Set has Measure Zero iff There is a Collection of Intervals I'm stuck on the following real analysis problem. 

Show that a set $E$ of real numbers has Lebesgue measure zero if and only if there is a countable collection $\{I_{k}\}_{k=1}^{\infty}$ of open intervals for which each point in $E$ belongs to infinitely many of the $I_{k}$'s and $\sum_{k=1}^{\infty}\ell(I_{k})<\infty$, where $\ell(I_{k})$ is the length of the interval $I_{k}$.

I think the converse implication $(\Leftarrow)$ should be trivial by the definition of the Lebesgue measure, but I'm not sure. I also think that I should be using the Vitali Covering Lemma somewhere (possibly in the forward implication?). Thanks in advance for any help!
 A: Unless I am missing something, the converse direction seems fairly straightforward to me.
As Andrés suggests, take $D = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty I_m$.  Then $x \in D$ iff $x$ is in infinitely many $I_m$, so $E \subseteq D$.  On the other hand $D$ is obviously measurable (even Borel), and by the Borel-Cantelli lemma (whose proof is elementary and works in any measure space) we have $m(D) = 0$, i.e. $D$ is a null set.  Now it is a standard fact that any subset of a measure-zero set is measurable and measure-zero (i.e. Lebesgue measure is complete), so we are done.
(One can show this last fact by noting that outer measure $m^*$ is monotone as an immediate consequence of its definition, so $m^*(E) \le m^*(D) = 0$.  And whatever definition of measurability you use, a set with outer measure zero is measurable.)
This kind of construction comes up a lot in probability.  There is a standard shorthand notation for $\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty I_m$: it's often denoted as $\limsup I_m$ or $\{I_m \text{ i.o.}\}$ ("infinitely often").
A: I'm posting 2 years late, unfortunately. I was brought to this post by some introductory remarks by Riesz and Szőkefalvi-Nagy in their book Functional Analysis that I didn't understand, namely their treatment of the forward direction of this statement. Here is the quote:

It will be sometimes be advantageous to give this definition the following form. A set $E$ is of measure zero if it can be covered by a sequence of [presumably countable] intervals of finite total length in such a way that every point of $E$ is an interior point of an infinite number of these intervals. The two definitions are equivalent [referring to the usual elementary definition of measure zero]. The second implies the first since, when all the points of E belong to an infinite number of intervals of finite total length, we can decrease this total length at will by supressing a finite number of these intervals.

