$E_{k} \subset [0,1]$ such that $\lim_{k \to \infty} m(E_{k}) = 1$ but $\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k} = \phi $ I am trying to find an example of collection $\left \{ E_{k} \right \}_{k=1}^{\infty}$ such that each $E_{k} \subset [0,1]$ satisfying $\lim_{k \to \infty} m(E_{k}) = 1$ but $\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k} = \phi$
What I tried was to construct a collection with monotonically increasing measure tending to 1. But such collections do not posses the property $\bigcap_{k=n}^{\infty}E_{k} = \phi$ for each positive integer n. How should I approach to construct such example?
 A: For each integer $n \geq 2,$ let ${\mathcal E}_n$ be the collection of the $n$ many sets $\;[0,1]-[0,\;\frac{1}{n}],\;\;$ $[0,1]-[\frac{1}{n},\;\frac{2}{n}],\;\;$  $[0,1]-[\frac{2}{n},\;\frac{3}{n}],\;\;$ $\ldots,\;$  $\;[0,1]-[\frac{n-1}{n},\;1].\;$
Now define the sequence $\{E_k\}$ as follows. First, put $E_1 = [0,1].$ Next, let $E_2$ and $E_3$ be the $2$ sets in ${\mathcal E}_2,$ ordered however you wish. Next, let $E_4$ through $E_6$ be the $3$ sets in ${\mathcal E}_3,$ ordered however you wish. Next, let $E_7$ through $E_{10}$ be the $4$ sets in ${\mathcal E}_4,$ ordered however you wish. Continue in this way. Clearly, the measure of $E_k$ approaches $1$ as $k \rightarrow \infty.$ Also, since infinitely often we're excluding every point from $[0,1],$ we have $\liminf_{n \rightarrow \infty} E_k = \emptyset.$
A: There's a classic example.  Let $E_k=[0,1]\setminus F_k$, where $F_k=\left[\sum_{n=1}^k\frac{1}{n},\sum_{n=1}^{k+1}\frac{1}{n}\right] \operatorname{mod}1$.  Mod 1 means to move points over by whole numbers until you're in $[0,1]$.
Clearly $m(E_k)=1-\frac{1}{k+1}\to1$.  Meanwhile, if there's any point, $x$, in $\cup\cap E_k$, then there's some tail for which $x+\mathbb{N}$ avoids all $F_k$, but the $F_k$ keep going forever covering all reals (since the summation diverges) without missing any numbers, so that couldn't happen.
