$\frac{\mu(B_{d}(x,r)\cap X_{m})}{\mu(B_{d}(x,r))}<\epsilon$

for all $0<r<r_{1}$. Let $r_{2}=\min\\{r_{1},R/(2m)\\}$. Also, let $K\subset
B(x,Rr_{0})$ satisfy

$\mu(B_{d}(x,Rr_{0})\setminus K)\leq\epsilon.$

Then for any $r_{2}\leq r<R_{0}$,

$\frac{\mu(B_{d/r}(x,R)\setminus
K)}{\mu(B_{d}(x,r))}=\frac{\mu(B_{d}(x,Rr)\setminus
K)}{\mu(B_{d}(x,r))}\leq\frac{\epsilon}{\mu(B_{d}(x,r_{2}))}.$

Since $K$ is compact, the metric spaces $(K,d/r)$ with $r_{2}\leq r<R_{0}$ are
uniformly totally bounded. On the other hand, by Lemma 2.9, if $0<r<r_{2}$
then

$B_{d/r}(x,R)\cap X_{m}=B_{d}(x,Rr)\cap X_{m}$

is covered by $m^{-4\log_{2}\epsilon}$ balls of the form
$B_{d/r}(w_{i},\epsilon)$. Thus

$\\{B_{d/r}(x,R):0<r<r_{0}\\}$

is uniformly totally bounded. Therefore, Theorem 4.15 proves (105).

Now let $x\in\operatorname{spt}\mu$ satisfy (105). By applying (105) to an
arbitrary sequence $r_{i}\to 0$ we see that $\operatorname{Tan}(\mu,x)$ is
non-empty. To see that it is compact, for each $j\in\mathbb{N}$ let
$(\nu_{j},y_{j})\in\operatorname{Tan}(\mu,x)$ and let $0<r_{j}<1/j$ be such
that

$d_{*}(T_{r_{j}}(\mu,x),(\nu_{j},y_{j}))<1/j.$

By (105) there exists a subsequence $r_{j_{k}}\to 0$ and a
$(\nu,y)\in\mathbb{M}_{*}$ such that

$d_{*}(T_{r_{j_{k}}}(\mu,x),(\nu,y))\to 0.$

In particular, $(\nu,y)\in\operatorname{Tan}(\mu,x)$ and, by the triangle
inequality, $(\nu_{j_{k}},y_{j_{k}})\to(\nu,y)$, as required. Finally, given
$\delta>0$, the existence of such an $r_{0}$ is given by the contrapositive to
(105).

######  Lemma 5.4.

Let $(X,d,\mu)$ be a metric measure space and suppose that
$\mathcal{S}\subset\mathbb{M}_{*}$. Then for any $R,\delta>0$,

$C_{R,\delta}(\mathcal{S}):=\\{x\in\operatorname{spt}\mu:d_{*}((T_{r}(\mu,d,x),\mathcal{S})\leq\delta\
\forall 0<r<R\\}$

is a closed subset of $X$. In particular, if $S\subset X$ is a Borel set of
points satisfying (105) and $\mathcal{C}\subset\mathbb{M}_{*}$ is closed,

$\mathcal{G}(\mathcal{C}):=\\{x\in
S:\operatorname{Tan}(\mu,d,x)\subset\mathcal{C}\\}$

is Borel.

Proof: Let $x\in X$ and $x_{j}\in C_{R,\delta}(\mathcal{S})$ with $x_{j}\to
x$. First note that, since $\mu\in\mathcal{M}_{\mathrm{loc}}(X)$, all but
countably many $r>0$ satisfy

$\mu(B(x,r)\setminus U(x,r))=0.$

For such an $r>0$, $T_{r}(\mu,d,x_{j})\to T_{r}(\mu,d,x)$. Indeed, eqs. 15 and
14 imply that $\mu(B(x_{j},r))\to\mu(B(x,r))$ and consequently Lemma 2.22
implies

$d_{\mathrm{pmGH}}\left(\left(\frac{\mu}{\mu(B(x_{j},r))},\frac{d}{r},x_{j}\right),\left(\frac{\mu}{\mu(B(x,r))},\frac{d}{r},x\right)\right)\to
0$

Therefore

$d_{*}(T_{r}(\mu,x),\mathcal{S})\leq\delta.$ (107)

For arbitrary $0<r<R$, let $r_{j}\downarrow r$ satisfy (107). Then
$\mu(B(x,r_{j}))\to\mu(B(x,r))$ and so

$d_{\mathrm{pmGH}}\left(\left(\frac{\mu}{\mu(B(x,r_{j}))},\frac{d}{r_{j}},x\right),\left(\frac{\mu}{\mu(B(x,r))},\frac{d}{r},x\right)\right)\to
0$

(for any $R>0$, the identity is a $(r_{j}-r)R$-isometry on $B(x,R)$).
Consequently (107) holds in this case too and hence
$C_{R,\delta}(\mathcal{S})$ is closed.

If $x$ satisfies (105) and $\operatorname{Tan}(\mu,d,x)\subset\mathcal{C}$
then by (106),

$\forall\delta>0,\ \exists r_{x}>0\text{ s.t.
}d_{*}(T_{r}(\mu,x),\mathcal{C})\leq\delta\ \forall 0<r<r_{x}.$ (108)

Conversely, if $\mathcal{S}$ is closed, (108) implies
$\operatorname{Tan}(\mu,x)\subset\mathcal{C}$. Thus,

$\mathcal{G}(\mathcal{C})=S\cap\bigcap_{\delta\in\mathbb{Q}^{+}}\bigcup_{R\in\mathbb{Q}^{+}}C_{R,\delta}(\mathcal{C}),$

a Borel set.

