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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.

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