Let $A$ denote the Ahlfors regularity constant for $E$, set $\ve = (2A)^{-1}$ and let $L = L(\ve)$ be the constant so that Lemma <ref> holds. Let $x \in E$, $0 < r< \diam(E)$, and let $F$ and $f : F \rightarrow \R^d$ be the set and $L$-bi-Lipschitz map, respectively, from Lemma <ref> for $x$ and $r.$ Additionally, define the function $h : F \rightarrow \R^d$ by setting $h(y) = \tfrac{f(y) - f(x)}{L}.$ In this way, $h$ is $L^2$-bi-Lipschitz and $h(F) \subseteq B_d(0,r).$ Since $h$ is $L^2$-bi-Lipschitz, it is invertible with $L^2$-bi-Lipschitz inverse $h^{-1} : h(F) \rightarrow H.$ By the Kirszbraun extension theorem, we can extend $h^{-1}$ to a Lipschitz function from $\R^d$ to $H$ (whose Lipschitz constant depends only on $L$). Then,

\begin{align}
\mathscr{H}^d(E \cap B(x,r) \cap h^{-1}(B_d(0,r))) &\geq \mathscr{H}^d(E \cap B(x,r) \cap F) \\
&\geq \mathscr{H}^d(E \cap B(x,r) ) - \mathscr{H}^d(E \cap B(x,r) \setminus F) \\
&\geq (2A)^{-1} r^d.
\end{align}

Since $A$ and $L$ are independent of $x$ and $r,$ this finishes the proof.

Let us now focus on proving Lemma <ref>. Let $\ve > 0$, $x \in E$ and $0 < r < \diam(E).$ Our first task is to define $F,$ which will be chosen by running a certain stopping-time argument, then choosing parts of $E \cap B(x,r)$ for which we do not stop too many times, and do not lie too near the boundaries of certain cubes.

Let $C_0 > 1,$ $\mathscr{D}$ be the Christ-David cubes for $E$, $k$ be such that $5\rho^{k+1} < r \leq 5\rho^k,$ and $\{Q_{0,i}\}_{i \in I}$ be the cubes in $\mathscr{D}_k$ that intersect $B(x,r).$ For each $i \in I,$ using the stopping-time construction for the cubes contained in $Q_{0,i}$, outlined at the beginning of Section <ref> (with $M = C_0$), we find a collection of stopping-time regions $\mathscr{S}_i$ so that the following holds. First,

\begin{align}\label{e:S_i}
\{Q \in \mathscr{D} : Q \subseteq Q_{0,i}\} = \bigcup_{S \in \mathscr{S}_i} S.
\end{align}

Second, by Lemma <ref>,

\[ \sum_{S \in \mathscr{S}_i } \ell(Q(S))^d \lesssim \ell(Q_{0,i})^d + \sum_{Q \subseteq Q_{0,i}} \beta^{d,1}_E(C_0B_Q)^2\ell(Q)^d \lesssim \mathscr{H}^d(Q_{0,i}) \lesssim \ell(Q_{0,i})^d, \]

where the penultimate inequality follows from Theorem <ref> and the fact that $E$ satisfies the BWGL, and the final inequality follows from Ahlfors regularity. Finally, for each $S \in \mathscr{S}_i$ which is not a singleton, there is a surface $\Sigma_S$ satisfying the properties of Lemma <ref>.

Let $R_0$ be the union of the cubes $Q_{0,i}$ and

\[ \mathscr{S} = \bigcup_{i \in I} \{S : S \in \mathscr{S}_i\}. \]

Using the same terminology as [9], we call a cube $Q \subseteq R_0$ a transition cube if it is the top cube, or a minimal cube of some $S \in \mathscr{S}$. Let $T$ denote the set of transition cubes. For $Q \in \mathscr{D},$ let $N(Q)$ denote the number of transition cubes containing $Q$ (this is the quantity $\ell(Q)$ in [9]). Let $\eta > 0$ (which will be chosen small momentarily) and define

\[ \sigma(Q) =  \{ x \in Q : \dist(x,E \setminus Q) \leq \eta \rho^k \}. \]

Let $N \geq 1$ (which will be chosen large) and define $F \subseteq E \cap B(x,r)$ by setting

\[ F = E \cap B(x,r) \cap \left(\bigcup_{Q \in T} \sigma(Q) \right)^c \cap\left( \bigcup_{\substack{Q \in T\\ N(Q) \geq N}} Q \right)^c. \]

Now that we have defined $F$, let us show that it takes up a large portion of $E \cap B(x,r).$ We first need the following.

We have

\begin{align}\label{e:T}
\sum_{Q \in T} \ell(Q)^d \lesssim r^d.
\end{align}

Since $E$ is doubling (by virtue of being Ahlfors $d$-regular), there are a bounded number of cubes in $R_0$, with constant depending only on $d$ and the Ahlfors regularity constant. In other words

\[ \#I \lesssim 1. \]

Using this with (<ref>), and the fact that $r \sim \ell(Q_{0,i})$ for each $i \in I$, we have

\begin{align}\label{e:S}
\sum_{S \in \mathscr{S}} \ell(Q(S))^d \lesssim \sum_{i \in I} \sum_{S \in \mathscr{S}_i} \ell(Q(S))^d \lesssim \sum_{i \in I} \ell(Q_{0,i})^d \lesssim r^d.
\end{align}

Since $T$ is the collection of all top cubes and minimal cubes from the stopping-time regions in $\mathscr{S}$, and every minimal cube has a child which is a top cube, (<ref>) follows from (<ref>).