We now prove three results that correspond results in Preiss MR890162 . The
following statement, and its proof, is essentially that of (MR890162, ,
Theorem 2.12) reformulated in our setting. See also MR3334235 ; MR2865538  for
adaptations to Gromov–Hausdorff tangent spaces and the appendix of MR3990192
for an adaptation to tangents of sets of finite perimeter in RCD spaces.

######  Proposition 5.5.

Let $(X,d,\mu)$ be an asymptotically doubling metric measure space. Then
$\mu$-a.e. $x\in\operatorname{spt}\mu$ has the following property:

$\forall(\nu,y)\in\operatorname{Tan}(\mu,x)\text{ and }\forall
z\in\operatorname{spt}\nu,\ (\nu,z)\in\operatorname{Tan}(\mu,x).$ (109)

Proof: For each $p\in\mathbb{N}$ let $A_{p}$ be the set of
$x\in\operatorname{spt}\mu$ for which there are
$(\nu_{x},y_{x})\in\operatorname{Tan}(\mu,x)$ and
$b_{x}\in\operatorname{spt}\nu_{x}$ such that

$d_{*}((\nu_{x},b_{x}),T_{r}(\mu,x))>\frac{1}{p}\quad\forall 0<r<1/p.$

Note that, if $x\not\in\cup_{p\in\mathbb{N}}A_{p}$, then $x$ satisfies the
required conclusion.

Suppose that, for some $p\in\mathbb{N}$, $\mu(A_{p})>0$. Then by the
separability of $(\mathbb{M}_{*},d_{*})$, there exists a $E\subset A_{p}$ with
$\mu(E)>0$ such that

$d_{*}((\nu_{x},b_{x}),(\nu_{y},b_{y}))<1/2p$

for every $x,y\in E$. By Theorem 2.10 there exists a Lebesgue density point
$x$ of $E$. Let $r_{k}\to 0$ be such that
$T_{r_{k}}(\mu,x)\to(\nu_{x},y_{x})$. By Proposition 2.20 there exists a
complete and separable pointed metric space $(Z,\zeta,z)$ and isometric
embeddings
$(\operatorname{spt}\mu,d/r_{k},x),(\operatorname{spt}\nu,y_{x})\to(Z,\zeta,z)$
with

$\frac{\mu}{\mu(B(x,r_{k}))}\to\nu_{x}.$

Let $\mu_{k}$ be the isometric copy of
$\mu\in\mathcal{M}_{\mathrm{loc}}(X,d/r_{k})$, let $E_{k}$ be the isometric
copy of $(E,d/r_{k})$ and let $c_{k}\in E_{k}$ be such that

$\zeta(b_{x},c_{k})<\zeta(b_{x},E_{k})+1/k.$

Note that

$\zeta(b_{x},E_{k})\to 0.$ (110)

Indeed, since $x$ is a density point of $E$, if $R=2\zeta(y_{x},b_{x})$ then

$\lim_{k\to\infty}\frac{\mu_{k}(E_{k}\cap
B(x,R))}{\mu_{k}(B(x,R))}=\lim_{k\to\infty}\frac{\mu(E\cap
B(x,Rr_{k}))}{\mu(B(x,Rr_{k}))}=1.$ (111)

On the other hand, suppose that there exists $\delta>0$ such that
$B(b_{x},\delta)\cap E_{k}=\emptyset$ for infinitely many $k$. Equations 14
and 15 imply

$\displaystyle\liminf_{k\to\infty}\frac{\mu_{k}(E_{k}\cap
B(x,R))}{\mu_{k}(B(x,R))}$ $\displaystyle\leq
1-\limsup_{k\to\infty}\frac{\mu_{k}(B(b_{x},\delta))}{\mu_{k}(B(x,R))}$
$\displaystyle\leq
1-\frac{\nu_{x}(B(b_{x},\delta/2))}{\nu_{x}(B(y_{x},2R))}<1,$

since $b_{x}\in\operatorname{spt}\nu_{x}$. This contradicts (111).

Now, (110) implies $\zeta(b_{x},c_{k})\to 0$. This and Lemma 2.22 imply

$d_{*}\left(\left(\frac{\mu}{\mu(B(x,r_{k}))},\frac{d}{r_{k}},c_{k}\right),(\nu_{x},b_{x})\right)\to
0.$

Therefore, there exists $k$ with $r_{k}<1/p$ such that

$d_{*}\left(\left(\frac{\mu}{\mu(B(x,r_{k}))},\frac{d}{r_{k}},c_{k}\right),(\nu_{x},b_{x})\right)<1/2p.$

Thus, since each $c_{k}\in E_{k}$,

$\displaystyle 1/p$
$\displaystyle<d_{*}\left(\left(\frac{\mu}{\mu(B(x,r_{k}))},\frac{d}{r_{k}},c_{k}\right),(\nu_{c_{k}},b_{c_{k}})\right)$
$\displaystyle\leq
d_{*}\left(\left(\frac{\mu}{\mu(B(x,r_{k}))},\frac{d}{r_{k}},c_{k}\right),(\nu_{x},b_{x})\right)+d_{*}((\nu_{x},b_{x}),(\nu_{c_{k}},b_{c_{k}}))$
$\displaystyle\leq 1/2p+1/2p,$

a contradiction. Therefore we must have $\mu(A_{p})=0$ for all
$p\in\mathbb{N}$.

