I study the book of Gamelin about uniform algebras. As a corollary of the following Lemma: "let $K$ be a (weak star) compact convex set consisting of positive measures and a (complex) measure $\mu$ is singular to all measures in $K$ then there is a set $E$ (being $F_{\sigma}$) such that $|\mu|(X \setminus E)=0$ and $m(E)=0$ for all $m \in K$" author states that if $K$ is weak star convex set consisting of positive measures and $\mu$ is a complex measure on $X$ (here $X$ is assumed to be compact Hausdorff) then there is a unique decomposition $\mu=\mu_a+\mu_s$ where $\mu_a$ is absolutely continous with respect to some measure in $K$ and $\mu_s$ is supported on some Borel set $E$ such that $\nu(E)=0$ for all $\nu \in K$.
Here is an outline of the proof: we put $c$ to be the supremum of $|\mu|(G)$ over all Borel sets $G$ such that the restriction $\mu_G$ is absolutely continuous with respect to some $\nu \in K$. Choose $\nu_n \in K$ and Borel sets $F_n$ such that $\mu_{F_n}$ are abs. cont. with respect to $\nu_n$ and $|\mu|(F_n) \to c$. Then put $F:=\bigcup_nF_n$ and $\nu_0:=\sum_n\frac{\nu_n}{2^n}$. Then $\nu \in K$. The measure $\mu_a:=\mu_F$ (restriction to $F$) is absolutely continuous with respect to $\nu$ and now comes the issue about which I want to ask:
Why $\mu_s:=\mu-\mu_a$ is singular to all measures in $K$?
From the choice of $F_n$ this seems to make sense but I don't see why it is true. I first tried to check this only for $\nu_0$: it is true that $\mu_s$ is supported on $X \setminus F$ but I didin't manage to prove that $\nu_0$ is supported on $F$ (which is the natural guess). And another thing:
Why this decomposition is unique?