For $\eta$ small enough and $N$ large enough, $\mathscr{H}^d(E \cap B(x,r) \setminus F) \leq \ve r^d$.

By Lemma <ref>(4) and (<ref>), for $\eta > 0$ small enough, we have

\[ \sum_{Q \in T} \mathscr{H}^d(\sigma(Q)) \leq C \eta^{\frac{1}{C}} \sum_{Q \in T} \mathscr{H}^d(Q) \leq C \eta^{\frac{1}{C}} \sum_{Q \in T} \ell(Q)^d \leq C\eta^\frac{1}{C} r^d \leq \ve r^d/2. \]

Similarly, for $L$ large enough, if

\[ A = \bigcup_{\substack{Q \in T \\ N(Q) \geq N }} Q \]

\[ \mathscr{H}^d(A) \leq N^{-1} \sum_{Q \in T} \mathscr{H}^d(Q) \leq C N^{-1} \sum_{Q \in T} \ell(Q)^d \leq C N^{-1}r^d \leq \ve r^d/2. \]

Combing the above estimates, we have

\begin{align}\label{e:bigFmeasure}
\mathscr{H}^d(E \cap B(x,r) \setminus F) \leq \ve r^d.
\end{align}

The next step in the proof of Lemma <ref> is to construct the bi-Lipschitz map from $F$ to $\R^d$, which will be done by stitching together the maps associated to each stopping-time region $S \in \mathscr{S}$. The amount of stitching is controlled by the constant $N$. We begin by defining a preliminary map $g$ on the set of cubes $Q$ contained in $R_0$.

There exists a map $g$ from the set of transition cubes $T$ so that for each such $Q$, $g(Q)$ is a cube in $\R^d$ with $\diam(g(Q)) = C_3^{-N(Q)} \ell(Q)$ for some large constant $C_3 > 1$, and for any $Q,Q' \in T$ satisfying $Q \subseteq Q'$ then $g(Q) \subseteq \tfrac{1}{2}g(Q').$

We define $g$ inductively, starting with the cubes $Q_{0,i}.$ Since $\#I \lesssim 1,$ we can find $C > 1$ large enough so that we can set $\{g(Q_{0,i})\}_{i \in I}$ to be a collection of cubes in $\R^d$ with diameters $5\rho^k$ and mutual distances between $5\rho^k$ and $5C\rho^k$ (with $k$ as defined after the statement of Proposition <ref>). Since $N(Q_{i,0}) = 0$ for each $i \in I,$ the cubes $g(Q_{i,0})$ have the correct diameter.

We now define $g(Q)$ for the remaining transition cubes. Suppose $g(Q)$ has been defined for some $Q \in T_m = T \cap \mathscr{D}_m$ with $m \geq k.$ Assume first of all that $Q$ is a minimal cube of a stopping-time region. We will define $g(R)$ for each child $R$ of $Q$. By definition $N(R) = N(Q) + 1$ for each child $R$ of $Q$. Since $E$ is doubling, $Q$ has a bounded number of children (with constant depending only on $d$ and the Ahlfors regularity constant). Thus, for $C_3> 1$ large enough, we can choose $g(R)$ to be a collection of cubes with diameters $C_3^{-N(R)}\ell(R)$ contained in $\tfrac{1}{2}g(Q)$ and so that

\begin{align}\label{e:g-estimate}
\dist(g(R),g(R')) \geq C_3^{-N(R)} \ell(R).
\end{align}

Assume now that $Q$ is the top cube of a stopping-time region $S \in \mathscr{S}$, which is not a singleton. We want to define $g(R)$ for the minimal cubes of $S$. Here, we deviate slightly from that written in [9]. Let $h = h_S$ be the bi-Lipschitz map from Lemma <ref> and let $\Sigma = \Sigma_S = h(\R^d) \cap MB_{Q}.$ Since $h$ is bi-Lipschitz with constant close to 1, $h^{-1}(\Sigma \cap B_Q)$ is contained in some ball of radius $2\ell(Q).$ Thus, since $\diam(g(Q)) = C_3^{-N(Q)}\ell(Q),$ we can find an affine map $\phi = \phi_Q$ so that

\begin{align}\label{e:in-g(Q)}
\phi(h^{-1}(\Sigma \cap B_Q)) \subseteq \frac{1}{3}g(Q)
\end{align}

\begin{align}\label{e:phi-est} C_4^{-1}C_3^{-N(Q)}|p-q| \leq |\phi(p) - \phi(q)| \leq C_3^{-N(Q)}|p-q|
\end{align}

for some constant $C_4 \geq 1$ and each $p,q \in \R^d.$

For $R \in \min(S)$, let $z_R \in \Sigma$ be the point closest to $x_R$ (the centre of $R$). Define $g(R)$ to be the cube centred at $\phi(h^{-1}(z_R))$ of size $C_3^{-N(Q)-1} \ell(R) = C_3^{-N(R)}\ell(R).$ Clearly then $g(R) \subseteq \tfrac{1}{2}g(Q).$

Continuing with the assumption that $Q$ is the top cube of a stopping-time region $S$, which is not a singleton, similarly to (<ref>), we will need to estimate the relative distance between $g(R),g(R')$ for $R,R' \in \min(S)$. This will facilitate our forthcoming estimates on the function $f$ (still to be defined).