######  Lemma 5.6.

Let $(\mu,x)\in\mathbb{M}_{*}$ with $x\in\operatorname{spt}\mu$ and

$M_{x}:=\limsup_{r\to 0}\frac{\mu(B(x,2r))}{\mu(B(x,r))}.$ (112)

For every $(\nu,y)\in\operatorname{Tan}(\mu,x)$ and $r>0$,

$\nu(B(y,2r))\leq M_{x}\nu(B(y,r))$

and

$1\leq\nu(B(y,1))\leq M_{x}.$

In particular, if $x$ satisfies (109), every
$(\nu,y)\in\operatorname{Tan}(\mu,x)$ is $M_{x}$-doubling.

Proof: By Propositions 2.20, 14 and 15, for any
$(\nu,y)\in\operatorname{Tan}(\mu,x)$ there exist $r_{i}\to 0$ such that, for
any $R>0$ and $\lambda>1$,

$\nu(B(y,2R))\leq\liminf_{i\to\infty}\frac{\mu(B(x,\lambda
2Rr_{i}))}{\mu(B(x,r_{i}))}$

and

$\nu(B(y,\lambda R))\geq\limsup_{i\to\infty}\frac{\mu(B(x,\lambda
Rr_{i})}{\mu(B(x,r_{i}))}.$

Consequently,

$\frac{\nu(B(y,2R))}{\nu(B(y,\lambda R))}\leq M_{x}.$

Since this is true for all $\lambda>1$ we also have

$\frac{\nu(B(y,2R))}{\nu(B(y,R))}\leq M_{x}.$ (113)

Also note that, by (15),

$\nu(B(y,1))\geq\limsup_{i\to\infty}\frac{\mu(B(x,r_{i}))}{\mu(B(x,r_{i}))}=1,$

and by (14),

$\nu(B(y,1))\leq\nu(U(x,2))\leq\liminf_{i\to\infty}\frac{\mu(U(x,2))}{\mu(B(x,1))}\leq\liminf_{i\to\infty}\frac{\mu(B(x,2))}{\mu(B(x,1))}\leq
M_{x}.$

######  Corollary 5.7.

Let $(\mu,d,x)\in\mathbb{M}_{*}$ and let $S\subset\operatorname{spt}\mu$ be
$\mu$-measurable with $x$ a density point of $S$. Then

$\operatorname{Tan}(\mu|_{S},x)=\operatorname{Tan}(\mu,x).$

Proof: Combine (84) and the definition of $\operatorname{Tan}$.

##  6 Proof of the main theorems

We first demonstrate how tangent measures supported on elements of
$\operatorname{biLip}(K)$ lead to a decomposition of the space into sets with
good tangential approximation.

Define the projection $\pi_{1}\colon\mathbb{M}_{pm}\to\mathbb{M}_{p}$ by

$(X,\mu,x)\mapsto(X,x)$

and $\pi_{2}\colon\mathbb{M}_{pm}\to\mathbb{M}_{*}$ by

$(X,\mu,x)\mapsto(\mu,x).$

Both of $\pi_{1},\pi_{2}$ are 1-Lipschitz.

For $M\geq 1$ let

$S_{1}(M)=\\{(X,\mu,x)\in\mathbb{M}_{pm}:M\geq\mu(U(x,1)),\ \mu(B(x,1))\geq
1/M\\}$

and $S_{2}(M)$ be the set of those $(X,\mu,x)\in\mathbb{M}_{pm}$ with

$\mu(U(y,r))\leq M\mu(B(y,2r))\quad\forall y\in X\text{ and }r>0.$ (114)

For $K\geq 1$ let

$\mathcal{K}(K,M)=\pi_{1}^{-1}(\operatorname{biLip}(K))\cap S_{1}(M)\cap
S_{2}(M).$

######  Lemma 6.1.

For any $K,M\geq 1$,

  1. 1.

$\operatorname{biLip}(K)$ is a compact subset of
$(\mathbb{M}_{p},d_{\mathrm{pGH}})$

  2. 2.

$S_{1}(M)$ and $S_{2}(M)$ are closed subsets of
$(\mathbb{M}_{pm},d_{\mathrm{pmGH}})$

  3. 3.

$\mathcal{K}(K,M)$ is a compact subset of
$(\mathbb{M}_{pm},d_{\mathrm{pmGH}})$ and hence a compact subset of
$(\mathbb{M}_{*},d_{*})$.

Proof: First note that the elements of $\operatorname{biLip}(K)$ are proper
and so do belong to $\mathbb{M}_{p}$. Moreover, for any $r>0$, the set

$\\{B(x,r):(X,x)\in\operatorname{biLip}(K)\\}$

is uniformly totally bounded and so, by Theorem 2.27,
$\operatorname{biLip}(K)$ is pre-compact.

