to \infty$, we have:


 * $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \bigcup_{k \mathop = n}^\infty A_k} = 0$

Sufficient Condition
Suppose that:


 * for each $\epsilon > 0$ there exists $\delta > 0$ such that for each $A \in \Sigma$ with $\map \mu A < \delta$, we have $\map \nu A < \epsilon$

Let $A \in \Sigma$ have $\map \mu A = 0$.

We aim to show that $\map \nu A = 0$.

Let $\epsilon > 0$.

Then there exists $\delta > 0$ such that for each $A \in \Sigma$ with $\map \mu A < \delta$, we have $\map \nu A < \epsilon$.

Since $\map \mu A = 0$, we have $\map \mu A < \delta$, and so:


 * $\map \nu A < \epsilon$

Since $\epsilon > 0$ was arbitrary and $\map \nu A \ge 0$, we have:


 * $\map \nu A = 0$

So:


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