Lebesgues Decomposition Theorem says that if $\mu$ and $\nu$ are two $\sigma$-finite measures, $\mu$ can be decomposed into $\mu=\mu_{ac}+\mu_{sing}$ where $\mu_{ac}$ is absolutely continuous with respect to $\nu$ and $\mu_{sing}$ is singular with respect to $\nu$.
By Radon-Nikodym Theorem $\mu_{ac}$ is absolutely continuous with respect to $\nu$, if $\mu_{ac}(B)=0$ whenever $\nu(B)=0$. In conclusion a measure that is not absolutely continuous must be a measure that fulfills $\mu_{*}(B)>0$, while $\nu(B)=0$ for some $B$.
One definition of a singular measure i´ve read (a first look at rigorous probability Therory by Rosenthal) definied it by a measure that satisfies $\nu(B)=0$ and $\mu_{sing}(B^c)=0$.
I don't really understand why there is no "third" kind of measure that satisfies $\mu_*(B)\in(0,1)$ while $\nu(B)=0$.
I thought maybe the case $\mu$*(B)$ \in(0,1)$ (while $\nu(B)=0$) is somehow impossible? Or is it maybe already implied by $\mu$*$(\tilde{B}^c)=0$ for some $\tilde{B}$?
I would appreciate any help and am sorry if anything is not clearly formulated since english is not my first language.
EDIT: I figured that the case $\mu$_*(B)$ \in(0,1)$ (while $\nu(B)=0$) implies that $\mu$ is of mixed type. Its possible to define $\mu_B(A)=\mathbb{P}(A|B)$ (obviously singular since $\mu_B(B)=1$). Then $\mu$ can be written as a sum of $\alpha\mu_B$ and $(1-\alpha)\mu_{B^c}$ with $\alpha\in(0,1)$. $\mu_{B^c}$ can be absolutely continuous wrt $\nu$ or also of mixed type.