Radon-Nikodym theorem on Borel measurable space

I want to prove the Radon-Nikodym theorem on Borel measurable space. Theorem is following:

$\mu$ is positive $\sigma$-finite Borel measure on $\sigma$-algebra $\mathcal{M}$, and $\lambda$ is complex Borel measure on $\mathcal{M}$. Then there is a unique $\lambda_a$ and $\lambda_s$ : measure on $\mathcal{M}$ such that $\lambda=\lambda_a+\lambda_s$, $\lambda_a\ll\mu$, and $\lambda_a\perp\mu$.

My proof is : Let $A$ be a set of support of $\mu$. Note that $A$ is measurable set since $\mu(A^c)=0$, $A^c$ is measurable. Define $\lambda_a=\lambda|_A$ and $\lambda_s=\lambda|_{A^c}$. Those two are well defined and also measure.

How's my proof? The main thing is because the space is Borel measurable, we can handle 'support'. (The last statement that defining $\lambda$'s is here) (And the original theorem that is cited is Rudin's RCA, theorem 6.10)

• Why is $\lambda|_A \ll \mu$? – Bungo Aug 8 '16 at 17:07
• @Bungo Isn't that clear by definition? If $\mu(E)=0$, then $E\subset A^c$, then $\lambda_a(E)=0$ – kayak Aug 8 '16 at 17:12
• It is not true that if $\mu(E) = 0$ then $E \subset A^c$. In other words, the support of $\mu$ can contain nonempty subsets of measure zero. Consider Lebesgue measure, which has support $\mathbb R$. – Bungo Aug 8 '16 at 17:20
• @Bungo Thanks for your comment! – kayak Aug 8 '16 at 17:35
• @Bungo Then is there any quick proof of the Lebesque-Radon-Nikodym theorem for special case? – kayak Aug 8 '16 at 17:35