To see that $\operatorname{biLip}(K)$ is closed, for each $i\in\mathbb{N}$ let
$(X_{i},x_{i})\in\operatorname{biLip}(K)$ and $(X,x)\in\mathbb{M}_{p}$ with

$d_{\mathrm{pGH}}((X_{i},x_{i}),(X,x))<1/i.$

For each $i\in\mathbb{N}$ let $f_{i}\colon(X_{i},x_{i})\to(X,x)$ be a
$2/i$-isometry given by Lemma 2.24. Also let
$\psi_{i}\colon\ell_{\infty}^{n}\to X_{i}$ be surjective and $K$-bi-Lipschitz
and define

$\phi_{i}=f_{i}\circ\psi_{i}\colon\ell_{\infty}^{n}\to X.$

Fix $R>0$. For any $y,z\in B(0,R)\subset\ell_{\infty}^{n}$,

$\displaystyle d(\phi_{i}(y),\phi_{i}(z))$
$\displaystyle=d(f_{i}(\psi_{i}(y)),f_{i}(\psi_{i}(z)))$ $\displaystyle\leq
d(\psi_{i}(y),\psi_{i}(z))+2/i$ $\displaystyle\leq K\|y-z\|_{\infty}+2/i.$
(115)

Thus the $\phi_{i}$ are equicontinuous on $B(0,R)$. Since the $\phi_{i}$ map
into $B(x,KR+2)$, a compact set, the Arzelà–Ascoli theorem gives a $\phi\colon
B(0,R)\to X$ such that, after passing to a subsequence, $\phi_{i}\to\phi$
uniformly on $B(0,R)$. Since this is true for all $R>0$, by taking a diagonal
subsequence, we see that there exists a $\phi\colon\ell_{\infty}^{n}\to X$
such that $\phi_{i}\to\phi$ uniformly on bounded sets. A similar estimate to
(115) shows that $\phi$ is $K$-bi-Lipschitz. Since each $\phi_{i}$ is
surjective and each $f_{i}$ is a $2/i$-isometry, $\phi$ is surjective. Thus
$\operatorname{biLip}(K)$ is closed, proving item 1.

Item 2 follows from Proposition 2.20 and eqs. 14 and 15.

Item 3 follows since $\mathcal{K}(K,M)$ is pre-compact and closed in
$(\mathbb{M}_{pm},d_{\mathrm{pmGH}})$ by Corollary 4.7 and items 1 and 2.

For $K\geq 1$ let

$\operatorname{biLip}(K)^{*}=\\{(\mu,x)\in\mathbb{M}_{*}:\operatorname{spt}\mu\in\operatorname{biLip}(K)\\}.$

######  Proposition 6.2.

Let $X$ be a complete metric space and $S\subset X$ a
$\mathcal{H}^{n}$-measurable set with $\mathcal{H}^{n}(S)<\infty$. Suppose
that, for $\mathcal{H}^{n}$-a.e. $x\in S$, $\Theta^{n}_{*}(S,x)>0$ and there
exists a $K_{x}\geq 1$ such that

$\operatorname{Tan}(\mathcal{H}^{n}|_{S},x)\subset\operatorname{biLip}(K_{x})^{*}.$

There exist compact $C_{i}\subset S$ with

$\mathcal{H}^{n}\left(S\setminus\bigcup_{i\in\mathbb{N}}C_{i}\right)=0$

and, for each $i\in\mathbb{N}$, an $\eta_{i}>0$ such that, for each $x\in
C_{i}$,

$\eta_{i}r^{n}\leq\mathcal{H}^{n}(B(x,r)\cap S)<2(2r)^{n}\ \forall
0<r<\eta_{i}$ (116)

and

$\operatorname{Tan}(\mathcal{H}^{n}|_{S},x)\subset\mathcal{K}(\eta_{i}^{-1},\eta_{i}^{-1}).$
(117)

Further, for each $i\in\mathbb{N}$ and any $\epsilon>0$, there exists
countably many closed $G_{j}\subset C_{i}$ with

$\mathcal{H}^{n}\left(C_{i}\setminus\bigcup_{j\in\mathbb{N}}G_{j}\right)=0,$

such that each $(C_{i},G_{j})$ satisfies
$GTA(\eta_{i},\eta_{i}^{-1},\epsilon,\min\\{\eta_{i},1/j\\})$.

Proof: For any $i\in\mathbb{N}$, Lemma 2.8 implies that

$S_{i}:=\\{x\in S:r^{n}/i\leq\mathcal{H}^{n}(B(x,r)\cap S)<2(2r)^{n}\ \forall
0<r<1/i\\}$

is $\mathcal{H}^{n}$-measurable. By assumption, the $S_{i}$ monotonically
increase to almost all of $S$ as $i\to\infty$.

Fix $i\in\mathbb{N}$. For $j\in\mathbb{N}$, let

$S_{i}^{j}:=\\{x\in
S_{i}:\operatorname{Tan}(\mathcal{H}^{n}|_{S},x)\subset\mathcal{K}(j,j)\\}.$

If $x\in S_{i}$ satisfies (109), Lemma 5.6 implies that every
$(\nu,y)\in\operatorname{Tan}(\mathcal{H}^{n}|_{S},x)$ is $4^{n+1}i$-doubling
with $1\leq\nu(B(y,1))\leq 4^{n+1}i$. Therefore, if
$\max\\{K_{x},4^{n+1}i\\}\leq j$ then $x\in S_{i}^{j}$ and so
$\mathcal{H}^{n}(S\setminus\cup_{i,j}S_{i}^{j})=0$. Lemma 6.1 implies that
$\mathcal{K}(j,j)$ is compact and so Lemma 5.4 implies each $S_{i}^{j}$ is
$\mathcal{H}^{s}$-measurable. Consequently, each can be decomposed as a
countable union of compact sets (up to a set of measure zero). Re-indexing
completes the proof of the first decomposition.