For $R,R' \in \min(S)$ there is $C_3 > 1$ large enough so that

\begin{align}\label{e:g-lip}
C_3^{-N(Q) -1}    [\ell(R) + \ell(R') + \dist(R,R')] &\leq \dist(g(R),g(R')) \\
&\hspace{2em}\leq   3C_3^{-N(Q)} [\ell(R) + \ell(R') + \dist(R,R')] .
\end{align}

Let $R_1,R_2 \in \min(S).$ Since $x_{R_i} \in c_0B_{R_i}$ for each $i=1,2,$ and $c_0B_{R_1} \cap c_0B_{R_2} = \emptyset$, we can find a constant $C_5$ so that

\begin{align}
C_5^{-1} [ \ell(R_1) + \ell(R_2) + \dist(R_1,R_2) ] \leq |x_{R_1} - x_{R_2}| \leq \ell(R_1) + \ell(R_2) + \dist(R_1,R_2).
\end{align}

Recalling the definition of $z_R$ (just below (<ref>)), by Lemma <ref> we know $|x_{R_i} - z_{R_i}| \leq C_6 \ve^\frac{1}{d+1} \ell(R_i)$ for some constant $C_6$ and for each $i =1,2.$ Recall also that, $h$ is $(1+C_7\ve)$-bi-Lipschitz for some constant $C_7.$ Choosing $\ve^\frac{1}{d+1} \leq \min\{(2C_5C_6)^{-1}, C_7^{-1}\},$ we have

\begin{align}
(2C_5)^{-1}  [\ell(R_1) + \ell(R_2) + \dist(R_1,R_2)]  &\leq C_5^{-1} [\ell(R_1) + \ell(R_2) + \dist(R_1,R_2)] \\
&\hspace{4em} - C_6 \ve^\frac{1}{d+1} \ell(R_1) - C_6 \ve^\frac{1}{d+1} \ell(R_2)   \\
&\leq |x_{R_1} - x_{R_2}| - C_6 \ve^\frac{1}{d+1} \ell(R_1) - C_6 \ve^\frac{1}{d+1} \ell(R_2) \\
&\leq |z_{R_1} - z_{R_2}| \leq (1+C_7\ve)|h^{-1}(z_{R_1}) - h^{-1}(z_{R_2})| \\
&\leq 2 |h^{-1}(z_{R_1}) - h^{-1}(z_{R_2})|.
\end{align}

After rearranging, this gives

\begin{align}\label{e:lower-h}
|h^{-1}(z_{R_1}) - h^{-1}(z_{R_2})| \geq (4C_5)^{-1}  [\ell(R_1) + \ell(R_2) + \dist(R_1,R_2)].
\end{align}

Similarly, one can also show

\begin{align}
|h^{-1}(z_{R_1}) - h^{-1}(z_{R_2})| \leq 2  [\ell(R_1) + \ell(R_2) + \dist(R_1,R_2)].
\end{align}

Let $y_{R_1} \in g(R_1)$ and $y_{R_2} \in g(R_2)$ be points so that $|y_{R_1}-y_{R_2}| = \dist(g(R_1),g(R_2)).$ Recalling that $g(R_i)$ is centred at $\phi(h^{-1}(z_{R_i}))$ and $\diam(g(R_i)) = C_3^{-N(Q)-1} \ell(R_i)$, choosing $C_3 \geq 8C_4C_5$, and using (<ref>) and (<ref>), we have

\begin{align}
\dist(g(R_1),g(R_2)) &= |y_{R_1}-y_{R_2}|  \\
& \geq |\phi(h^{-1}(z_{R_1})) - \phi(h^{-1}(z_{R_2}))| - |y_{R_1}- \phi(h^{-1}(z_{R_1}))| - |y_{R_2} - \phi(h^{-1}(z_{R_2}))| \\
& \geq (4C_4C_5)^{-1}C_3^{-N(Q)} [\ell(R_1) + \ell(R_2) + \dist(R_1,R_2)] \\
&\hspace{4em} - C_3^{-N(Q)-1} \ell(R_1) - C_3^{-N(Q)-1}\ell(R_2) \\
& \geq C_3^{-N(Q)-1} [ \ell(R_1) + \ell(R_2) + \dist(R_1,R_2) ],
\end{align}

The reverse inequality follows similarly;

\begin{align}
\dist(g(R_1),g(R_2)) &= |y_{R_1}-y_{R_2}|  \\
& \leq |\phi(h^{-1}(z_{R_1})) - \phi(h^{-1}(z_{R_2}))| + |y_{R_1}- \phi(h^{-1}(z_{R_1}))| + |y_{R_2} - \phi(h^{-1}(z_{R_2}))| \\
& \leq 2C_3^{-N(Q)} [\ell(R) + \ell(R') + \dist(R,R')] + C_3^{-N(Q)-1} \ell(R) + C_3^{-N(Q)-1}\ell(R') \\
& \leq 3C_3^{-N(Q)} [ \ell(R) + \ell(R') + \dist(R,R') ].
\end{align}

Now that we have defined $g,$ we can use it to define a function $f$. By definition, for each $x \in F$ there is a minimal transition cube, which we denote by $Q(x),$ so that $x \in Q(x).$ It must be that $Q(x)$ is top cube of a stopping-time region $S(x)$ (which is not a singleton), otherwise $x$ would be contained in a smaller transition cube. Furthermore, $x$ is contained in arbitrarily small cubes from $S(x)$ and so $x \in F \cap \Sigma_{S(x)}$ by (<ref>). Since $x \in \Sigma_{S(x)},$ $h_{Q(x)}^{-1}(x)$ makes sense, and we can define

\[ f(x) = \phi_{Q(x)}(h^{-1}_{Q(x)}(x)). \]

Notice, by (<ref>), $f(x) \in g(Q(x)).$ The proof of Lemma <ref> is finished once we show the following.

