Characterization of Absolutely Continuous Measures

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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 finite measure on $\struct {X, \Sigma}$.


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

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


Proof

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 $\map \nu A \ge \epsilon$.

We aim to show that:

there exists $B \in \Sigma$ such that $\map \mu B = 0$ but $\map \nu B \ne 0$.

At which point we have:

$\nu$ is not absolutely continuous with respect to $\mu$.

Fix one such $\epsilon$.

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

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

and:

$\map \nu {A_k} \ge \epsilon$

We have:

\(\ds \sum_{k \mathop = 1}^\infty \map \mu {A_k}\) \(<\) \(\ds \sum_{k \mathop = 1}^\infty \frac 1 {2^k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac 1 {2^k} - 1\)
\(\ds \) \(=\) \(\ds \frac 1 {1 - \frac 1 2} - 1\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(<\) \(\ds \infty\)

So by the Borel-Cantelli Lemma, we have:

$\ds \map \mu {\limsup_{n \mathop \to \infty} A_n} = \map \mu {\bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty A_j} = 0$

Since $\nu$ is finite, we also have:

$\map \nu {B_1}$ is finite.

We have:

$\ds \bigcup_{k \mathop = n}^\infty A_k = A_n \cup \bigcup_{k \mathop = n + 1}^\infty A_k$

From Set is Subset of Union, we obtain:

$\ds \bigcup_{k \mathop = n + 1}^\infty A_k \subseteq \bigcup_{k \mathop = n}^\infty A_k$

So:

the sequence $\ds \sequence {\bigcup_{k \mathop = n}^\infty A_k}_{n \in \N}$ is increasing.

and from the definition of the limit of an decreasing sequence of sets, we have:

$\ds \bigcup_{k \mathop = n}^\infty A_k \uparrow \bigcap_{n \mathop = 1}^\infty \bigcup_{k \mathop = n}^\infty A_k$

Write:

$\ds B_n = \bigcup_{k \mathop = n}^\infty A_k$

for each $n$ and:

$\ds B = \bigcap_{n \mathop = 1}^\infty \bigcup_{k \mathop = n}^\infty A_k$

We have:

$\ds \map \mu {B_1} \le \paren {\frac 1 2}^{1 - 1} = 1$

in particular:

$\map \mu {B_1}$ is finite.

Applying Measure of Limit of Decreasing Sequence of Measurable Sets on $\nu$ we obtain:

$\ds \map \nu B = \lim_{n \mathop \to \infty} \map \nu {B_n}$

We have, from Set is Subset of Union:

$\ds A_n \subseteq \bigcup_{k \mathop = n}^\infty A_n = B_n$

So, from Measure is Monotone, we have:

$\map \nu {A_n} \le \map \nu {B_n}$

So from Limits Preserve Inequalities:

$\ds \lim_{n \mathop \to \infty} \map \nu {B_n} \ge \lim_{n \mathop \to \infty} \map \nu {A_n}$

Recall that we have:

$\map \nu {A_n} \ge \epsilon$

for each $n$, so:

$\ds \lim_{n \mathop \to \infty} \map \nu {B_n} \ge \epsilon$

from Lower and Upper Bounds for Sequences.

We therefore have:

$\map \nu B \ge \epsilon$

so:

$\map \nu B \ne 0$

So:

$\nu$ is not absolutely continuous with respect to $\mu$.

$\Box$


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 $\mu$.

$\blacksquare$


Sources

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