Consider two singular measures $ m$ and $v$ and $v$ is absolutely continuous with respect to $m$ show that $v=0$

Question

Consider two singular measures $m$ and $v$ on a measurable space $(X,\mathcal{A})$ and $v$ is absolutely continuous with respect to $m$, i.e., $(v<<m)$. Show that $v=0$.

Answer 1

$m$ and $v$ are singular means there exist disjoint $A$ and $B$ such that $A\cup B = X$ and $m(A) = 0, v(B) = 0$.

Then for any $C \in \mathcal{A}$, we have

$$v(C) = v(C\cap A) + v (C\cap B) = v(C\cap A)$$

then since $m(C\cap A) =0$ and $v$ is absolutely continuous with respect to $m$, we have $v(C\cap A) =0$

Answer 2

Let $A, B\in\mathcal A$ be a partition of $X$ such that $m(C) =0$ for any $C\subset B$ with $C\in \mathcal A$ and $v(D) =0$ for any $D\subset A$ with $D\in\mathcal A $. Let $E\in\mathcal A$. Then E is the disjoint union of $E\cap A$ and $E\cap B$. Since $E\cap A\subset A$ with $E\cap A\in\mathcal A$, we have $v(E\cap A) =0$. Similarly, $m(E\cap B) =0$, and as $v\ll m$, we have $v(E\cap B) =0$. It follows from finite additivity of measures that $v(E) =0$.