The map $f: F \rightarrow \R^d$ is $L$-bi-Lipschitz, where $L$ depends on $\ve.$

Let $x,y \in F$ be distinct points. Let $Q_1$ be the maximal cube contained in $R_0$ so that $x \in Q_1$ but $y \not \in Q_1.$ Assume to begin with that $Q_1$ is contained in some stopping-time region $S$, but that $Q_1$ is not the top cube of $S.$ By maximality, any cube containing $Q_1$ contains both $x$ and $y$, in particular, this holds true for $Q(S).$ Assume further that there are cubes $R_x,R_y \in \min(S)$ so that $x \in R_x$ and $y \in R_y.$ This implies, first, that

\begin{align}\label{e:R_x-low}
|x-y| \leq \ell(R_x) + \ell(R_y) + \dist(R_x,R_y).
\end{align}

Secondly, since $x \not \in \sigma(R_x)$ and $y \not\in \sigma(R_y)$ (by the definition of $F$), we also have

\begin{align}\label{e:R_x-up}
|x-y| \gtrsim  \ell(R_x) + \ell(R_y) + \dist(R_x,R_y).
\end{align}

The estimates on $f$ then follow from the fact that $f(x) \in g(R_x),$ $f(y) \in g(R_y),$ $N(Q) \leq N$ and the estimates (<ref>), (<ref>), and (<ref>), since

\begin{align}
|x -y| &\leq  \ell(R_x) + \ell(R_y) + \dist(R_x,R_y) \lesssim_L \dist(g(R_x),g(R_y)) \leq |f(x) - f(y)| \\
&\leq \dist(g(R_x),g(R_y)) + \diam(R_x) + \diam(R_y) \lesssim_N \ell(R_x) + \ell(R_y) + \dist(R_x,R_y) \\
&\lesssim |x-y|.
\end{align}

If it is the case that $x$ is not contained in any cube from $\min(S)$, then $f(x) = \phi_{Q(S)}(h^{-1}_{Q(S)}(x))$. By similar arguments to (<ref>), it can be shown that

\[ C_3^{-N(Q) -1}    [ \ell(R_y) + \dist(x,R_y)] \leq \dist(x,g(R_y)) \leq   3C_3^{-N(Q)} [\ell(R_y) + \dist(x,R_y)], \]

and the estimates follow in the same way as above. Swapping the roles of $x$ and $y$, the same is true if $y$ is not contained in any cube from $\min(S).$ If neither $x$ nor $y$ are contained in minimal cubes, the estimates follow directly from an argument similar to that which established (<ref>).

Now we are left with the case that $Q_1$ is the top of a stopping-time region. Let $Q_2$ be the parent of $Q_1.$ It must be that $Q_2$ is a transition cube, and by maximality, $x,y \in Q_2.$ Since $x,y \in Q_2$, $x \in Q_1\setminus \sigma(Q_1)$ and $y \not\in Q_1$ we have $|x-y| \sim \ell(Q_1).$
To estimate the distance between $f(x)$ and $f(y)$, first note that since $x,y \in Q_2,$ we have $f(x),f(y) \in g(Q_2)$ and so

\[|f(x) - f(y)| \leq \diam(g(Q_2)) \lesssim \ell(Q_2) \sim \ell(Q_1) \sim |x-y| .\]

Next, let $Q_1'$ be the sibling of $Q_1$ so that $y \in Q_1'.$ By (<ref>) we have

\[ |f(x) - f(y)| \geq \dist(g(Q_1),g(Q_1')) \gtrsim \ell(Q_1) \sim |x-y| \]

and this finishes the proof of the bi-Lipschitz estimates.

Now for the converse to Proposition <ref>.

Suppose $E$ is UR, then $E$ satisfies the $\BWGL.$

We will need the following two lemmas.

Let $A \subseteq \R^d$ be compact and $f : A \rightarrow H$ be an $L$-bi-Lipschitz function. Then there exists constant $m$ and $M$ (depending only on $d$ and $L$), and an extension $g : \R^d \rightarrow \tilde{H} \coloneqq H \times \R^m$ of $f$ which is $M$-bi-Lipschitz.

The above lemma is essentially <cit.>. The proof is identical up to notational changes (swapping $\R^n$ for $H$). Additionally, the constant $M$ in <cit.> is actually independent of $n$.

