Sets of null (Lebesgue)-measure and sigma compacts

Question

Let $E\subset{\mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure.

¿It is possible to find $F\subset{\mathbb R}$ such that $E\subset F$, $m(F)=0$ and $F$ is $\sigma$-compact (that is, numerable union of compact sets)?

Obviously if $E$ is numerable, this is trivial.

Moreover, if $m(\overline{E})=0$ (where $\overline{E}$ is the clousure of $E$), it is also true. Just take $F=\overline{E}$ and $\overline{E}=\bigcup_{n\in\mathbb N} \overline{E}\cap[-n,n]$.

IS there a null-set $E$ such that $m(\overline{E})>0$ but $E$ cannot be covered by a null-sigma compact set?

I am also interested when $E\subset\mathbb T$, where $\mathbb T$ is the unit circle.

Answer 1

The answer is no. There exists counterexample. The detailed answer for the related question is here.

A short answer from Andreas Blass:

A closed set of Lebesgue measure zero has empty interior. So a countable union of such sets, a $_$ of measure zero, is meager (also called "first Baire category), and so are all its subsets. But there are Lebesgue null sets that are not meager, for example, the set of those numbers in $[0,1]$ whose binary expansion does not have asymptotically half zeros and half ones (i.e., those numbers whose binary expansions violate the strong law of large numbers).