Royden 4Ed. Question 6.2.9 I am reading Royden's real analysis 4th edition. There is one problem given by 6.2.9. 
Show that a set $E$ of real numbers has measure zero if and only if there is a countable collection of open intervals $\{I_k\}_{k=1}^\infty$ for which each point in $E$ belongs to infinitely many of the $I_k$'s and $\sum_k l(I_k) <\infty$.
One direction may be immediate from Borel-Cantelli lemma, since $E$ is at most a subset of $\lim\inf I_k$. But how does one construct $I_k$ from given $E$?
 A: Let's assume that $\lambda(E) = 0$. Let $k \in \Bbb N$. By a possible definition of the Lebesgue measure we find a countable sequence $\{ I_n^k \}_{k=1}^\infty$ of open intervals of $\Bbb R$ with 
$$ E \subset \bigcup_{n=1}^\infty I_n^k \qquad \text{and} \qquad \sum_{n=1}^\infty \lambda(I_n^k) < 2^{-k} \; .$$
Note that each $x \in E$ is contained is some $I_n^k$. Doing this for each $k \in \Bbb N$, we get a countable collection $\{ I_n^k \}_{k,n=1}^\infty$, such that each $x \in E$ is contained in infinitaly many $I_n^k$. Furthermore, we have 
$$ \sum_{k=1}^\infty \sum_{n=1}^\infty \lambda(I_n^k) \leq \sum_{k=1}^\infty 2^{-k} \leq 1 < \infty \; .
$$
Let $\{ I_k \}_{k=1}^\infty$ be a sequence of open intervals, such that each $x \in E$ is contained in infinitaly many $I_k$ and 
$$ \sum_{k=1}^\infty \lambda(I_k) < \infty \; $$
Then we have 
$$ E \subset \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty I_n \; . $$
Since $E \subset \bigcup_{n=k}^\infty I_n$, we have 
$$ \lambda(E) \leq \sum_{n=k}^\infty \lambda(I_n) \to 0 \quad (k \to \infty) \; ,
$$
and it follows that $\lambda(E) = 0$.
A: If $l(E) = 0,$ then for each $m\in \mathbb {N},$ there exist intervals $I_{m1},I_{m2},\dots $ that cover $E$ such that $\sum_n l(I_{mn})< 2^{-m}.$ The collection $\{I_{mn}\}$ then has the desired property.
Suppose now $\{I_k\}$ has the property. Then
$$\int_E (\sum \chi_{I_k}) = \int_E \infty = l(E)\cdot \infty \le \int_{\mathbb {R}} (\sum \chi_{I_k}) =  \sum\int_{\mathbb {R}}  \chi_{I_k} = \sum l(I_k) < \infty.$$
It follows that $l(E) = 0.$
