To prove the isoperimetric inequality for compact sets ( K \subset \mathbb{R}^n ) using Hausdorff measure, we proceed as follows:

Key Definitions and Setup

• Minkowski Content: For a compact set ( K ), the exterior Minkowski content is defined as
  [ \mu_+(K) = \liminf_{h \to 0} \frac{| { x : 0 < \text{dist}(x, K) \leq h } |}{h}. ]
  This measures the "rate" at which the v...

Yes. You can keep the sharp constant from the Brunn–Minkowski proof and dispense with the finite-perimeter machinery by using a general (and fairly elementary) comparison between the outer Minkowski content and the Hausdorff measure of the topological boundary:

Fact. For every compact K ⊂ R^n, μ+(K) ≤ H^{n−1}(∂K).

Granted this, your inequality follows immediately from the Brunn–Minkowski/Minkows...

Here is the shortest proof one can give that avoids nobody’s favourite heavy machinery but still really uses the basic ideas of “perimeter = Hausdorff‐measure of the reduced boundary” and the usual isoperimetric‐inequality for finite‐perimeter sets. In fact one shows

(i) If (H^{n-1}(\partial K)<\infty) then (K) has finite perimeter and
[ P(K);=;H^{n-1}(\partial^*K)...

The inequality you want to prove

|K|^{\frac{n-1}{n}}\le \frac{1}{n,\omega _n^{1/n}};H^{,n-1}(\partial K) \qquad (1)

for every compact set K ⊂ ℝⁿ is nothing other than the usual isoperimetric inequality written with the geometric – i.e. Hausdorff – surface measure instead of the Minkowski (outer) perimeter μ₊(K).
At first sight it therefore looks as if one should only have to show

μ₊...

<think> We are asked to provide a simple proof of the isoperimetric inequality for an arbitrary compact set in R^n, using the Hausdorff measure on the boundary. The standard proof uses the theory of sets of finite perimeter, which is non-elementary. The goal is to find an elementary proof.

The inequality to prove is: |K|^{(n-1)/n} ≤ n^{-1} ω_n^{-1/n} H^{n-1}(∂K) for any compact set K ⊂ R^n, whe...

This is a great question that touches on the heart of why geometric measure theory (GMT) is both powerful and necessary. You're asking for a "simple" proof of the isoperimetric inequality for a general compact set : starting from the version involving Minkowski content:

Yes.
Neither currents nor the whole theory of sets of finite perimeter are really needed.
Starting from the Brunn–Minkowski form of the isoperimetric inequality one only has to compare the outer Minkowski content of the set with the Hausdorff measure of its topological boundary. The comparison can be done with two standard (and elementary) tools:

• the Lebesgue–Besicovitch density theorem fo...

This is an excellent question about whether there's a simpler proof of the isoperimetric inequality using Hausdorff measure instead of Minkowski content.

The short answer is: there is no known elementary proof of this result that avoids the machinery of geometric measure theory, at least not one that works for general compact sets.

Here's why this is challenging and what the obstacles are:

...

Isoperimetric Inequality with Hausdorff Measure for General Compact Sets

The question asks for a simple proof of the isoperimetric inequality for general compact sets in terms of the Hausdorff measure:

This is a challenging request because there's a significant gap between the standard Minkowski con...