Absolute Continuity of Measures 📂Measure Theory

Absolute Continuity of Measures

Definition ¹

Let’s assume given a measurable space $( \Omega , \mathcal{F} )$ . If measures $\nu$ , $\mu$ satisfy $\mu (A) = 0 \implies \nu (A) = 0$ for all $A \in \mathcal{F}$ , then $\nu$ is said to be absolutely continuous with respect to $\mu$ and is denoted by $\nu \ll \mu$ .

Explanation

As the notation $\nu \ll \mu$ suggests, $\mu$ has a strong sense of ‘dominating’ over $\nu$ . The question is why we call this ‘absolute continuity’. I have looked for a good explanation for a long time, but for learners at the level of studying real analysis, there is no easier way to understand than proving the following equivalent condition.

Theorem

$\nu \ll \mu$ $\iff$ $\forall \varepsilon > 0$ , $\exists \delta > 0 : F \in \mathcal{F}, \mu ( F ) < \delta \implies \nu (F) < \varepsilon$

Proof

Let’s assume that for all $n \in \mathbb{N}$ there exists a sequence $\left\{ F_{n} \right\}_{n \in \mathbb{N}} \subset \mathcal{F}$ that satisfies $\displaystyle \mu ( F_{n} ) < {{1} \over {2^n}}$ and $\nu (F_{n}) > \varepsilon$ .

If we set $\displaystyle A : = \bigcap_{n \in \mathbb{N}} F_{n}$ , then $\mu (A) = 0$ , but $\nu (A) \ne 0$ , hence, there is a contradiction with $\mu (A) = 0 \implies \nu (A) = 0$ .

$(\Leftarrow)$

If we set $\forall \varepsilon > 0$ about $\delta = \varepsilon$ , $\mu (A) = 0 \implies \nu (A) = 0$

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