Let $X$ compact Hausdorff space in wich every $\left\{x\right\}$ is a Baire set and let $\mu$ be a Baire measure. Define $\mu_{pp}(A)=\sum_{x\in A}\mu(\left\{x\right\})$
(a) Prove that $\mu_{pp}$ is a pure point measure. I already proves!
(b) Prove that $\mu_{c}:=\mu-\mu_{pp}$ is a nonnegative continuous measure. I already prove!
(c)If $\mu=\nu_c+\nu_{pp}$ with $\nu_{pp}$ and $\nu_{c}$ continuous, prove that $\nu_{pp}=\mu_{pp}$ and so $\nu_{c}=\mu_{c}$ (Correction. $\nu_{pp}$ is a pure point)
How prove (c)?
Let $C=\{x:\mu \{x\} \neq 0\}$ and $D=\{x:\nu_{pp} \{x\} \neq 0\}$, Verify the following for any measurable set $E$: $$\nu_c(E)=\nu_c(E\setminus C\cup D)=\mu (E\setminus C\cup D)-\nu_{pp}(E\setminus C\cup D)$$ $$=\mu_c (E\setminus C\cup D)+\mu_{pp} (E\setminus C\cup D)-\nu_{pp}(E\setminus C\cup D)$$ $$=\mu_c (E\setminus C\cup D)+0+0=\mu_c(E).$$ This proves that $\nu_c=\mu_c$ and it follows automatically that $\mu_{pp}=\nu_{pp}$.