The limit of the measure of a sequence of sets is zero. Let $(X, M, μ)$ be a measure space. I am looking for a sequence of measurable sets such that the limit of the measure of a sequence of sets is zero, but the measure of the limit superior of the sets is not zero. I think the following classic construction will serve my purpose.(I copied it from Wikipedia) Let's define:
$$ E_n=  \left[\frac {j}{2^k}, \frac {j+1}{2^k} \right]$$
where, 
$$ k=\lfloor \log_2 n\rfloor$$
$$ j=n-2^k$$
Intuitively I can feel that the limit superior of the sequence
$\{E_n\}$ is the closed inteval $[0,1]$. However, I am struggling to prove it. Could you please help me? Also, my guess might be incorrect. Thanks so much. Have a great day.
 A: For each $n\in \Bbb N$ let  $D(n)$ be a  finite family of Lebesgue-measurable subsets of $[0,1]$ such that $\cup D(n)=[0,1]$ and such that $0<\min \{\mu(d):d\in D(n)\}\le\max \{\mu(d):d\in D(n)\}\le 2^{-n},$ where $\mu$ is Lebesgue measure.
For example $D(n) =\{[\,k2^{-n},(k+1)2^{-n}\,]: 0\le k<2^{-n};\, k\in \Bbb Z\}.$
Now $D(n)$ has at least $2^n$ members so $\cup_{n\in \Bbb N}D(n)$ is countably infinite. Let $f:\Bbb N\to \cup_{n\in \Bbb N}D(n)$ be any bijection.
Given $\epsilon>0,$ take $m\in \Bbb N$ with $2^{-m}<\epsilon$ and let $n_{\epsilon}=\max f^{-1}(\,\cup_{j=1}^m D(j)\,).$ Now if $n>n_{\epsilon}$ then $f(n)\in D(k)$ for some $k>m,$ so $n>n_{\epsilon}\implies \mu(f(n))<2^{-m}<\epsilon.$
Therefore $\lim_{n\to \infty}\mu(f(n))=0.$
Given $x\in [0,1]$ and $n\in \Bbb N:$ For $n\ge m\in \Bbb N$ choose $g(m)\in \Bbb N$ such that $f(m)\in D(g(m)).$ Let $n'=\max \{g(m):m\le n\}.$
Let $r=\frac {1}{2}\min \{ \mu(d):d\in \cup_{j\le n'}D(j)\}.$ 
Let $n''\ge n+n'$ where $n''$ is large enough that $\max \{\mu(d):d\in D(n'')\}\le r.$ Now  for some $n_0$ we have $x\in f(n_0)\in D(n'')$...(because $\cup D(n'')=[0,1]$ ). Observe that $k\le n \implies \mu(f(k))\ge 2r,$ but $\mu(f(n_0)\le r. So $n_0>n.$
So  $\forall n\in \Bbb N\,\exists n_0>n\,(x\in f(n_0).$
Therefore $\{j\in \Bbb N: x\in f(j)\}$ is infinite so $x\in \lim \sup_nf(n).$
