$\nu = \nu_a + \nu_s, \nu_a \ll \mu \implies \nu_s \perp \mu$

Let $\mu$ and $\nu$ be positive measure. Can we claim that $\nu = \nu_a + \nu_s, \nu_a \ll \mu \implies \nu_s \perp \mu$?

When we prove the uniqueness part of Radon-Nikodym theorem, $$ \nu_a + \nu_s = \nu_a'+ \nu_s' $$ where $\nu_a, \nu_a' \ll \mu$, $\nu_s, \nu_s' \perp \mu$ is assumed to use the fact that $$ \nu \ll \mu, \nu \perp\mu \implies \nu = 0$$ But in my case, it's not applicable since $\nu_s' \perp \mu$ is not assumed. Can we still claim that?