They immediately discuss the forward direction as well, but this is discussed in appreciable detail in Andrés's answer.
I was confused about this statement for a long while, so I turned to the wonderful world of StackExchange for help.
After reading the proof of the Borel-Cantelli Lemma, I realized that what the two authors were saying above was completely correct. But it lacks detail which I wanted to fill for this answer. Namely, there is absolutely no need to introduce the set
$$D = \bigcap_{n = 1}^{\infty}\bigcup_{m = n}^{\infty}I_m,$$
as introduced in the other two answers.
As given, we have a countable collection $\{I_{k}\}_{k=1}^{\infty}$ of open intervals for which each point in $E$ belongs to infinitely many of the $I_{k}$'s and $\sum_{k=1}^{\infty}\ell(I_{k})<\infty$, where $\ell(I_{k})$ is the length of the interval $I_{k}$.
But since the sequence of partial sums converges, the sequence of tails of this sum 
$$\sum_{k = n}^{\infty}\ell(I_k)$$
also converges to $0$. The key observation is that the countable collection $\{I_{k}\}_{k=n}^{\infty}$ will cover every point in $E$, since every point in $E$ is covered infinitely many times, and the new collection is just the old collection without finitely many of the intervals. So by truncating finitely many terms of the sum we can find a subcovering for $E$ which is as small as we would like.
A: This is a nice problem, it turned out more interesting than I first thought.
Suppose first that $E$ is a measure zero set, and let's show that there is such a sequence of intervals. Since $E$ is measure zero, for each $\epsilon>0$ there is a sequence $\mathcal I_\epsilon$ of open intervals, the sum of whose lengths adds up to less than $\epsilon$. Consider these sequences for all $\epsilon$ of the form $1/2^n$, $n=0,1,\dots$. Since the countable union of countable sets is countable, we can rearrange the collection of all these intervals in a single sequence, and the sum of their lengths is bounded above by $\sum_{n=0}^\infty1/2^n<\infty$. It remains to argue that this sequence is as desired, that is, each point belongs to infinitely many of the intervals in it. To see this, consider a point $a\in E$. For each $n$ pick some interval $I_n\in\mathcal I_{1/2^n}$ with $a\in I_n$. For any given $n$, this interval has some length $\epsilon_n>0$, and we can find $m>n$ such that $1/2^m<\epsilon_n$, which means that the interval $I_m$ is different from $I_n$, since its length is strictly smaller. Since this holds for each $n$, we can easily find a sequence $n_0<n_1<\dots$ such that the intervals $I_{n_0},I_{n_1},\dots$ are all distinct. Since $a$ is in all of them, and $a\in E$ was arbitrary, we are done.
The converse requires one idea, which I confess is what made the problem attractive to me. 
We are given a sequence of open intervals $(I_n)_{n\ge0}$, the sum of whose lengths is finite, and a set $E$ with the property that all its points belong to infinitely many intervals in the sequence. I will argue by contradiction that $E$ has measure zero. Accordingly, suppose that this is not the case. Let's start with a simple observation. Suppose first that $E$ is compact. For each $n$, since all points of $E$ belong to infinitely many of these intervals, we have that $E\subseteq \bigcup_{k>n}I_k$. Since $E$ is compact, we can then find $m>n$ such that in fact $E\subseteq_{k=n+1}^m I_k$. From this we conclude that there is a sequence $n_0<n_1<\dots$ such that for all $k$, $$E\subseteq\sum_{n_k<m<n_{k+1}}I_m$$ and therefore, denoting by $\ell(I)$ the length of the interval $I$ and letting $m$ denote Lebesgue measure, we have that $$\sum_n \ell(I_n)\ge\sum_k\sum_{n_k<m<n_{k+1}}\ell(I_m)\ge\sum_k m(E)=+\infty.$$
This is a contradiction, thus showing that, if $E$ is compact, it must indeed have measure zero, as wanted.
Of course, compact sets are very special, but it follows at once that the result holds not just when $E$ is compact but even if all we know is that it is Lebesgue measurable. The reason is that, by inner regularity of Lebesgue measure, if $E$ is Lebesgue measurable and has positive measure, then it contains a compact subset $F$ of positive measure, and the argument above gives us a contradiction. 
It is here that an idea is needed because, for all we know, it could be that not only is $E$ not null, but in fact it could even fail to be measurable. In order to deal with this case, we replace $E$ with a "well-behaved" larger set. Namely, let $D=\{x\in\mathbb R:$ there are infinitely many $n$ such that $x\in I_n\}$. Clearly, $E\subseteq D$. But $D$ is nicer, because it is Borel and therefore Lebesgue measurable. Indeed, $$ D=\bigcap_{n\ge 0}\bigcup_{m>n}I_m: $$ A point $x\in\mathbb R$ is in the set in the right-hand side of this equation if and only if for all $n$ there is an $m>n$ such that $x\in I_m$, that is, if and only if $x$ is in infinitely many of the intervals, which is to say that $x\in D$. And it follows that $D$ is Borel because, by design, the set in the right-hand side is Borel (in fact, $G_\delta$).
We are done: By the argument above, since $D$ is measurable, and all its points belong to infinitely many of the $I_n$ (by definition), then $D$ has measure zero Since $E\subseteq D$, we conclude that $E$ also has measure $0$.

This characterization of measure zero sets is useful in practice. I had encountered a variant before, in the study of strong measure zero sets, see here.

