Set of measure zero and limsup of its covering Show that a subset $E$ of $R$ has measure zero iff there exists a sequence of intervals $\{I_{n}\}$ such that $a)\sum\ m(I_{n})<\infty; $ and $b)$ $x $ in $E$ implies $x$ lies in infinitely many $I_{n}$'s .
I am sorry for formatting, I am newcomer to this forum, most of the things seems pretty new to me.
I am comfortable proving one direction from right to left, where I use the Borel-Cantelli lemma to get the result.
While attempting left to right, we can easily get the covering of the set $E$ which has measure less than any chosen epsilon ( by def of outer measure). My problem is showing  for each $x$ in $E$ lies in infinitely many of $I_{n}$'s. That is essentially proving $x$ belong to $\limsup I_{n}$ (I think). I can imagine $x$ lying in many of $I_{n}$ but I do not know how can I be rigorous on that point. 
 A: ($\Rightarrow$): Suppose $mE = 0$. For any $\epsilon>0$, there exists a countable collection of intervals $\{I_n^\epsilon\}_n$ such that $E \subset \cap_n I_n^\epsilon$, and $\sum_n m I_n^\epsilon < \epsilon$. By choosing $\epsilon = \frac{1}{2^m}$, we have a countable collection of intervals for each $m$. The collection of intervals $\{I_n^{\frac{1}{2^m}}\}_{n,m}$ is countable, so we can relabel them as $\{J_n\}$. By construction, each element $x \in E$ lies in infinitely many $J_n$, and we have $\sum m J_n \leq \sum_m \frac{1}{2^m} = 1$.
($\Leftarrow$): Suppose a) & b) are true. Let $\phi(x) = \sum 1_{I_n}(x)$. By assumption, if $x \in E$, we have $\phi(x) = \infty$. However, $\int \phi \leq \sum m I_n < \infty$, and since $\infty \cdot m E \leq \int \phi$, we see that $m E = 0$.
Notes: 


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*For the 'only if' direction, note that for each $m$, the collection $\{I_n^{\frac{1}{2^m}}\}_{n}$ is a covering of $E$. The collection may be overlapping, and it is possible that the collection is the same for different values of $m$ (but the total interval length is $< \frac{1}{2^m}$).

*Here is an explicit formula for relabeling the $\{I_n^{\frac{1}{2^m}}\}_{n,m}$: If $k$ is a positive integer, let $p(k) = \lceil \sqrt{2k+\frac{1}{4}} - \frac{1}{2} \rceil$, and then let $n(k) = p(k)-(k-p(k))$, $m(k) = k-p(k)+1$. Then the mapping $k \mapsto (n(k),m(k))$ is a bijection between $\mathbb{N}$ and $\mathbb{N}^2$. (To see where the mapping $p$ came from, it is loosely an inverse of the mapping $n \mapsto \frac{1}{2}n(n+1)$.) So we let $J_k = I_{n(k)}^{\frac{1}{2^{m(k)}}}$.

