Lebesgue decomposition theorem and fundamental theorem of calculus

On page 42 of 'Evans-Gariepy, Measure theory and fine properties of functions' it's stated and proved this theorem: let $\mu$ and $\nu$ be Radon measures on $\mathbb{R}^n$. Then $\nu=\nu_{ac}+\nu_s$ where the first is absolutely continuous respect to $\mu$ and the second is singular respect $\mu$.

Furthermore the derivative $D_{\mu}\nu=D_{\mu}\nu_{ac}$ and $D_{\mu}\nu_s=0 \; \mu-a.e$. Up to this everything is ok and proved, but then is stated a consequence not proved (because it seems obvious) but I don't understand: $\nu(A)=\int_{A}D_{\mu}\nu_{ac} d\mu+\nu_s(A)$. Why is there the term $\nu_s(A)$?

I thought this: $\nu(A)=\int_{A}D_{\mu}\nu_{} d\mu =\int_{A}D_{\mu}\nu_{ac} d\mu+\int_{A}D_{\mu}\nu_{s}d\mu=\int_{A}D_{\mu}\nu_{ac} d\mu$

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You almost have the answer already. $\nu(A) = \nu_{ac}(A) + \nu_{s}(A)$, and $\nu_{ac}(A) = \int_A D_\mu \nu_{ac} d\mu$. You wrote $\nu_s(A) = \int_A D_\mu \nu_s d\mu$, but this is not in general true, because the fundamental theorem of calculus (equality) doesn't hold for $\nu_s$ singular with respect to $\mu$. You can only say that $\int_A D_\mu \nu_s d\mu \leq \nu_s(A)$. –  Nicholas Stull Apr 20 '12 at 12:40
Ok, thanks i got it –  balestrav Apr 20 '12 at 12:44
No problem. I wasn't sure if I had explained it well, but this was the argument that came to mind. –  Nicholas Stull Apr 20 '12 at 12:46

Think about $\nu_{ac}$ the Lebesgue measure and $\nu_s$ the counting measure.
For the computation $\int_{A} d nu = \int_A d \nu_s + \int_A d \nu_s = \int_A D_\mu \nu_{ac} d \mu + \nu_s(A)$, but
In general $$\int_A d \nu_s \neq \int_A D_\mu \nu_{s} d \mu = 0.$$