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!

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").
