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For every x ∈ H we set ρ(x) := dist (x, PW (G ∩ B 2σ (0))) . 2 The proof of the Marstrand–Mattila rectifiability criterion Observe that ρ(x) ≤ ησ . 25) Indeed, if this were false, we would have B ησ (x) ∩ G = ∅. 17). Using the 5r -Covering Lemma, we can find a countable set {x i }i∈I ⊂ H ∩ Dσ/4 (0) such that, if we set ρi := ρ(x i ), we find: • The disks {D20ρi (x i )} cover H ∩ Dσ/4 (0); • The disks {D4ρi (x i )} are pairwise disjoint. Since H k (H ∩ Dσ/4(0)) = H k (Dσ/4 (0)) = ωk (σ/4)k , we conclude that ωk ρik = i∈I 1 20k ωk σ k H k (H ∩ Dσ/4 (0)) = .

Then µ is an m-dimensional rectifiable measure. Note that the cases m = 0 and m = n are trivial. In the case m = 1 and n = 2, the Theorem was first proved by Besicovitch in his pioneering work [2]. More precisely, Besicovitch proved it for measures of the form H 1 E, when E is a Borel set with H 1 (E) < ∞ and his proof was later extended to planar Borel measures by Morse and Randolph in [24]. In [23], Moore extended the result to the case m = 1 and arbitrary n. The general case was open for a long time until Preiss solved it completely in [25].

Let us fix a test function ϕ ∈ C c (Rm ) and recall that 1 rk ϕ(x) dµ0,r (x) = 1 rk x r ϕ dµ(x) . 34) We now use the Area Formula to write ϕ x r dµ(x) = G ϕ F(z) r J F(z) d L k (z) . 35) Let C > 0 be such that ϕ ∈ C c (BC (0)). Then we have G ϕ F(z) r J F(z) d L k (z) = G∩BCr/(Lip F) (0) ϕ F(z) r J F(z) d L k (z) . 8 Recall that • 0 is a point of density 1 for G and therefore r −k L k (BCr (0) \ G) vanishes as r ↓ 0; • d F is Lebesgue continuous at 0, and therefore lim r −k r↓0 BCr (0) |J F(z) − J F(0)| d L k (z) = 0 ; • F is differentiable at 0 and hence F(z) lim sup ϕ r↓0 z∈BCr (0) r d F0 (z) r −ϕ = 0.