If set $A_n$ converges to $A$ in $L^1$, then is it also true for the $r$-neighborhood? Let $A$ be a bounded simply connected set in $\mathbb{R}^2$, and $A^r$ denote the $r$-neighborhood of $A$, that is, $A^r:=\{x: d(x,A) \le r\}$. Suppose there exists $A_n \rightarrow A$ in $L^1$, that is, $|A \Delta A_n| \rightarrow 0$, then prove $A_n^r \rightarrow A^r$ in $L^1$. 

It seems to be very intuitive based on the picture, but strangely I've no idea how to make a rigorous proof. Any comments will be really appreciated.



Thanks to the answer provided by @zhw, I realized I had asked a trivial question. Later I asked another question here On the $r$-neighborhood of a set in $\mathbb{R}^2$. to fully make sense of the background and my concern.
No, here's a counterexample: Let $A = D(0,1),$ and $A_n = D(0,1) \cup \{(x,y): |y| &lt; (e^{-x^2}) /n\}.$ Then $A_n \to A$ in $L^1$ (in your sense), but $|A_n^r| = \infty$ for every $r$ and $n.$
No, here's a counterexample: Let $A = D(0,1),$ and $A_n = D(0,1) \cup \{(x,y): |y| &lt; (e^{-x^2}) /n\}.$ Then $A_n \to A$ in $L^1$ (in your sense), but $|A_n^r| = \infty$ for every $r$ and $n.$