Let $A$ be a simply connected open set in $\mathbb{R}^2$ with smooth boundary. Define $$A^r := \{x \in \mathbb{R}^2: d(x, A) \le r \},$$ where $d$ is the distance function. Let $P$ denote the perimeter function. Show that $$\lim_{r \rightarrow 0} \frac{P(A^r)-P(A)}{r}= 2\pi.$$
It is certainly true for the case $A$ is convex. Indeed, one can easily show for any convex set $A$, $$P(A^r)-P(A)=2 \pi r, \quad \forall r>0. $$ Generally speaking, if $A$ is just a simply conncected set with smooth or C^1 boundary, I cannot even find a formula to describe $P(A^r). I think this is essentially a calculus problem, a very natural question. Can anyone give me any ideas? Thanks in advance.
Let $\gamma$ be an arclength parametrization of $\partial A$, oriented counterclockwise. Then $\gamma'$ is the unit tangent vector. It's convenient to use complex notation here, so that $-i\gamma'$ is the unit normal pointing to the right of the tangent. The boundary of $A^r$ is traced by $\gamma_r(t) = \gamma(t) - ir \gamma'(t)$. Hence, $$\gamma_r'(t) = \gamma'(t) - i r\gamma''(t)$$ Since $\gamma'$ is unit tangent, $\gamma''$ is a unit normal times curvature $k(t)$. Multiplication by $-i$ aligns it with $\gamma'$ again; thus, $$\gamma_r'(t) = (1+rk(t))\gamma'(t)$$ It remains to integrate $|\gamma_r'|$ to get the length of $\partial (A^r)$; this leads to $$\int |1+rk(t)|\,dt= \int (1+rk(t))\,dt = P(A) +r\int k(t)\,dt $$ when $r$ is small enough. The total curvature of a simple closed curve is $2\pi$.