Now fix $i\in\mathbb{N}$ and let $C=C_{i}$ and $\eta=\eta_{i}$. Since
$\mathcal{H}^{n}(C)<\infty$, Lemma 2.3 implies $\Theta^{*,n}(C,x)\leq 1$ for
$\mathcal{H}^{n}$-a.e. $x\in C$ and since $\mathcal{H}^{n}(S\setminus
C)<\infty$, $\Theta^{n}_{*}(C,x)\geq\eta$ for $\mathcal{H}^{n}$-a.e. $x\in C$.
Thus $\mathcal{H}^{n}|_{C}$ is asymptotically doubling. Therefore, by
Corollary 5.7, Theorems 2.10 and 117,

$\operatorname{Tan}(\mathcal{H}^{n}|_{C},x)\subset\mathcal{K}(\eta^{-1},\eta^{-1})\quad\text{for
}\mathcal{H}^{n}\text{-a.e. }x\in C.$ (118)

For any $\epsilon>0$ the sets

$C_{j,\epsilon}:=C_{1/j,\epsilon}(\mathcal{K}(\eta^{-1},\eta^{-1}))$

defined in Lemma 5.4 are closed and $\cup_{j}C_{j,\epsilon}$ contains $C$. For
any $j\in\mathbb{N}$, any $x\in C_{j,\epsilon}$ and any $0<r<1/j$, let
$(\nu,y)\in\mathcal{K}(\eta^{-1},\eta^{-1})$ with

$d_{*}(T_{r}(\mathcal{H}^{n}|_{C},x),(\nu,y))<2\epsilon.$

Apply Theorem 4.11 to obtain a $K\subset C\cap B(x,r/2\epsilon)$ and
$K_{\nu}\subset B(y,1/2\epsilon)$ with

$\max\left\\{\frac{\mathcal{H}^{n}(C\cap B(x,r/2\epsilon)\setminus
K)}{\mathcal{H}^{n}(C\cap B(x,r))},\nu(B(y,1/2\epsilon)\setminus
K_{\nu})\right\\}<2\epsilon$ (119)

and

$d_{\mathrm{pmGH}}(T_{r}(K,\mathcal{H}^{n}|_{K},x),(K_{\nu},\nu|_{K_{\nu}},y))<6\epsilon.$
(120)

Note that (114) implies $(Y,\nu)$ is $2\eta^{-1}$-doubling. Equation 119
allows us to apply Lemma 4.17 with $k=\log_{2}2\eta^{-1}$,
$R=\epsilon^{-\frac{1}{2k}}$ and $\delta=1$ to obtain

$d_{\mathrm{pmGH}}((K_{\nu},\nu|_{K_{\nu}},y),(\operatorname{spt}\nu,\nu,y))\leq
9\epsilon^{\frac{1}{2k}}.$

The triangle inequality and (120) then give

$d_{\mathrm{pmGH}}(T_{r}(K,\mathcal{H}^{n}|_{K},x),(\operatorname{spt}\nu,\nu,y))\leq
18\epsilon^{\frac{1}{2k}}.$

In particular,

$d_{\mathrm{pGH}}((K,d/r,x),\operatorname{biLip}(\eta^{-1}))\leq
18\epsilon^{\frac{1}{2k}}.$

Also, eqs. 119 and 116 imply

$\mathcal{H}^{n}(B(x,r)\setminus K)<2\epsilon\mathcal{H}^{n}(B(x,r))\leq
4\cdot 4^{n}\epsilon r^{n}=\eta 4(4(\epsilon/\eta)^{1/n}r)^{n}.$

That is, $C_{j,\epsilon}$ satisfies
$GTA(\eta,\eta^{-1},\epsilon^{*},\min\\{\eta,1/j\\})$, for some $\epsilon^{*}$
depending upon $\eta$ such that $\epsilon^{*}\to 0$ as $\epsilon\to 0$. Since
$\epsilon>0$ was arbitrary, setting $G_{j}=C_{1/j,\epsilon^{\prime}}$ for each
$j\in\mathbb{N}$ and appropriate $\epsilon^{\prime}$ completes the proof.

To prove that (3) implies (1) in Theorem 1.2 we require the following result
(perturbations, , Theorem 1.1). It concerns the set of all 1-Lipschitz
functions $f\colon X\to\ell_{2}^{n}$, equipped with the supremum norm.

######  Theorem 6.3.

Let $X$ be a complete metric space and $m\in\mathbb{N}$. Suppose that
$S\subset X$ is purely $n$-unrectifiable with $\mathcal{H}^{n}(S)<\infty$ and
satisfies $\Theta_{*}^{n}(S,x)>0$ for $\mathcal{H}^{n}$-a.e. $x\in S$. The set
of all 1-Lipschitz $f\colon X\to\ell_{2}^{m}$ with $\mathcal{H}^{n}(f(S))=0$
is residual and hence dense.

