This question assumes familiarity with combinatorial cardinal characteristics of the continuum.


Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It is known that
$$\operatorname{cov}(\mathcal{N})\cdot\operatorname{cov}(\mathcal{M})\le 
\operatorname{cov}(\mathcal{E})\le 
\operatorname{cov}(\mathcal{N})\cdot \mathfrak{d},$$
and that the first inequality is consistently strict
([Bartoszynski-Shelah](http://ac.els-cdn.com/016800729290001G/1-s2.0-016800729290001G-main.pdf?_tid=25592cb0-a207-11e5-a909-00000aacb35e&acdnat=1450058908_add3397e9133b524d167bbf5b4ee89be)).


Is the second inequality consistently strict?


I conjecture that the answer is positive, and known, but did not find it in the cited paper (or elsewhere, but I didn't search thoroughly).


**Update:** This problem is solved below by Ashutosh, and the solution suggests a [follow-up question](https://mathoverflow.net/questions/226159/how-many-closed-measure-zero-sets-are-needed-to-cover-the-real-line-really?lq=1).



How many closed measure zero sets are needed to cover the real line?
Since we can cover the reals by at most $\mathfrak{r}$ (reaping number) closed null sets, we can do a countable support iteration of rational perfect set forcing over a model of CH to get a model of $\omega\_1 = \mathfrak{r} < \mathfrak{d} = \omega\_2$. 


To see that $\operatorname{cov}(\mathcal{E}) \leq \mathfrak{r}$, just note that for every infinite $A \subseteq \omega$, $i \in \{0, 1\}$, the set $N\_{A, i} = \{x \in 2^{\omega} : (\forall n \in A) (x(n) = i)\}$ is closed null.


$\mathfrak{d} = \omega\_2$ in this model because Miller real is not dominated by any ground model real. Also the ground model p-points are preserved so $\mathfrak{r} = \mathfrak{u} = \omega\_1$. The proof of this fact can be found in chapter 23 of Halbeisen's Combinatorial set theory.