As Nate pointed out, the converse direction can be solved more simply and in greater generality if we start directly with the observation that $E\subseteq D$,  $D$ is measurable, and it suffices to show that $D$ has measure 0. For this, simply note that for any $k$, $$ m(D)=m\left(\bigcap_{n\ge 0}\bigcup_{m>n}I_m\right)\le m\left(\bigcup_{m>k}I_m\right)\le\sum_{m>k}\ell(I_m)\to_{k\to\infty}0,$$ since the tails of a convergent series converge to 0, and we are done. 
As Nate pointed out as well, this direction is a (consequence of a) particular case of the Borel-Cantelli lemma, which is what I just proved. Note this proof does not use that $m$ is Lebesgue measure or that the $I_m$ are intervals. All we need is that $m$ is a positive measure on some measure space, and that $\sum_n m(I_n)<+\infty$.
A: Let $O$ be the set of open real intervals. Let $m$ denote Lebesgue measure.
$(1).$ If $E\subset \cup S$ where (i) $\;S= \{I_n\}_{n\in \Bbb N}\subset O$ with $ \sum_{n=1}^{\infty}\ell(I_n)\in \Bbb R,$ and (ii) $\;$ every $x\in E$ belongs to infinitely many members of $S,$ then:
For $m\in \Bbb N$ let $B(m)=\sum_{n=m}^{\infty}\ell(I_n).$ Then $\lim_{m\to \infty} B(m)=0$ because $\sum_{n=1}^{\infty}\ell(I_n)$ converges.
For every $m\in \Bbb N$ we have $E\subset \cup_{n\ge m}I_n$ so the outer measure $\ell^o(E)$ satisfies $$\ell^o(E)\le \ell(\cup_{n\ge m}I_n)\le \sum_{n=m}^{\infty}\ell(I_n)=B(m).$$
Since $\lim_{m\to \infty}B(m)=0,$ therefore $\ell^o(E)=0.$ Hence $E$ is measurable with $m(E)=0.$
(2a). If $m(E)=0$ and $E\ne \emptyset:$
We recursively define a family $\{C_n\}_{n\in \Bbb N},$ where each $C_n$ is a  countable collection of non-empty bounded members of $O,$ as follows:
Let $U_1$ be open with $ m(U_1)<1/2$ and $E\subset U_1=\cup C_1$   where $C_1$ is a countable pairwise-disjoint subset of $O\setminus \{\emptyset\}.$ So $2^{-1}>m(U_1)=\sum_{s\in C_1}\ell(s).$
For $n\in \Bbb N$ let $U_{n+1}$ be open with $E\subset U_{n+1}$ and $m(U_{n+1})<2^{-n-2}.$ Then $U_{n+1}=\cup D_{n+1}$ where $D_{n+1}$ is a countable  pairwise-disjoint subset of $O\setminus \{\emptyset\},$ so $2^{-n-2}>m(U_{n+1})=\sum_{s\in D_{n+1}}\ell(s).$ 
Let $P(n)$ be  the set of endpoints of the members of $\cup_{i=1}^nC_i.$  Each $s\in D_{n+1}$ is bounded and non-empty, while $P(n)$ is only countable, so for each $(a,b)\in D_{n+1}$ take $a',b'\in \Bbb R$ with $a'\le a$ and $b\le b',$ such that $b'-a'\le 2(b-a)$ and such that $\{a',b'\}\cap P(n)=\emptyset.$
Now let $C_{n+1}=\{(a',b'):(a,b)\in D_{n+1}\}.$ We have $E\subset \cup C_{n+1}$ with $\sum_{s\in C_{n+1}}\ell(s)<2^{-n-1}$ and $C_{n+1}\cap C_i=\emptyset$ for all $i\le n.$
Finally let $\{I_n\}_{n\in \Bbb N}=\cup_{n\in \Bbb N} C_n.$ The remaining details are left to the reader.
(2b). The case $E=\emptyset$ is trivial.
