A Cantor-type set Let $A=\{\sum_{k=1}^\infty \frac{a_k}{k!} : a_k\in\{0,\,1\}\}$.
We can prove that $A$ is a closed set with $\operatorname{int}(A)=\emptyset$ and Lebesgue measure of $0$.
Is there a $m \in \mathbb{N}$ s.t. $A+A+\cdots+A$ ($m$ times) is of positive Lebesgue measure?
(Note that for the Cantor set $C$ we have $C+C=[0,\,2]$.)
 A: Here is a reference...
Erdős, Pál; Volkmann, B., Additive Gruppen mit vorgegebner Hausdorffscher Dimension [Additive groups with given Hausdorff dimension], J. Reine Angew. Math. 221, 203-208 (1966). ZBL0135.10202.
For each $\alpha \in (0,1)$, they define an additive [sigma-compact] subgroup $G(\alpha)$  of $\mathbb R$ with Hausdorff dimension $\alpha$.  [For more detail, see below.]  For every $\alpha$, your set $A$ is a subset of $G(\alpha)$.  Since $G(\alpha)$ is a group, $A+A+\dots +A \subseteq  G(\alpha)$.  So $A+A+\dots+A$ has Hausdorff dimension at most $\alpha < 1$, and thus contains no interval.
[Of course, this is true for all $\alpha \in (0,1)$, so $A+A+\dots+A$ has Hausdoff dimension $0$.]

More detail from the Erdős & Volkmann paper.  Each real number $x$ has a unique "Cantor expansion" of the form
$$
x = \lfloor x \rfloor + \sum_{k=2}^\infty \frac{a_k(x)}{k!}
\tag1 $$
where $a_k(x) \in \mathbb Z$ and $0 \le a_k(x) < k$.  Fix $\alpha \in (0,1)$.  Let $G(\alpha)$ be the set of all reals $x$ such that:
there exists $\kappa = \kappa(x) > 0$ and $k_0 \in \mathbb N$ such
that, for all $k \ge k_0$
\begin{align}
\text{either}\qquad & a_k(x) \le \kappa\;k^\alpha
\\ \text{or}\qquad & a_k(x) \ge k - \kappa\;k^\alpha .
\tag 2\end{align}
Satz 1. $G(\alpha)$ is an additive group with $\dim G(\alpha) = \alpha$.
The proof that $G(\alpha)$ has dimension $\alpha$ is relatively easy.  The more involved part is the proof that $G(\alpha)$ is a group.  They have to keep track of "carrying" when you add two expansions of the form $(1)$ with restrictions $(2)$.

Plug
My paper with Miller,
Edgar, G. A.; Miller, Chris, Borel subrings of the reals, Proc. Am. Math. Soc. 131, No. 4, 1121-1129 (2003). ZBL1043.28003.
We show a Borel set $\subseteq \mathbb R$ which is a subring is either all of $\mathbb R$ or else has Hausdorff dimension zero.  That was a question asked by Volkmann in 1960.
