Suppose $A$ is a simply connected open set in $\mathbb{R}^2$ with rectifiable Jordan boundary. Let $\gamma:[0,1] \rightarrow \mathbb{R}^2$ be the parametrization of $\partial A$. Let $A^r$ denote the $r$-neighborhood of $A$, that is, $A^r:=\{x: d(x,A) \le r\}$. Let $0=t_0<t_1< \cdots <t_n=1$ be the partition of $[0,1]$ such that $P(A_n) \rightarrow P(A)$ where $P$ is the perimeter function and $A_n$ is the polygon with verticies $\gamma(t_i), i=1,2,...n$. The construction of $A_n$ can be realized since $\partial A$ is rectifiable and $P(A)=sup_{0=t_0<t_1< \cdots <t_n=1} \sum_{I=1}^{n} |\gamma(t_i)-\gamma(t_{i-1})|$. We can also make sure $A_n \Delta A \subset (\partial A)^{\frac{1}{n}}$ and thus $A_n \rightarrow A$ in $L^1$.
My question is, based on the previous construction, can we conclude $A_n^r \rightarrow A^r$ in $L^1$ and $P(A_n^r) \rightarrow P(A^r)$? If one of them is true then I will be very happy. However even if the picture is very intuitive, I cannot come up with a rigorous proof. Especially I think the first claim is so intuitive, but strangely I still have no idea how to give a rigorous proof.
The question comes from my random doodling in a geometric measure theory class. By the way, I have proved $P(A^r) \le P(A)+2\pi r$ for any simply connected polygon $A$ and any $r>0$.
Any comments or ideas will be really appreciated.