If $v \ll \mu$ and $v \perp \mu$, then $v=0$

Question

Let $\mu$ and $v$ be two signed measures defined in the measure space (X,Q). We say that $\mu$ and $v$ are mutually singular, and write $\mu \perp v$, if there exist measurable sets A and B such that

$X=A \cup B$, and $A \cap B = \emptyset$

and

$|\mu|(A)=0$, $|v|(B)=0$

We also say that $\mu$ is singular relative to $v$ (or that $v$ is singular relative to $\mu$)

How do I test this?

"If $v \ll \mu$ and $v \perp \mu$, then $v=0$"

At the moment I have not achieved anything, I would appreciate your help.

Answer 1

Let $(X, \mathcal{F}) $ be a measurable space.

$v \perp \mu$ implies $\exists A\in\mathcal{F} $ such that $\mu(A) =0$ and $v(A^c) =0$

$v<<\mu$ and $\mu(A) =0$ implies $v(A) =0$ Thus, $v(X) =v(A) +v(A^c) =0$.

Answer 2

This is very similar to Sourav Ghosh's answer, but I just want to flesh it out a bit more.

Let $(X, \mathcal F)$ be a measurable space, and let $\mu$ and $\nu$ be measures on $\mathcal F$. (I am using $\nu$ instead of $v$.)

• $\nu$ is absolutely continuous with respect to $\mu$ (or dominated by $\mu$) if for each $A \in \mathcal F$ where $\mu(A) = 0$, then $\nu(A) = 0$. We write $\nu \ll \mu$.
• $\nu$ and $\mu$ are mutually singular if there is a set $A \in \mathcal F$ such that $\nu(A) = 0$ and $\mu(A^\complement) = 0$. We write $\nu \perp \mu$.

Now, we want to prove that if $\nu \ll \mu$ and $\nu \perp \mu$, then $\nu$ is the zero measure. By mutual singularity, there is a set $A \in \mathcal F$ such that $\nu(A) = 0$ and $\mu(A^\complement) = 0$. By absolute continuity, we know that $\nu(A^\complement) = 0$. Thus,

\begin{align*} \nu(X) &= \nu(A \cup A^\complement)\\ &= \nu(A) + \nu(A^\complement)\\ &= 0 + 0\\ &= 0. \end{align*}

Finally, by monotonicity of measures, we conclude that for any $B \in \mathcal F$, we have $\nu(B) = 0$, as desired.