Show that if $A$ has measure zero in $R^n$, the sets $\overline{A}$ and $\mbox{Bd}(A)$ need not have measure zero.
Attempt at a solution: I know we could use a counter example but I'm trying to practice proof writing and I was wondering if someone could tell me if my proof works and how I could improve upon it.
Let $a_i \in A \setminus \mbox{Bd}(A)$ and $a_i{'}$ $\in \mbox{Bd}(A)$ If $A$ has measure zero we can cover $A$ with rectangles $Q_1, Q_2, \ldots $ such that $\sum_{i=1}^\infty v(Q_i) < \epsilon$. Let $\mbox{Bd}(A) = B_(a',\epsilon)$. Then adjoin to $Q_i$ the set $Q$ that covers $\mbox{Bd}(A)$. We know from construction of $\mbox{Bd}(A)$ that $V(Q) \ge \epsilon$. Then the total volume of $Q_i + Q$ $\ge \epsilon $ and this implies that $\mbox{Bd}(A)$ and $\overline{A}$ are not necessarily of measure zero.