Proof of Theorem $7.13$ in Rudin's RCA Theorem $7.13$, Walter Rudin's Real and Complex Analysis.

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*Why does it suffice to prove for the case $\mu\ge 0$? $\mu = \mu^+ - \mu^-$ is the Jordan decomposition of $\mu$, where $\mu^+,\mu^-$ are positive measures. How does the general case follow from $\mu\ge 0$ and the Jordan decomposition?


*Why is $\overline D\mu$ a Borel function? I see that $\sup_{0<r<1/n} Q_r\mu(x)$ decreases as $n$ increases, and is a lower-semicontinuous function for each $n$ (following the reasoning in Section $7.2$.). However, I don't see how this implies that $\overline D\mu$ is a Borel function. I also know that any lower-semicontinuous function is Borel, so it'd suffice to prove lower-semicontinuity, if we can.


*Why is it true that for every $x\in K^c$, $$(\overline D\mu)(x) = (\overline D\mu_2)(x)$$
Proof attached for reference:


Thank you!
 A: The fact that $\sup_{0<r<1/n} Q_r\mu(x)$ decreases as $n$ increases is used to asser existence of ehn limit in $(\overline D \mu )(x)$. Limits of Borel measurable functions are Borel measurable.
$\mu_1(E)=\mu(E\cap K)=0$ for any Borel set $E$ contained in $K^{c}$. So $\mu (E)=\mu_2(E)$ for such sets. From this and the definition of $(\overline D \mu )(x)$ and $(\overline D \mu_2 )(x)$ it follows that $(\overline D \mu )(x)=(\overline D \mu_2 )(x)$ for $x \in K^{c}$.
[Note that if $x \in K^{c}$ then $B(x,\frac 1 n)$ is contained in $K^{c}$ for $n$ sufficiently large].
A: Here's an answer for (1). For a complex measure $\mu$, we say $\mu \perp m$ if $\mu_a \perp m$ and $\mu_b \perp m$, where $\mu_a, \mu_b$ are the real and imaginary parts of $\mu$, respectively. By the Jordan decomposition theorem, we can write
$$\mu_{k} = \mu_{k}^+ - \mu_{k}^-$$
for $k = a,b$, where $\mu_k \geq 0$. That is, any complex measure is the linear combination of four (finite) positive measure. Moreover, you can prove that a signed measure $\nu$ is singular to $m$ if and only if $\nu^+ \perp m$ and $\nu^- \perp m$.
Thus, if you can prove the result for $\mu \geq 0$, you can use the above decomposition to immediately get: $$\lim_{i\to \infty} \frac{\mu(E_i(x))}{m(E_i(x))} = \lim_{i\to \infty} \frac{\mu_a^+(E_i(x)) - \mu_a^-(E_i(x)) + i\left[\mu_b^+(E_i(x)) - \mu_b^-(E_i(x)) \right]}{m(E_i(x))} = 0$$
With the limit holding for $m$-almost everywhere.