Let $E \subseteq H$ and suppose there are constants $C$ and $\theta > 0$ so that for each $x \in E$ and $r \in (0,\diam E),$ there exists an Ahlfors $d$-regular set $\tilde{E}$ with regularity constant at most $C$, so that

\[ \mathscr{H}^d(E \cap B(x,r) \cap \tilde{E}) \geq \theta r^d, \]

and so that $\tilde{E}$ satisfies the $(2,2)$-geometric lemma, meaning

\[ \sum_{\substack{R \in \mathscr{D}^{\tilde{E}} \\ R \subseteq Q}} \beta_{\tilde{E}}^{d,2}(C_0B_R)^2\ell(R)^d \lesssim \ell(Q)^d \]

for all $C_0 >1$ and $Q \in \mathscr{D}^{\tilde{E}}.$ Then $E$ also satisfies the $(2,2)$-geometric lemma.

The above is a consequence of <cit.>, whose proof is found in <cit.> and remains valid for subsets of $H$. For the definition of the general $(p,q)$-geometric lemma, see <cit.>. This is stated in terms of a Carleson measure condition, however, the above formulation is equivalent and is similar to how we formulated the BWGL, see after Definition <ref>.

Let $A \geq 1$ and $\ve >0$. We need to show $\BWGL(Q,A,\ve) < C(A,\ve)\ell(Q)^d$ for all $Q \in \mathscr{D}.$ Let $C_0 \geq 2\rho^{-1}.$ By Theorem <ref>, with $p = 2,$ we know

\[ \BWGL(Q,A,\ve) \lesssim_{A,\ve} \ell(Q)^d + \sum_{\substack{R \in \mathscr{D} \\ R \subseteq Q}} \beta^{d,2}_E(C_0B_R)^2\ell(R)^d \]

for all $Q \in \mathscr{D}.$ Thus, we are done if we can show that

\begin{align}\label{e:letsshow0}
\sum_{\substack{R \in \mathscr{D} \\ R \subseteq Q}} \beta^{d,2}_E(C_0B_R)^2\ell(R)^d \lesssim \ell(Q)^d
\end{align}

for all $Q \in \mathscr{D}.$ This will follow from the fact that $E$ satisfies the $(2,2)$-geometric lemma, which we verify by Lemma <ref>. First, by <cit.>, $E$ has big pieces of bi-Lipschitz images of $\R^d,$ meaning there are constants $L,\theta>0$ such that for all $x \in E$ and $r \in (0,\diam E),$ there is a compact subset $A \subseteq \R^d$ and an $L$-bi-Lipschitz map $f:A \rightarrow H$ such that

\begin{align}\label{e:BPBLI}
\mathscr{H}^d(E \cap B(x,r) \cap f(A)) \geq \theta r^d.
\end{align}

Let $x \in E$, $r \in (0,\diam E)$, and let $A = A_{x,r}$ and $f = f_{x,r}$ be as above. By Lemma <ref> (and identifying $\tilde{H}$ with $H$), we can extend $f$ to a bi-Lipschitz function $g: \R^d \rightarrow H$, and we set $\tilde{E} = g(\R^d).$ Clearly

\[ \mathscr{H}^d(E \cap B(x,r) \cap \tilde{E}) \geq \theta r^d. \]

Additionally, $\tilde{E}$ satisfies the $(2,2)$-geometric lemma. To see this, notice that $\tilde{E}$ satisfies the BWGL by Lemma <ref>, and so, by Theorem <ref>,

\[ \sum_{\substack{Q \in \mathscr{D}^{\tilde{E}} \\ R \subseteq Q}} \beta_{\tilde{E}}^{d,2}(C_0B_R)^2\ell(R)^d \lesssim \mathscr{H}^d(Q) \lesssim \ell(Q)^d \]

for all $C_0 >1$ and $Q \in\mathscr{D}^{\tilde{E}}.$ Equation (<ref>) and hence the proposition now follows by Lemma <ref>.

§ PARAMETERISATION IN HILBERT SPACE

In this section, we prove some of the main results in Theorem <ref>. The main objective is to prove the local Lipschitz description of the surfaces $\Sigma_k$ – Theorem <ref> (7). We will prove this and all the results necessary preliminary results, which include Theorem <ref> (10) and (11). In addition, we will prove Theorem <ref>(13). The results Theorem <ref> (2) and (4)-(6) are immediate from the definition in Section <ref>. For the remaining points of Theorem <ref> ((1),(3),(8),(9) and (12)), we direct the reader to the proofs contained in [12], which are self-contained given the results we prove in this section. We shall on comment on this after we are finished in (7).

§.§ Preliminaries
We collect a series of preliminary lemmas, to aid the proof of Theorem <ref> (7). Included here is the proof of Theorem <ref> (11).

