Approaching the circumference of a circle

We had a lengthy discussion yesterday, on how to prove that the circumference of a circle is $2\pi r$. By using google, the most commonly found proof starts in the following way.

"Consider the regular n-gon inside the circle touching the circle at its vertices, and the regular n-gon outside the circle touching the circle at its edges. Then the circumference of the inner n-gon is smaller than the circumference of the circle, which is smaller than the circumference of the outer n-gon."

While this apparently is true for n-gons, it is not true for arbitrary geometric shapes inside and outside the circle. So we conjectured it was the convexity of the n-gon that is of importance. Which leads to the question:

Let $C$ be a circle with radius $r$, and $A_n$ be a family of convex sets with $A_n \subset C$ for each $n$. If, for $n \rightarrow \infty$, the volume of these sets converges to $\pi r^2$ (the volume of $C$), will the circumference necessarily converge to $2\pi r$?

Yes.

If you don't mind, I'll do the problem in $\mathbb R^d$. Write $B_2^d$ for the unit Euclidean ball (centred at the origin, radius 1). Let $A_n$, as before, be a sequence of convex subsets of $B_2^d$ with $V(A_n)\to V(B_2^d)$, where $V(\cdot)$ denotes volume. Write $$ V_1(K,B_2^d) = V(\underbrace{K,\dotsc,K}_{d-1},B_2^d) $$ for that mixed volume. I will use the following facts about mixed volume:

1. $V_1(K,K) = V(K)$
2. $V_1(K,L)$ is increasing in both arguments.
3. The surface area of a convex body $K$ is $$ S(K) = \lim_{\epsilon\downarrow 0} \frac{V(K+\epsilon B_2^d) - V(K)}{\epsilon} = dV_1(K,B_2^d) $$ (where $K+\epsilon B_2^d$ is the Minkowski sum).
4. Minkowski's first inequality: $$ V(K)^{(d-1)/d} V(L)^{1/d} \le V_1(K,L) $$

Using (4), then (2) (we assumed $A_n\subset B_2^d$), then (1) yields $$ V(A_n)^{(d-1)/d} V(B_2^d)^{1/d} \le V_1(A_n,B_2^d) \le V_1(B_2^d,B_2^d) = V(B_2^d) $$ Sending $n\to\infty$ yields, by the squeeze law, $$ \lim_{n\to\infty} V_1(A_n,B_2^d) = V(B_2^d) = V_1(B_2^d,B_2^d) $$ Multiplying by $d$ and using (3) twice yields $$ \lim_{n\to\infty} S(A_n) = S(B_2^d) $$ as desired.