Taking limits of both sides of inequality in Royden and Fitzpatrick

A theorem in Royden and Fitzpatrick's "Real Analysis" relies on taking limits of both sides.

First the theorem: Let $E$ be a measurable set of finite outer measure. Then for each $\varepsilon > 0$, there is a finite disjoint collection of open intervals $(I_k)$ for which if $O = \cup_{k=1}^{n} I_k$, then $m^*(E\setminus O) + m^*(O\setminus E) < \varepsilon$ where $m^*$ denotes the set function outer measure.

So in the proof, there's this part:

$\sum_{i=1}^{n} l(I_k) = m^*(\cup_{k=1}^{n} I_k) \leq m^*(\cup_{k=1}^{\infty} I_k) < \infty$, where l denotes length.

Since "the right-hand side of this inequality is independent of $n$," R and F have concluded that

$\sum_{i=1}^{\infty} l(I_k) < \infty$.

$H_n < \infty$, yet $\lim_{n \to \infty} H_n = \infty$
where $H_n$ is the nth Harmonic number $= \sum_{i=1}^{n} \frac{1}{i}$ ?
• What the book meant was if $H_n \leq C < \infty$ then $\lim_n H_n < \infty$. – Xiao Apr 4 '14 at 13:02
• The intervals are disjoint, and open intervals are measurable. Then, restricting $m^*$ to the collection of measurable sets (as in Caratheodory's criterion) gives a measure. Then the given conclusion follows from countable additivity of measures. – Nicholas Stull Apr 4 '14 at 13:02
• $m^*(\bigcup_1^\infty I_k)$ is the bound of your sums. – user42761 Apr 4 '14 at 13:19