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