Let $P,P'$ be $d$-planes in $H$, $x \in P$ and $B$ a ball centred on $x.$ Let $\ve >0$ and suppose

\begin{align}\label{e:d_B1}
d_B(P,P') \leq \ve.
\end{align}

Let $P''$ be the plane parallel to $P'$ such that $x \in P''.$ Then for any $A >0$ we have

\begin{align}\label{e:d_B2}
d_{AB}(P,P'') \lesssim \ve
\end{align}

\begin{align}\label{e:d_B3}
d_{AB}(P,P') \lesssim \ve (1 + 1/A).
\end{align}

Let $A >0$ and $P''$ be as above. Since $P'$ and $P''$ are parallel, $x \in P''$ and $\dist(x,P') \leq \ve r_B$, we have

\begin{align}\label{e:dP'P''}
\dist(P',P'') \leq \ve r_B.
\end{align}

So, for each $y \in P' \cap AB,$ we can find $z \in P'' \cap AB$ such that $|y-z| \lesssim \ve r_B.$ Similarly, for each $z \in P'' \cap AB$, we can find $y \in P' \cap AB$ such that $|y-z| \lesssim \ve r_B.$ This implies

\begin{align}
r_{AB}d_{AB}(P,P') \lesssim \ve r_B + r_{AB}d_{AB}(P,P'')
\end{align}

and so (<ref>) will follow once we show (<ref>). Furthermore, since $x \in P \cap P''$ we have

\[ d_{\frac{1}{2}B}(P,P'') = d_{AB}(P,P''),\]

so it suffices to show

\[ d_{\frac{1}{2}B}(P,P'') \lesssim \ve.\]

This will be the goal for the rest of the proof. Let $z \in P'' \cap \tfrac{1}{2}B.$ By (<ref>) there exists $z' \in P'$ such that

\begin{align}\label{e:zz'}
|z - z'| \leq \ve r_B.
\end{align}

It follows that $z' \in P' \cap (\tfrac{1}{2} + \ve)B$ which by (<ref>). implies the existence of a point $y' \in P \cap B$ such that

\begin{align}\label{e:z'y'}
|z' - y'| \leq \ve r_B.
\end{align}

By (<ref>), since $z' \in (\tfrac{1}{2} + \ve)B,$ it actually follows that $y' \in P \cap (\tfrac{1}{2}+2\ve)B$. Since $B$ is centred on $x$ and $x \in P$ there exists a point $y \in P \cap \tfrac{1}{2}B$ such that

\begin{align}\label{e:y'y}
|y' - y| \leq 2\ve r_B.
\end{align}

Combing the above, we have

\begin{align}
\dist(z, P \cap \tfrac{1}{2}B) \leq |z - y| \leq |z - z'| + |z' - y'| + |y'-y| \leq 4\ve r_B.
\end{align}

Since $z$ was an arbitrary point in $P'' \cap \tfrac{1}{2}B$, this gives

\[ \sup_{y \in P'' \cap \frac{1}{2}B} \dist(y, P \cap \tfrac{1}{2}B) \leq 4\ve r_B. \]

Swapping the roles of $P$ and $P''$ we also get

\[ \sup_{y \in P \cap \frac{1}{2}B} \dist(y, P'' \cap \tfrac{1}{2}B) \leq 4\ve r_B \]

which finishes the proof.

Suppose $P,P'$ are two $d$-planes in $H$ which contain the origin and satisfy $d_{B(0,1)}(P,P') \leq \ve.$ Then for $y \in P'$, we have

\begin{align}\label{e:proj}
|\pi^\perp_P(y)| \lesssim \ve |y|,
\end{align}

where $\pi_P^\perp$ denotes the orthogonal projection on $P^\perp,$ the linear space orthogonal to $P$

Let $B = B(0,1).$ Since $0 \in P \cap P'$ and $d_{B}(P,P') \leq \ve,$ we have $d_{AB}(P,P') \leq \ve$ for all $A >0.$ So, taking $A = |y|,$ we get

\begin{align}
|\pi^\perp_P(y)| = |y - \pi_P(y)| = \dist(y,P) \leq |y|d_{B(0,|y|)}(P,P') \leq \ve |y|.
\end{align}

Let $B$ be a ball and suppose $P,P'$ are $d$-planes in $X$. For each $\alpha >0$ there exists $\ve > 0$ such that if

\[ d_B(P,P') \leq \ve \]

\[  P \cap (1-\alpha)B \subseteq \pi_P( P' \cap B). \]

The proof of Lemma <ref> is similar to the topological argument used to prove Lemma <ref>, we shall omit the details. The following is Lemma <ref> (11).

$\emph{(11)}$ holds.

Let $y \in \Sigma_k \cap V_k^8$. By Lemma <ref> this mean that

\[\sum_{j \in J_k} \theta_{j,k}(y) = 1.\]

Then, if $i \in J_k$ is such that $y \in 10B_{i,k}$, we have

\begin{align}
|\sigma_k(y) - \pi_{i,k}(y)| &= \big|y + \sum_{j \in J_k} \theta_{j,k}(y)[\pi_{j,k}(y)-y] - \pi_{i,k}(y) \big|   \\
&=\big| \sum_{j \in J_k} \theta_{j,k}(y)[\pi_{j,k}(y) - \pi_{i,k}(y)] \big| \leq \sum_{j \in J_k} |\theta_{j,k}(y)| |\pi_{j,k}(y) - \pi_{i,k}(y)|.
\end{align}

Since $\supp \theta_{j,k} \subseteq 10B_{j,k}$, if $j \in J_k$ is such that $\theta_{j,k}(y) \not=0$, then $y \in 10B_{j,k}$. In this case $|x_{j,k} - x_{i,k}| \leq 20r_k.$ This, along with the definition of $\ve_k(y)$, gives

\begin{align}\label{e:sig-pi1}
d_{x_{i,k},100r_k}(P_{i,k},P_{j,k}) \leq \ve_k(y)
\end{align}

which implies

\begin{align}\label{e:sig-pi2}
|\pi_{j,k}(y) - \pi_{i,k}(y)| \leq C\ve_k(y) r_k.
\end{align}

As shown in the proof of Lemma <ref>, the balls $10B_{j,k}, \ j \in J_k$ have bounded overlap, hence

\[ |\sigma_k(y) - \pi_{i,k}(y)| \leq \sum_{j \in J_k} |\theta_{j,k}(y)| C\ve_k(y) r_k \leq C \ve_k(y) r_k  \]

as required.

