I have a question about the following proof by Royden.
What I do not understand is the part where it says $\Sigma \Sigma l(I_{k,i}) < \Sigma m^*(E_k)+\epsilon / 2^k$
Why is there a strict inequality there?
As far as I know, $a_n < b_n$ does not necessarily imply that $\Sigma a_n < \Sigma b_n$.
It does. If $a_n \le b_n$ then you have $\sum_n a_n \le \sum_n b_n$ (assuming they both converge, of course, this holds in the example above).
Now suppose there is some index such that $a_k < b_k$. Then we can make a new sequence $a_n'$ that is equal to $a_n$ everywhere except $a'_k = b_k$. Then we have $a'_k \le b_k$ and so $\sum_n a'_n \le \sum_n b_n$. Since $\sum_n a_n = a_k-b_k+\sum_n a'_n$, and $a_k < b_k$, we have $\sum_n a_n < \sum_n a'_n \le \sum_n b_n$.