# Isoperimetric inequality and geometric measure theory

The following version of the isoperimetric inequality can be easily deduced from the Brunn-Minkowski inequality:

Theorem. If $$K\subset\mathbb{R}^n$$ is compact, then $$|K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}\mu_+(K),$$ where $$\omega_n$$ is the volume of the unit ball and
$$\mu_+(K)=\liminf_{h\to 0} \frac{|\{x:\, 0<{\rm dist}\, (x,K)\leq h\}|}{h}$$ is the Minkowski content.

If $$K$$ is the closure of a bounded set with $$C^2$$ boundary, then $$\mu_+(K)=H^{n-1}(\partial K)$$ (Hausdorff measure) so we have $$|K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K).$$ However, the above inequality is true for any compact set without assuming anything about regularity of the boundary. My question is:

## Is there a simple proof of the following result?

Theorem. If $$K\subset\mathbb{R}^n$$ is compact, then $$|K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K).$$

This result can be proved using the machinery of the geometric measure theory. If $$H^{n-1}(\partial K)=\infty$$, the inequality is obvious. If $$H^{n-1}(\partial K)<\infty$$, then $$K$$ has finite perimeter and the isoperimetric inequality for sets of finite perimeter yields $$|K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}P(K)= n^{-1}\omega_n^{-1/n} H^{n-1}(\partial^* K)$$ where $$P(K)$$ is the perimeter of $$K$$ and $$\partial^*K\subset\partial K$$ is the reduced boundary. For details see [1].

Unfortunately, this argument is very far from being elementary.

[1] Ambrosio, L., Fusco, N., Pallara, D.: Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000.