Let $\lambda( \cdot )$ denote Lebesgue measure on $[0,1]$. Let $(A\_n)\_{n=1}^\infty$ be a decreasing sequence of Borel subsets of $[0,1]$ such that $\bigcap\_{n=1}^\infty A\_n = \emptyset$. Given $\epsilon > 0$ does there exist an open set $U \subseteq [0,1]$ such that 


(i) $\lambda(A\_n \setminus U) = 0$ for some $n$,


(ii) $\lambda(U) \leq \epsilon$, and


(iii) $\lambda(\partial U) = 0$,


where $\partial U$ is the boundary of $U$?


I don't have a good intuition whether this should be true or not. One idea might be to cover some $A\_n$ by an open set with measure much smaller than $\epsilon$, then hope that we can expand that open set somewhat to make its boundary null.



Is every set of small measure contained in an open set of small measure with null boundary?
No, such a $U$ does not exist in general. You can have (i)+(ii), but (iii) is too much to ask. The problem is that the $A\_n$ might all be dense in $[0,1]$ so that $\partial U = \overline{U} \setminus U =[0,1]\setminus U$ has measure $\geq 1-\epsilon$ by (ii).


