Let $\mu$ be a Baire measure and suppose $\mu=\lambda_{ac}+\lambda_s=\nu_{ac}+\nu_{s}$ where $\nu_{s}\perp \mu,\ \lambda_{s}\perp \mu,\ \nu_{ac}<<\nu$ and $\lambda_{ac}<<\mu$.
(a) Prove if $A,B$ are such that $\mu(A)=\mu(B)=0,\ \lambda_{s}(X\setminus A)=0=\nu_{s}(X\setminus B)$, then $\lambda_{s}(C)=\lambda_{s}(C\cap A\cap B)=\nu_{s}(C\cap A\cap B)=\nu_{s}(C)$, so $\nu_s=\lambda_s$
Why $\lambda_{s}(C)=\lambda_{s}(C\cap A\cap B)$? I dont see this.
pd: Apparently the question must be changed $C\cap A\cap B$ by $C\setminus (A\cup B)$
Thanks to Kavi Rama Murthy I have this: $\lambda_s\perp \mu\Rightarrow \exists A\subset X: \lambda_s(A)=0$ and $\mu(X\setminus A)=0.$
$\nu_s\perp \mu\Rightarrow \exists A\subset X: \nu_s(A)=0$ and $\mu(X\setminus A)=0$.
Now, $\lambda_s(C)=\lambda_s(C\setminus (A\cup B)) +\lambda_s(C\cap (A\cup B))$ Now, $\lambda_s(C\cap (A\cup B))\leq \lambda_s(C\cap A)+\lambda_c(C\cap B)=0+0$ then $\lambda_s(C)=\lambda_s(C\setminus(A\cup B))$
Affirmation. $\lambda_s(C\setminus(A\cup B))=\nu_s(C\setminus(A\cup B))$
Indeed. $\lambda_s(C\setminus(A\cup B))+\lambda_{ac}(C\setminus(A\cup B))=\nu_s(C\setminus(A\cup B))+\nu_{ac}(C\setminus(A\cup B))$
Because $\mu(X\setminus A)=0$ and $\mu(X\setminus B)=0$ then $\mu(X\setminus (A\cup B)=0$. And this implies $\mu(C\setminus (A\cup B))=0$.
Now, $\lambda_{ac}$ and $\nu_{ac}$ are a.c. then $\lambda_{ac}(C\setminus (A\cup B))=\nu_{ac}(C\setminus (A\cup B))=0$
therefore $\lambda_{s}(C)=\nu_{s}(C)$ and this implies $\lambda_{ac}(C)=\nu_{ac}(C)$.
Therefore decomposition is unique.
Thanks!
In your second line $C\cap A\cap B$ should be replaced by $C\setminus A\cup B$ at both places. Similarly, in the first line you should have $\lambda_s(A)=\lambda_s(B)=0$. Alternatively, you can make everything right by changing $\mu (A)=\mu (B)=0$ to $\mu (A^c))=\mu (B^c)=0$