Looking for a set like the fat Cantor set? Problem:

Assume $0 < \epsilon < 1$ and $\mu_{\mathcal L}$ is the Lebesgue measure. Find a measurable set $A \subset [0, 1]$ such that the closure of $A$ is $[0, 1]$ and $\mu_{\mathcal L}(A) = \epsilon$.

My Attempt:
I quickly started reading about the fat Cantor set, whose essential feature (as far as I know) is that its measure can be any such $\epsilon$. But I think the fatal flaw is that the fat Cantor set is closed, whereas I need a set whose closure is $[0, 1]$.
I don't know where to look for a solution. Can I fiddle with fat Cantor sets and solve this problem or do I need a fundamentally different approach?
Thanks.
 A: The simplest example is something like
$$A = [0,\varepsilon] \cup \left( \mathbb{Q} \cap [0,1] \right). $$
That is, pick your favorite set of measure $\varepsilon$, then throw in the rational numbers.  The rationals are dense in $\mathbb{R}$ (and therefore dense in $[0,1]$), but have zero measure.  Thus
$$\varepsilon = \mu([0,\varepsilon]) \le \mu(A) \le \mu([0,\varepsilon]) + \mu(\mathbb{Q}) = \varepsilon + 0 = \varepsilon, $$
but $\overline{A} = [0,1]$, which seems to be the desired result.

Another interesting set is something like the following:

Let $\{q_j\}_{j=1}^{\infty}$ be an enumeration of the rational numbers in $[0,1]$, fix $\varepsilon > 0$, and let
$$ B_j = B(q_j, \varepsilon 2^{-j-1}) $$
denote the (open) ball of radius $2^{-j-1}$ (i.e diameter $2^{-j}$) centered at $q_j$.  Take $A$ to be the union of these balls, i.e.
$$ A = [0,1] \cap \bigcup_{j=1}^{\infty} B_j. $$
Observe that, thanks to the countable subadditivity of the measure, the measure of $A$ is bounded above by $\varepsilon$:
$$ \mu(A)
= \mu\left( \bigcup_{j=1}^{\infty} B_j \right)
\le \sum_{j=1}^{\infty} \mu(B_j)
= \sum_{j=1}^{\infty} \varepsilon 2^{-j}
= \varepsilon.$$
Moreover, $A$ contains all of the rational numbers in $[0,1]$, so its closure is $[0,1]$.

This example does not quite meet the requirements of the problem (we cannot really bound the measure of the set from below, as the balls may overlap, or may not be entirely contained in $[0,1]$), but it is (I think) an interesting example of what can happen—this is an open set which has measure as small as we like, which is nevertheless dense in $[0,1]$.
