TITLE: Is this theorem in Rudin's Real and Complex Analysis wrong as stated?
QUESTION [9 upvotes]: I have forgotten all of my measure theory and will now ask a very dumb question. 
Consider the following theorem, which I have produced word for word from Rudin's Real and Complex Analysis, third edition. It is theorem $7.13$ in chapter $7$ and can be found on page $142$. 

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
$$\lim _{i\rightarrow \infty} \frac{\mu(E_i(x))}{m(E_i(x))} =0\
 \text{a.e.} \ [m].$$

Consider the measure $\mu$ defined by giving a point mass to each point in $\mathbb Q$. This is a singular measure with respect to the Lebesgue measure. In other words, the measure of a set is the number of rational points it contains. 
Let $x=0$ in $\mathbb R$, and consider the intervals $(-1/n, 1/n)$ that shrink nicely to $0$. The $\mu$ measure of these is infinite. So each term in the limit is infinite, and the limit itself must be infinite. This seems to contradict the theorem. 
Where have I gone wrong in this reasoning? 
(This question is closely related.)

REPLY [6 votes]: Your measure $\mu$ is not complex-valued, as it admits the value $\infty$, so your construction does not contradict the theorem.