Is a Borel finite measure on $\mathbb{R}^2$ bounded by "width" absolutely continuous? Consider a finite Borel measure $\mu$ on $\mathbb{R}^2$ such that for each stripe* $S$ of width $\leq \varepsilon$ (with $\varepsilon\geq 0$) one has $\mu(S)\leq \varepsilon$. Show that $\mu$ is absolutely continuous with respect to the Lebesgue measure, or find a counter example.

So far I've only managed to show that $\mu(A) = 0$ for numerable sets or for sets whose projection in one of the two components is zero.  

*A set $S$ is a stripe $S$ of width $\leq \varepsilon$ if there exists a continuous function $f: 
\mathbb{R} \rightarrow \mathbb{R} $ such that $S \subseteq \{(x,y) \in \mathbb{R}^2 | f(x)\leq y \leq f(x)+\varepsilon\} $ or  $ S \subseteq \{(x,y) \in \mathbb{R}^2 | f(y)\leq x \leq f(y)+\varepsilon\}$.
Show that a square with side $\epsilon$ has measure less than $\epsilon$. Then, the bounded circle with radius $\frac{\epsilon}{\sqrt{2}}$ has also measure less than $\epsilon$. 

For each set with Lebesgue measure zero, cover it with a finite union of balls such that the sum of areas is less than some small number- the measure $\mu$ of the cover would be bounded by that small number times $\sqrt{2}$. So by the exterior measure property, the measure $\mu$ of the Lebesgue measure zero set, is also zero.
Show that a square with side $\epsilon$ has measure less than $\epsilon$. Then, the bounded circle with radius $\frac{\epsilon}{\sqrt{2}}$ has also measure less than $\epsilon$. 

For each set with Lebesgue measure zero, cover it with a finite union of balls such that the sum of areas is less than some small number- the measure $\mu$ of the cover would be bounded by that small number times $\sqrt{2}$. So by the exterior measure property, the measure $\mu$ of the Lebesgue measure zero set, is also zero.