######  Corollary 6.4.

Let $X$ be a complete metric space and let $S\subset X$ be
$\mathcal{H}^{n}$-measurable with $\mathcal{H}^{n}(S)<\infty$ and
$\Theta_{*}^{n}(S,x)>0$ for $\mathcal{H}^{n}$-a.e. $x\in S$. For $L\geq 1$ and
$\delta>0$ suppose that

  1. 1.

$\iota\colon[0,r]^{n}\to X$ is continuous with

$\mathcal{H}^{n}_{\infty}(\iota([0,r]^{n})\setminus S)<\mathcal{H}^{n}([\delta
r,(1-\delta)r]^{n})/L.$

  2. 2.

$f\colon\iota([0,r]^{n})\to[0,r]^{n}\subset\ell_{2}^{n}$ is $L$-Lipschitz with

$\|f(\iota(x))-x\|_{2}<\delta r\quad\forall x\in\partial[0,r]^{n}.$

Then $S$ is not purely $n$-unrectifiable.

Proof: Suppose that $S$ is purely $n$-unrectifiable and let $\epsilon>0$ be
such that

$\mathcal{H}^{n}_{\infty}(\iota([0,r]^{n})\setminus
S)<\mathcal{H}^{n}([(\delta+\epsilon)r,(1-(\delta+\epsilon))r]^{n})/L.$

Let $g\colon\iota([0,r]^{n})\to\ell_{2}^{n}$ be given by Theorem 6.3 such that

  * •

$g$ is $L$-Lipschitz;

  * •

$\|f-g\|_{\infty}<\epsilon r$;

  * •

$\mathcal{H}^{n}(g(S))=0$.

In particular,

$\|g(\iota(x))-x\|<(\delta+\epsilon)r\quad\forall\ x\in\partial[0,r]^{n}$
(121)

and, since $\mathcal{H}^{n}_{\infty}=\mathcal{H}^{n}$ in $\ell_{2}^{n}$,

$\displaystyle\mathcal{H}^{n}(g(\iota([0,r]^{n})))$
$\displaystyle\leq\mathcal{H}^{n}(g(\iota([0,r]^{n})\setminus
S))+\mathcal{H}^{n}(g(S))$ $\displaystyle\leq
L\mathcal{H}^{n}_{\infty}(\iota([0,r]^{n})\setminus S)+0$
$\displaystyle<\mathcal{H}^{n}([(\delta+\epsilon)r,(1-(\delta+\epsilon))r]^{n}).$
(122)

Note that (121) implies that

$g(\iota([0,r]^{n}))\supset[(\delta+\epsilon)r,(1-(\delta+\epsilon))r]^{n}$

(this is a consequence of Brouwer’s fixed point theorem; See (perturbations, ,
Lemma 7.1) for a proof involving the unit ball). Consequently,

$\mathcal{H}^{n}(g(\iota([0,r]^{n})))\geq\mathcal{H}^{n}([(\delta+\epsilon)r,(1-(\delta+\epsilon))r]^{n}),$

contradicting (122)

Recall the definition of $\operatorname{biLip}(K)^{*}$ preceding Proposition
6.2.

######  Theorem 6.5.

Let $(X,d)$ be a complete metric space, $n\in\mathbb{N}$ and $E\subset X$ a
$\mathcal{H}^{n}$-measurable set with $\mathcal{H}^{n}(E)<\infty$. Suppose
that, for $\mathcal{H}^{n}$-a.e. $x\in E$, $\Theta^{n}_{*}(E,x)>0$ and there
exists a $K_{x}\geq 1$ such that

$\operatorname{Tan}(X,d,\mathcal{H}^{n}|_{E},x)\subset\operatorname{biLip}(K_{x})^{*}.$
(123)

Then $E$ is $n$-rectifiable.

Proof: By the Kuratowski embedding, we may identify $X$ with an isometric copy
contained in $\ell_{\infty}$.

Suppose that $E$ is not $n$-rectifiable. Then by Lemma 2.2 there exists a
compact purely $n$-unrectifiable $S\subset E$ with $\mathcal{H}^{n}(S)>0$. By
Lemma 2.3, $\Theta_{*}^{n}(S,x)>0$ for $\mathcal{H}^{n}$-a.e. $x\in S$. In
particular, $\mathcal{H}^{n}|_{S}$ is asymptotically doubling and so by
Corollary 5.7,

$\operatorname{Tan}(\mathcal{H}^{n}|_{S},x)\subset\operatorname{biLip}(K_{x})^{*}$

for $\mathcal{H}^{n}$-a.e. $x\in S$. Then $S$ satisfies the hypotheses of
Proposition 6.2. Since $\mathcal{H}^{n}(S)>0$, Proposition 6.2 implies the
existence of a compact $C\subset S$ with $\mathcal{H}^{n}(C)>0$ and a
$0<\eta<1$ such that, for any $\epsilon>0$, there exist countably many compact
$G_{j}\subset C$ with $\mathcal{H}^{n}(C\setminus G_{j})=0$ such that each
$(C,G_{j})$ satisfies $GTA(\eta,\eta^{-1},\epsilon,\min\\{\eta,1/j\\})$.

