Lebesgue-Radon-Nikodym representation

Question

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!

Answer 1

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 $|f\,dm|=|f|\,dm$.

Answer 2

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 $$m(F) \leq \frac{1}{\alpha}\epsilon < \epsilon$$

Answer 3

Here is a naive proof of $d|\nu| = d|\lambda|+|f|dm$. I am currently learning this area so hopefully this is correct.

Let $d\rho = fdm$ and $\mu = |\lambda|+|\rho|$,

$$\lambda \ll |\lambda|, \rho \ll |\rho| \implies \lambda \ll \mu,\ \rho \ll \mu.$$

Since $\nu = \lambda + \rho$,

$$\nu \ll \mu.$$

Due to the above, there exists $f_1, f_2, f_3 \in L_1(\mu)$ such that,

$$d\nu = f_1 d\mu,\ d\lambda = f_2d\mu,\ d\rho=f_3d\mu.$$

Again, since, $d\nu = d\lambda + d\rho$,

$$f_1 = f_2 + f_3, \mu\text{-a.e.} \implies |f_1| = |f_2 + f_3|, \mu\text{-a.e.}$$

Since $\lambda \perp m$, there exist $E$ and $F$ such that $E \cap F = \phi$ and $E \cup F = \mathbb{R}^n$ such that $E$ is null for $\lambda$ and $F$ is null for $m$. Since $\rho \ll m$, $F$ is null for $\rho$ too. So, we can replace $f_2$ by $f_2 \chi_F$ and $f_3$ by $f_3\chi_E$ without affecting $d\lambda$ and $d\rho$ (this part should probably be made more precise).

So, $|f_1| = |f_2| + |f_3|, \mu\text{-a.e.}$ and the result ($d|\nu| = d|\lambda|+d|\rho|$) follows by using the fact that $d|\nu| = |f_1|d\mu$, $d|\lambda| = |f_2|d\mu$ and $d|\rho| = |f_3|d\mu = |f|dm$.