Mutually singular measures

Remarks

Example: on the Borel sets of $\R$, let $\lambda$ be Lebesgue measure and let $\delta_0$ be the Dirac measure at $0$. Take $A = \R \setminus \{0\}$ and $B = \{0\}$. Then $\lambda(B) = 0$ and $\delta_0(A) = 0$, so $\lambda \perp \delta_0$.

Non-example: let $\mu$ be Lebesgue measure restricted to $[0, 1]$ and $\nu$ Lebesgue measure restricted to $[0, 2]$. If $\mu(B) = 0$, then $B \cap [0, 1]$ has length $0$, so $A$ contains almost all of $[0, 1]$ and $\nu(A) \ge 1$. So clause 3 fails for every choice, and $\mu$ and $\nu$ are not mutually singular.

The relation is symmetric. For signed or complex measures, $\mu \perp \nu$ means that the total variations $\abs{\mu}$ and $\abs{\nu}$ are mutually singular. A measure $\nu$ is called singular with respect to $\mu$ when $\nu \perp \mu$.