Fix $0<\gamma<1$ and $0<\delta<\eta/4\sqrt{n}$ and let $m_{0}\in\mathbb{N}$ be
given by Theorem 3.4 for these parameters and $K=\eta^{-1}$. Let $m\geq m_{0}$
be such that

$20\sqrt{n}\eta^{-2}2^{-m\gamma/2}+\sqrt{n}2^{-m}<\delta$ (124)

and set $\epsilon=4^{-m}$. Since $\mathcal{H}^{n}(C)>0$, there exists
$j\in\mathbb{N}$ for which $G_{j}$ obtained in the previous paragraph
satisfies $\mathcal{H}^{n}(G_{j})>0$. Set $G=G_{j}$ and
$R_{0}=\min\\{\eta,1/j\\}$, so that $(C,G)$ satisfies
$GTA(\eta,\eta^{-1},4^{-m},R_{0})$. Let $\Lambda>0$ be obtained from Theorem
3.16. Fix $x$ a density point of $G$ with $\Theta^{*,n}(X,x)\leq 1$ and let
$0<r<R_{0}$ be such that

$\Lambda\mathcal{H}^{n}(B(x,20\eta^{-1}r)\cap X\setminus G)<\delta r^{n}.$
(125)

Then Theorem 3.4 gives an $\iota\colon[0,r]^{n}\subset\ell_{\infty}^{n}\to
B(x,20\eta^{-1}r)\subset\ell_{\infty}$ satisfying Theorem 3.4 items 1, 2 and
3.

Since $\iota|_{\mathcal{D}(r,m)}^{-1}$ is $2\eta^{-1}$-Lipschitz and takes
values in $[0,r]^{n}$, an application of the second conclusion of Theorem 2.6
extends it to a $\sqrt{n}2\eta^{-1}$-Lipschitz map
$f\colon\iota([0,r]^{n})\to[0,r]^{n}\subset\ell_{2}^{n}$. By Theorem 3.4 item
2, for any $p,q\in[0,r]^{n}$ with $\|p-q\|_{\infty}\leq 2^{m}r$ we have

$\displaystyle\|f(\iota(p))-f(\iota(q))\|_{2}$
$\displaystyle\leq\sqrt{n}2\eta^{-1}\|\iota(p)-\iota(q)\|_{\infty}$
$\displaystyle\leq 20\sqrt{n}\eta^{-2}2^{-m\gamma/2}r.$

In particular, for any $p\in[0,r]^{n}$, pick $q\in\mathcal{D}(r,m)$ with
$\|p-q\|_{\infty}\leq 2^{-m}r$. Then

$\displaystyle\|f(\iota(p))-p\|_{2}$
$\displaystyle\leq\|f(\iota(p))-f(\iota(q))\|_{2}+\|f(\iota(q))-p\|_{2}$
$\displaystyle\leq 20\sqrt{n}\eta^{-2}2^{-m\gamma/2}r+\|q-p\|_{2}$
$\displaystyle\leq 20\sqrt{n}\eta^{-2}2^{-m\gamma/2}r+\sqrt{n}2^{-m}r$
$\displaystyle<\delta r$

by (124). Thus, $f$ satisfies Corollary 6.4 item 2.

Combining Theorem 3.4 item 3 and (125) implies that $f$ satisfies Corollary
6.4 item 1. An application of Corollary 6.4 shows that $S$ is not purely
$n$-unrectifiable, a contradiction.

Finally we prove that (1) implies (2) in Theorem 1.2.

######  Theorem 6.6.

Let $(X,d)$ be a complete metric space, $n\in\mathbb{N}$ and $E\subset X$ an
$n$-rectifiable set with $\mathcal{H}^{n}(E)<\infty$. For any $x\in E$ that
satisfies the conclusion of Theorem 2.4 and Lemma 2.3 item 1,

$T_{r}(X,d,\mathcal{H}^{n}|_{E},x)\to(\mathbb{R}^{n},\|\cdot\|_{x},\mathcal{H}^{n}/2^{n},0)$
(126)

as $r\to 0$. In particular, for $\mathcal{H}^{n}$-a.e. $x\in E$,

$\operatorname{Tan}(X,d,\mathcal{H}^{n}|_{E},x)=\\{(\mathbb{R}^{n},\|\cdot\|_{x},\mathcal{H}^{n}/2^{n},0)\\}.$

Proof: Let $x\in E$ be as in the hypotheses and $f_{x},\|\cdot\|_{x},C_{x}$ as
in the conclusion of Theorem 2.4. For each $r>0$, let

$\epsilon_{r}=\sup\left\\{\left|1-\frac{\|f_{x}(y)-f_{x}(z)\|_{x}}{d(y,z)}\right|:y\neq
z\in C_{x}\cap B(x,\sqrt{r})\right\\},$

so that $\epsilon_{r}\to 0$ as $r\to 0$. Also let $f_{r}=f_{x}/r$ and

$C_{r}=C_{x}\cap B(x,\sqrt{r}).$

First note that, for any $y,z\in C_{r}$ with
$d(y,z)/r\leq\min\\{r^{-1/2},\epsilon_{r}^{-1/2}\\}$,

$\displaystyle\left|\frac{d(y,z)}{r}-\|f_{r}(y)-f_{r}(z)\|_{x}\right|$
$\displaystyle=\frac{|d(y,z)-\|f_{x}(y)-f_{x}(z)\|_{x}|}{r}$
$\displaystyle\leq\frac{\epsilon_{r}d(y,z)}{r}$
$\displaystyle\leq\sqrt{\epsilon_{r}}.$ (127)

Now observe that $f_{r}\colon(C_{r},d/r)\to(\mathbb{R}^{n},\|\cdot\|_{x})$ is
$L_{r}$-bi-Lipschitz, for

