$E_{k} \subset [0,1]$ 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?

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.