Let $k \geq 0$, $y,z \in \Sigma_k \cap V_k^8$ and $i \in J_k$ so that $y,z \in 10B_{i,k},$ then

\[ |(\sigma_k(y) - \sigma_k(z)) - (\pi_{i,k}(y) - \pi_{i,k}(z))| \lesssim \ve_k(y) |y-z|  \]

Using the fact that $y,z \in V_k^8$ and so $\sum_{j \in J_k} \theta_{j,k}(y) = 1$, with the analogous quantity holding for $z,$ we have

\begin{align}
|(\sigma_{k}(y) - \sigma_k(z)) - (\pi_{i,k}(y) - \pi_{i,k}(z))| &\leq  \sum_{j \in J_k} |\theta_{j,k}(y)| | \pi_{j,k}(y-z) - \pi_{i,k}(y-z)|  \\
&\hspace{4em} +  \sum_{j \in J_k} |\theta_{j,k}(y) - \theta_{j,k}(z)||\pi_{j,k}(z) - \pi_{i,k}(z)| \\
&= S_1 + S_2.
\end{align}

We start by estimating $S_1.$ By (<ref>), for each $j \in J_k$ such that $\theta_{j,k}(y) \not= 0,$ we have

\[d_{x_{i,k},100r_k}(P_{i,k},P_{j,k}) \leq \ve_k(y). \]

Then, Lemma <ref> implies

\begin{align}
|\pi_{j,k}(y-z) - \pi_{i,k}(y-z)| &\leq |\pi_{j,k}(y-z) - \pi_{i,k}(\pi_{j,k}(y-z)) | + |\pi_{i,k}(\pi_{j,k}(y-z)) - \pi_{i,k}(y-z)| \\
&=| \pi_{i,k}^\perp(\pi_{j,k}(y-z))| + | \pi_{i,k}(\pi_{j,k}^\perp(y-z))| \\
&\lesssim \ve_k(y) |y-z|.
\end{align}

Since the ball $10B_{j,k}, \ j \in J_k,$ have bounded overlap, we conclude that $S_1 \lesssim \ve_k(y) |y-z|.$ We also see that $S_2 \lesssim \ve_k(y) |y-z|$ by the bounded overlap of the balls $10B_{j,k},$ (<ref>), and Lemma <ref>, since

\begin{align}
|\theta_{j,k}(y) - \theta_{j,k}(z)||\pi_{j,k}(z) - \pi_{i,k}(z)| \lesssim r_k^{-1}|y-z| \ve_k(z) r_k \lesssim \ve_k(z) |y-z|.
\end{align}

§.§ Local Lipschitz description

We can now begin the proof of Theorem <ref>(7), which follows from the proposition below. This is the analogue of <cit.>.

There are constant $\{C_i\}_{i=1}^7$ such that the following holds for all $k \geq 0$ and $j \in J_k.$ First, there is a function $A_{j,k} : P_{j,k} \cap 49B_{j,k} \rightarrow P_{j,k}^\perp$ such that

\begin{align}\label{e:Anorm}
|A_{j,k}(x_{j,k})| \leq C_1\ve r_k.,
\end{align}

\begin{align}\label{e:Alip}
|DA_{j,k}(z)| \leq C_2 \ve, \quad \text{for all} \ z \in P_{j,k} \cap 49B_{j,k}
\end{align}

and if $\Gamma_{A_{j,k}}$ is the graph of $A_{j,k}$ over $P_{j,k}$ then

\begin{align}\label{e:graphA}
\Sigma_{k} \cap D(x_{j,k},P_{j,k},49r_k) = \Gamma_{A_{j,k}} \cap D(x_{j,k},P_{j,k},49r_k).
\end{align}

Next, there is a function $F_{j,k} : P_{j,k} \cap 40B_{j,k} \rightarrow P_{j,k}^\perp$ such that

\begin{align}\label{e:Fnorm}
|F_{j,k}(x_{j,k})| \leq C_3\ve r_k,
\end{align}

\begin{align}\label{e:Flip}
|DF_{j,k}(z)| \leq C_4\ve  \quad \text{for} \ z \in P_{j,k} \cap 40B_{j,k},
\end{align}

\begin{align}\label{e:Flip1}
|DF_{j,k}(z)| \leq C_5\ve \quad \text{for} \ z \in P_{j,k} \cap 7B_{j,k},
\end{align}

and if $\Gamma_{F_{j,k}}$ is the graph of $F_{j,k}$ over $P_{j,k}$ then

\begin{align}\label{e:graphF}
\Sigma_{k+1} \cap D(x_{j,k},P_{j,k},40r_k) = \Gamma_{F_{j,k}} \cap D(x_{j,k},P_{j,k},40r_k).
\end{align}

For a given plane $P$, a ball $B$, and a differentiable function $f:P \rightarrow H$, it is not difficult to show that $|Df(z)| \leq C$ for all $z \in P \cap B$ if and only if $|f(x) - f(y)| \leq C|x-y|$ for all $x,y \in P \cap B.$ In the following proof, we will make use of the second characterization.