$L_{r}=\frac{1+\epsilon_{r}}{1-\epsilon_{r}^{2}}\to 1.$

Therefore

$\frac{1}{L_{r}^{n}}\mathcal{H}^{n}_{x}|_{f_{r}(C_{r})}\leq(f_{r})_{\\#}(\mathcal{H}^{n}_{r}|_{C_{r}})\leq
L_{r}^{n}\mathcal{H}^{n}_{x}|_{f_{r}(C_{r})},$ (128)

where $\mathcal{H}^{n}_{r}$ denotes $\mathcal{H}^{n}$ on the metric space
$(C_{r},d/r)$ and $\mathcal{H}^{n}_{x}$ denotes $\mathcal{H}^{n}$ on
$(\mathbb{R}^{n},\|\cdot\|_{x})$. However,

$(f_{r})_{\\#}(\mathcal{H}^{n}_{r}|_{C_{r}})=(f_{r})_{\\#}\left(\frac{\mathcal{H}^{n}|_{C_{r}}}{r^{n}}\right)=\frac{(f_{r})_{\\#}(\mathcal{H}^{n}_{d}|_{C_{r}})}{r^{n}},$

where $\mathcal{H}^{n}_{d}$ denotes $\mathcal{H}^{n}$ on $(X,d)$. Also
observe, for any $R>0$, by lemmas 2.5 and 9.

$\displaystyle\mathcal{H}^{n}_{x}(B(0,R)\setminus f_{r}(C_{r}))$
$\displaystyle\leq\mathcal{H}^{n}_{x}(B(0,R)\setminus f_{r}(C_{r}\cap
B(x,Rr/L_{r})))$ (129)
$\displaystyle\leq(2R)^{n}-\mathcal{H}^{n}_{x}(f_{r}(C_{r}\cap
B(x,Rr/L_{r})))$
$\displaystyle\leq(2R)^{n}-\frac{L_{r}^{n}}{r^{n}}\mathcal{H}^{n}_{d}(C_{r}\cap
B(x,Rr/L_{r}))$ $\displaystyle\to(2R)^{n}-(2R)^{n}=0$

Let $\|\cdot\|_{\mathcal{M}}$ denote the total variation of elements of
$\mathcal{M}((B(0,R),\|\cdot\|_{x}))$. Then eqs. 129 and 128 imply

$\displaystyle\left\|\frac{(f_{r})_{\\#}(\mathcal{H}^{n}_{d}|_{C_{r}})}{r^{n}}-\mathcal{H}^{n}_{x}\right\|_{\mathcal{M}}$
$\displaystyle\leq\left\|\frac{(f_{r})_{\\#}(\mathcal{H}^{n}_{d}|_{C_{r}})}{r^{n}}-\mathcal{H}^{n}_{x}|_{f_{r}(C_{r})}\right\|_{\mathcal{M}}$
$\displaystyle\qquad+\|\mathcal{H}^{n}_{x}|_{f_{r}(C_{r})}-\mathcal{H}^{n}_{x}\|_{\mathcal{M}}$
$\displaystyle\leq(L_{r}^{n}-1)\mathcal{H}^{n}_{x}(B(0,R))+\mathcal{H}^{n}_{x}(B(0,R)\setminus
f_{r}(C_{r}))\to 0.$

That is,

$\frac{(f_{r})_{\\#}(\mathcal{H}^{n}_{d}|_{C_{r}})}{(2r)^{n}}\to\frac{\mathcal{H}^{n}_{x}}{2^{n}}\quad\text{strongly
on }B(0,R).$ (130)

Equation 130 has two consequences. Firstly, it shows that, for any
$R,\delta>0$,

$B(f_{r}(C_{r}),\delta)\supset B(0,R)$

for sufficiently small $r>0$. Together with (127), this shows that there exist
$\epsilon^{\prime}_{r}\to 0$ such that

$f_{r}\colon(C_{r},d/r)\to(\mathbb{R}^{n},\|\cdot\|_{x})\quad\text{is an
}\epsilon^{\prime}_{r}\text{-isometry.}$ (131)

Equation 130 also implies that

$\frac{(f_{r})_{\\#}(\mathcal{H}^{n}|_{C_{r}})}{r^{n}}\to\mathcal{H}^{n}_{x}\quad\text{in
}\mathcal{M}_{\mathrm{loc}}(\mathbb{R}^{n},\|\cdot\|_{x}).$ (132)

Corollaries 4.4, 131 and 132 then imply

$d_{\mathrm{pmGH}}(T_{r}(C_{r},d,\mathcal{H}^{n}|_{C_{r}},x),(\mathbb{R}^{n},\|\cdot\|_{x},\mathcal{H}^{n}/2^{n},0))\to
0$

and so eqs. 9 and 4.11 imply (126).

Theorems 6.5, 6.6 and 7 complete the proof of Theorem 1.2. To prove Theorem
1.1, note that, by the translation and scale invariance of $\mathbb{R}^{n}$,
one can replace (3) by

$\frac{d_{\mathrm{pGH}}((Y_{r}\cap B(0,r),0),(E_{r},x))}{r}\to 0$ (133)

and obtain an equivalent hypothesis. Equations 133 and 2 imply (123) (by using
Theorem 4.11) and so Theorem 6.5 implies Theorem 1.1.

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