Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer

up vote 1 down vote accepted

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 $$ \int_A d \nu_s \neq \int_A D_\mu \nu_{s} d \mu = 0.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.