Covering measure one sets by closed null sets (The following question arose in a joint research with Adam Przeździecki and Boaz Tsaban.)
For a $\sigma$-ideal $\mathcal{I}$ of subsets of the unit interval 
$[0,1]$, define
$$\newcommand{\card}[1]{\left|#1\right|}\newcommand{\cov}{\operatorname{cov}}\newcommand{\sub}{\subset} \cov(\mathcal{I}):=\min\{\card{\mathcal{A}}: \mathcal{A}\sub \mathcal{I}, \bigcup\mathcal{A}=[0,1]\}.
$$
Let $\mathcal{E}$ be the family of all $F_\sigma$ Lebesgue null subsets of the unit interval, and $\mathcal{N}$ be the family of all Lebesgue null subsets of the unit interval.


*

*Let $\kappa_\mathcal{E}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{E}$ elements of $\mathcal{E}$.

*Similarly, let $\kappa_\mathcal{N}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{N}$ elements of $\mathcal{N}$.


We have $\kappa_\mathcal{E}\leq \cov(\mathcal{E})$, and $\kappa_\mathcal{N}=\cov(\mathcal{N})$.
Question 1: Is it provable that $\kappa_\mathcal{E}=\cov(\mathcal{E})$?
Question 2: Does the cardinal number $\cov(\mathcal{E})$ change if we work in a closed positive subset of the unit interval instead of the whole interval?
A negative answer for Question 2 implies a positive answer for Question 1.
 A: The answer to the first question is yes: it is always true that $\kappa_{\mathcal E} = \mathrm{cov}(\mathcal E)$. The answer to the second question is no.
As Piotr points out, a negative answer to the second question implies a positive answer to the first. [This is because if $A \subseteq [0,1]$ is a set of measure one, then it contains a closed $K \subseteq [0,1]$ of positive measure (indeed, we can get the measure of $K$ as close to $1$ as we like). A negative answer to the second question means that it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $K$; since $K \subseteq A$, it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $A$ too.]
So it suffices to show that the answer to the second question is no: the $\mathcal E$-covering number is the same for any positive-measure closed subsets of $[0,1]$.
To see this, fix a closed $K \subseteq [0,1]$ of positive measure $\alpha$. Let $\mathcal F$ be a family of sets in $\mathcal E$ that covers $K$. Our goal is to cover $[0,1]$ with a family of sets in $\mathcal E$ of size $|\mathcal F|$. This shows that the $\mathcal E$-covering number for $[0,1]$ is no bigger than the analogous number for $K$. It is obvious that the $\mathcal E$-covering number for $[0,1]$ is no smaller than the analogous number for $K$, since any cover of $[0,1]$ by sets in $\mathcal E$ is also a cover of $K$. 
Define a map $\phi: K \rightarrow [0,1]$ by
$$\phi(x) = \frac{1}{\alpha} \cdot m(K \cap [0,x])$$
for all $x \in K$, where $m$ denotes the Lebesgue measure. Using the fact that $K$ is closed, one can show that $\phi$ is surjective. In particular,
$$\phi(\mathcal F) = \{\phi[A \cap K] : A \in \mathcal F\}$$
is a family of sets that covers $[0,1]$. To finish the proof, it is enough to show that the map $A \mapsto \phi[A \cap K]$ sends sets in $\mathcal E$ to sets in $\mathcal E$: then $\phi(\mathcal F)$ is the desired $\mathcal E$-cover of $[0,1]$.
From the definition of $\phi$, we see
$$m(\phi[(0,b)]) = \frac{1}{\alpha} \cdot m(K \cap (0,b))$$
and taking complements we get
$$m(\phi[(a,1)]) = \frac{1}{\alpha} \cdot m(K \cap (a,1)).$$
and taking intersections, we see that for any interval $(a,b) \subseteq [0,1]$,
$$m(\phi[(a,b) \cap K]) = \frac{1}{\alpha}(m((a,b) \cap K)).$$
That is, $\phi$ can increase the measure of an interval by at most a factor of $\frac{1}{\alpha}$. With a little more (routine) work, this shows that the map $A \mapsto \phi[A \cap K]$ sends null sets to null sets.
Observe that $\phi$ is continuous (the inverse image of an open interval in $[0,1]$ is an open interval in $K$). Therefore $\phi$ maps compact sets to compact sets. Therefore the map $A \mapsto \phi[A \cap K]$ sends compact sets to compact sets. Since $[0,1]$ is compact, this means that it also sends $F_\sigma$ sets to $F_\sigma$ sets. So we have showed that the map $A \mapsto \phi[A \cap K]$ sends null $F_\sigma$ sets to null $F_\sigma$ sets, and we are done.
