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In Rudin Real and Complex Analysis, the proof of theorem 7.13 (page 142 3rd ed), why is $\mu$ regular?

If I understand correctly: prove that a singular complex Borel measure is regular. The reasoning Rudin gives (Theorem 2.18) doesn't help me.


for reference:


(1) Borel measure = measure defined on the sigma-algebra of all Borel sets.


(2) A measure is outer regular if it obeys $\mu(E)=\inf\{μ(V):E⊂V,V \mbox{ open}\}$. for any measurable set $E$.


(3) A measure if inner regular if it obeys $\mu(E)=\sup\{\mu(K):K\subset E,K\mbox{ compact}\}$ for every open set $E$, and for every measurable set $E$ such that $\mu(E)\lt\infty$.


(4) A measure if regular if it both inner regular and outer regular.


(5) A measure is singular is if it concentrated on a set disjoint to the set the Lebesgue measure is concentrated on.


(6) A measure is concentrated on a set if it is zero on any set that doesn't intersect the set it is concentrated on.


The theorem: " Associate to each $x\in\mathbb{R}^k$ a sequence $\{E_i(x)\}$ that "shrinks to $x$ nicely." If $\mu$ is a complex Borel measure and $\mu \perp m$ then: $$ \lim_{i\to\infty}\frac{\mu(E_i(x))}{m(E_i(x))} = 0 $$ m almost everywhere.

Proof:

The Jordan decomposition theorem shows that it suffices to prove the theorem under the additional assumption that $\mu\geq0$. In that case, arguing as in the proof of Theorem 7.10, we have: $$ \frac{\alpha(x)\mu(E_i(x))}{m(E_i(x))}\leq\frac{\mu(E_i(x))}{m(B(x,r_i)}\leq\frac{\mu(B(x,r_i))}{m(B(x,r_i))} $$ Hence the theorem is a consequence of the special case $(D\mu)(x)=0$ a.e. [m], which will now be proved. The upper derivative $\bar{D}\mu$, defined by: $$ (\bar{D}\mu)(x)=\lim_{n\to\infty}\left[\sup(\left\{(Q_r\mu)(x):r\in(0,1/n)\right\})\right]\,\forall x\in\mathbb{R}^k $$ is a Borel function, because the quantity in brackets decreases as $n$ increases, and is, for each $n$, a lower semicontinuous function of $x$; the reasoning used in Sec. 7.2 proves this.

Choose $\lambda>0,\varepsilon>0$. Since $\mu\perp m$, $\mu$ is concentrated on a set of Lebesgue measure $0$. The regularity of $\mu$ (Theorem 2.18) shows therefore that there is a compact set $K$, with $m(K)=0,\,\mu(K)>\mu(\mathbb{R}^k)-\varepsilon$.

...." continues...

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  • $\begingroup$ Could you give the definitions of the involved concepts? $\endgroup$ – Davide Giraudo Jul 5 '13 at 11:58
  • $\begingroup$ Yes: (1) Borel measure = measure defined on the sigma-algebra of all Borel sets. (2) A measure is outer regular if it obeys $\mu(E)=\inf({\mu(V):E\subset V,V open})$. for any measurable set E.(3) A measure if inner regular if it obeys $\mu(E)=\sup({\mu(K):K\subset E, K compact})$ for every open set E, and for every measurable set E such that $\mu(E)<\infty$. (4) A measure if regular if it both inner regular and outer regular. A measure is singular is if it concentrated on a set disjoint to the set the Lebesgue measure is concentrated on. $\endgroup$ – PPR Jul 5 '13 at 16:04
  • $\begingroup$ Thank you. Can you include this in the OP? And if you can't write all the proof, can you give at least the most relevant elements? $\endgroup$ – Davide Giraudo Jul 5 '13 at 16:21
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The theorem is correct. A complex measure is finite by definition, so the measure is finite on compact sets. Then theorem $2.18$ applies and gives regularity.

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  • $\begingroup$ doh. That was tricky because he first assumed (via Jordan decomposition) that this is a non-negative measure. But of course the earlier restriction of complex measure holds... Thank you! $\endgroup$ – PPR Jul 5 '13 at 20:11

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