Uniqueness of decomposition in the Radon-Nikodym theorem

I am having some problems trying to understand why the decomposition is unique in Radon-Nikodym. I'll write the theorem first:

Let $\mu$ and $\nu$ be two finite measures over the measurable space $(X,\Sigma)$. Then, there is a unique pair of measures $\nu_a,\nu_s$ over $\Sigma$ such that:

(1) $\nu=\nu_a+\nu_s$, $\space \nu_a << \mu, \nu_s \perp \mu$

(2) there is a unique function $h \in L^1(\mu)$ that verifies $$\nu_a(E)=\int_E h(x) d\mu$$

For the function $h$, since in the proof of existence one deduces that $h \geq 0$, then if there are $h,h'$ that satisfy (2), we would have $\int_X h-h' d\mu=0$ (without loss of generality I am assuming that $h$ and $h'$ are finite everywhere), but then $h=h'$ a.e.

I can't show that the pair $(\nu_a,\nu_s)$ is unique. In Rudin's textbook, the author says that if there were two pairs $(\nu_a,\nu_s)$,$(\nu_a',\nu_s')$, then $\nu_a-\nu_a'=\nu_s'-\nu_s$, and we would have $\nu_a-\nu_a' << \mu$ and $\nu_s'-\nu_s \perp \mu$. If a measure $\nu$ satisfies $\nu \perp \mu$ and $\nu << \mu$, then $\nu=0$, so from this it would follow uniqueness.

So, I don't know how to show that if $\nu_s \perp \mu$ and $\nu_s' \perp \mu$ then $\nu_s'-\nu_s \perp \mu$.

I would really appreciate if someone could help me to understand why that last statement is true.

if $\nu_s \perp \mu$ and $\nu_s' \perp \mu$ then $\nu_s'-\nu_s \perp \mu$.
There is a set $E$ of full measure for $\mu$ which is null for $\nu_s$. Similarly, there is a set $F$ of full measure for $\mu$ which is null for $\nu_s'$.
Hence, $E\cap F$ is a set of full measure for $\mu$ which is null for $\nu_s-\nu_s'$.