Is the upper limit of sets empty? Let $E_k^n$ be measurable subset of $[0,1]$ for any natural numbers $k$ and $n$.
For a fixed $n$ we have
$$E_1^n\supseteq E_2^n\supseteq
E_3^n\supseteq\ldots
E_{k}^n\supseteq E_{k+1}^n\supseteq\ldots,$$
$$\bigcap_{k=1}^{\infty}E_{k}^n=\emptyset.$$
My question is how can be found such $\sigma: N\to N$ that
$$\limsup_{n\to\infty}E_{\sigma(n)}^n=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_{\sigma(k)}^k=\emptyset.\,?$$
I think that such $\sigma$ is exists, but can't find it. I can take $\sigma$ such that $mes E_{\sigma(n)}^n < 2^{-n}$, and I'll get
$$mes\left(\limsup_{n\to\infty}E_{\sigma(n)}^n \right)=0,$$
but I needn't it.
Thanks for help.
 A: There need not be any such $\sigma$.  To construct a counterexample, begin with some set $C\subseteq[0,1]$ that has measure zero but has the cardinality of the continuum.  Since the number of functions $\sigma:N\to N$ is also the cardinality of the continuum, fix a bijection $b$ from $C$ to the set of all such functions $\sigma$.  So, for any $x\in C$ and any $n\in N$, we have a natural number $b(x)(n)$.  Now define 
$$
E^n_k=\{x\in C\mid b(x)(n)\geq k\}.
$$
Notice first that all these sets $E^n_k$ are measurable, because they are subsets of $C$ and therefore have measure zero.  It is clear from the definition that $E^n_k\supseteq E^n_{k+1}$.  For any fixed $n$, the intersection $\bigcap_{k=1}^\infty E^n_k$ consists of those $x\in C$ for which the natural number $b(x)(n)$ is greater than or equal to every natural number $k$; that's absurd, so there is no such $x$, and the intersection is empty.  So all your hypotheses are satisfied by these $E^n_k$'s, and it remains to check that there is no $\sigma$ as in your desired conclusion.
So consider any $\sigma:N\to N$.  It is $b(x)$ for some (unique) $x\in C$.  For this $x$ and arbitrary $n\in N$, we have $b(x)(n)\geq \sigma(n)$ (in fact, of course, we have equality, as $b(x)=\sigma$), and so $x\in E^n_{\sigma(n)}$.  Since we have this for all $n$, we have in particular that $x\in\limsup_nE^n_{\sigma(n)}$.
