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In Folland,

$\textbf{3.22 Theorem.}$ Let $\nu$ be a regular signed or complex Borel measure on $\mathbb{R}^n$, and let $d\nu = d \lambda + f dm$ be its Lebesgue-Radon-Nikodym representation. Then for $m$ almost every $x\in\mathbb{R}^n$, \begin{align*} \lim_{r\rightarrow 0} \frac{\nu(E_r)}{m(E_r)}=f(x) \end{align*} for every family $\{E_r\}_{r>0}$ that shrinks nicely to x.

In proof, I need to prove $d\vert\nu\vert=d\vert \lambda \vert +\vert f \vert dm$.

Could you give some hints??..

Any help is appreciated..

Thank you!

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  • $\begingroup$ Presumably this is covered within the proof (or a preceeding lemma)...? $\endgroup$ – Math1000 May 8 '18 at 16:22
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    $\begingroup$ Apply Lebesgue radon nikodym for positive measure... $\endgroup$ – user251257 May 8 '18 at 16:23
  • $\begingroup$ Consider the Jordan decomposition of the measure. $\endgroup$ – Kanu Kim May 8 '18 at 16:48
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Because $\lambda$ and $f\,dm$ are mutually singular, one can easily check that $$ |\nu|=|\lambda|+|f\,dm|. $$ And writing $f=h\,|f|$, with $|h|=1$, it is apparent from the definition of total variation that $|f\,dm|=|f|\,dm$.

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Let $f\in L^1_{loc}$ by Theorem 3.21 it suffices to show that if $\nu\perp \mu$ and $\nu$ is regular then $$\lim_{r\rightarrow 0}\frac{\nu(B(r,x))}{m(B(r,x)} = 0 \ \text{for} \ m \ \text{a.e.} \ x$$ Since $\nu\perp \mu$ then we can find $G,H\in M$ with $G\cup H = \mathbb{R}^n$,$G\cap H = \emptyset$, and $\nu(G) = m(H) = 0$. We need $$\lim_{r\rightarrow 0}\frac{\nu(B(r,x))}{m(B(r,x))} = 0 \ \text{for a.e.} \ x\in G$$ It suffices to show that for each $\lambda,\epsilon > 0$ that $$m\left(\{x\in G: \lim_{r\rightarrow 0}\frac{\nu(B(r,x))}{m(B(r,x))} > \lambda\}\right) < \epsilon$$ by regularity we can find an open $U$ where $G\subset U$ with $\nu(U) < \frac{\epsilon \lambda}{3^n \alpha}$ for each $x\in f$, we can find $r_x$ such that $$\frac{\nu(B(r_x,x))}{m(B(r_x,x))} > \lambda$$ and $B(r_x,x)\subset U$. Let $F\subset \bigcup_{x\in F}B(r_x,x)$ so by theorem 3.15 $ c < m(F)$ and we can find $\{B(r_{x_j},x_j\}_{1}^{n}$ pairwise disjoint such that $$\sum_{1}^{n}B(r_{x_j},x_j) > 3^{-n}c$$ then $$\frac{\epsilon \lambda}{3^n \alpha} > \nu(U) \geq \sum_{1}^{n}\nu(B(r_{x_j},x_j) \geq \sum_{1}^{n} \lambda m(B(r_{x_j},x_j) \geq \lambda 3^{-n}c$$ So $$\frac{\epsilon \lambda}{3^n \alpha } \geq \lambda 3^{-n}m(F)$$ so $$m(F) \leq \frac{1}{\alpha}\epsilon < \epsilon$$

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