Let $A$, $B\subset \mathbb R^N$ be given such that $A\subset B$. Assume that $\mathcal H^{N-1}(B\setminus A)<\epsilon$ where $\epsilon>0$ is a fixed constant and $\mathcal H^{N-1}$ is the $N-1$ dimensional Hausdorff measure (so we may think $A$ and $B$ are two curves embedded in $\mathbb R^N$). Moreover, we know that $\mathcal H^{N-1}(\overline {B}\setminus B)<\epsilon$ where $\overline{B}$ denotes the closure of set $B$, and $\mathcal H^{N-1}(\bar A\setminus A)=0$.
My question: do we have $$ \mathcal H^{N-1}(\overline {B\setminus A})<2\epsilon $$ hold?
Update: I added an assumption on $A$ such that $A$ is compact, i.e., $$ \mathcal H^{N-1}(\overline {A}\setminus A^{\circ})=0 $$ where by $A^\circ$ we mean the interior of $A$. That is, I assume that $\mathcal H^{N-1}(\partial A)=0$.
Hence, I may write \begin{align} \mathcal H^{N-1}(\overline {B\setminus A})\leq\mathcal H^{N-1}(\overline {\overline {B}\setminus A^\circ}) = \mathcal H^{N-1}( {\overline {B}\setminus A^\circ})\\ \leq\mathcal H^{N-1}( {\overline {B}\setminus B})+\mathcal H^{N-1}( {{B}\setminus A})+\mathcal H^{N-1}( {{A}\setminus A^\circ})\leq 2\epsilon. \end{align}
PS: I understand that $A^\circ$ might be ill-defined... I am trying to work out a fix.
Hint:
The answer is no because by formula between closure and boundary $$ \overline {A}=A\cup (\partial A-A) $$ We have $$ \overline {B-A}=(B-A)\cup (\partial (B-A)-(B-A)) $$ Given $\mathcal H^{N-1}(B-A)<\epsilon$, plus $\mathcal H^{N-1}(\overline {B}\setminus B)<\epsilon$ and $\mathcal H^{N-1}(\bar A\setminus A)=0$, it can not prevent measure of $\partial (B-A)$ from getting big, particular if $B-A$ is dense like subset of $\Bbb{Q}$.
Edit:
Here is a fix. Just assume $\mathcal H^{N-1}(\partial (B-A))<\epsilon$. Then you can prove it.