We prove Proposition <ref> by induction. Since $\Sigma_0 = P_0$ it is clear that (<ref>)-(<ref>) hold for $k=0$, which establishes the base case. The induction will be carried out in two stages.

§.§ Stage 1
In this section, we show the following.

If (<ref>)-(<ref>) hold for $k$, then (<ref>)-(<ref>) hold for $k+1.$

Let $j \in J_{k+1}.$ Our goal is to define a map $A_{j,k+1}$ so that (<ref>)-(<ref>) hold. We need some preliminary results. First, by Definition <ref> (1) there is $i \in J_k$ such that

\begin{align}\label{e:2B_ik}
x_{j,k+1} \in 2B_{i,k}.
\end{align}

By translating we may assume $x_{i,k} = 0,$ so that $P_{i,k}$ is a linear subspace. To simplify notation we shall write $V = P_{i,k}$ and $W=P_{j,k+1}$. By Definition <ref> (5), we have

\begin{align}\label{e:dVW}
d_{0,20r_k}(V,W) \leq \ve,
\end{align}

so we know that there exists $p \in W \cap B(0,20r_k)$ satisfying

\begin{align}\label{e:p}
|p| \leq 20\ve r_k.
\end{align}

Let $W'$ denote the linear subspace parallel to $W$. By Lemma <ref> we have for any $A >0$,

\begin{align}\label{e:dVW1}
d_{0,20Ar_k}(V,W') \lesssim \ve \quad \text{and} \quad d_{0,20Ar_k}(V,W) \lesssim \ve(1+1/A).
\end{align}

Let $\pi$ denote the orthogonal projection onto $V$, $\pi_W$ denote the orthogonal projection on $W$, $\pi_W^\perp$ denote the orthogonal projection on $W^\perp$ (the linear subspace orthogonal to $W'$) and $\pi_{W'}$ denote the orthogonal projection onto $W'$. In this way,

\begin{align}\label{e:defpiw}
\pi_{W}(x) = \pi_{W}^\perp(p) + \pi_{W'}(x).
\end{align}

Since $W^\perp$ and $W'$ are linear subspaces, and orthogonal projections on linear subspaces are bounded linear operator with norm at most 1, we have

\begin{align}\label{e:pi_W}
| \pi_{W}(y) | \leq |\pi_{W}^\perp(p)| + |\pi_{W'}(y)| \leq |p| + |y| \leq 20\ve r_k + |y|.
\end{align}

Let $F = F_{i,k}$ be the map satisfying (<ref>)-(<ref>) for $k$ and $i \in J_k$ and define a map $f: W \cap 39B_{i,k} \rightarrow W$ by setting

\[ f(y) = \pi_W ( \pi(y) + F(\pi(y))). \]

See Figure <ref>. We have the following.

Definition of $f$.

The map $f$ is well-defined and has a Lipschitz inverse $f^{-1}: W \supseteq U \rightarrow W \cap 39B_{i,k}$ such that

\begin{align}\label{e:fprop}
\emph{Lip}(f^{-1}) + ||f^{-1}|| \leq 3.
\end{align}

We start by considering the action of $\pi$ on elements of $W \cap 39B_{i,k}$. Along the way, we show $f$ is well-defined. Let $y \in W \cap 39B_{i,k}$. Recall that since $x_{i,k} = 0$ we have $|y| \leq 39r_k.$ By (<ref>) (with $A = |y|/20r_k$) we have

\begin{align}\label{e:pi-y}
|\pi(y) - y| &= \dist(y,V) \leq |y|d_{B(0,|y|)}(V,W) \stackrel{\eqref{e:dVW1}}{\leq} C\ve|y|( 1 + 20r_k/|y|) \\
&= C\ve(|y| + 20r_k) \lesssim \ve r_k.
\end{align}

Hence, if $\ve$ is small enough $\pi(y) \in V \cap 40B_{i,k} = \text{Dom}(F)$ for all $y\in W \cap 39B_{i,k}$, and so $f$ is well-defined. Next, we establish some bi-Lipschitz estimates for $\pi$. Let $y,z \in W.$ Since $\pi$ is linear and $y-z \in W'$, by (<ref>) and Lemma <ref>, we have

\begin{align}\label{e:pi-yz}
|(\pi(y) - \pi(z)) - (y-z)| = |\pi(y-z) - (y-z)| \leq C\ve |y-z|.
\end{align}

Now that we have information for $\pi,$ we can prove estimates similar to (<ref>) and (<ref>) for $f$. Notice, if $y,z \in W \cap 39B_{i,k}$ then

\begin{align}\label{e:fdist}
|f(y) -y| &= |\pi_W( \pi(y) - y + F(\pi(y)))| \\
&\leftstackrel{\eqref{e:pi_W}}{\leq} 20\ve r_k + |\pi(y) - y + F(\pi(y))| \\
&\leq 20\ve r_k +  |\pi(y) -y | + |F(\pi(y))| \stackrel{\eqref{e:pi-y} \atop \eqref{e:Fnorm}}{\leq} (C + C_3)\ve r_k
\end{align}

\begin{align}\label{bilipf}
|f(y) - y - (f(z) - z)| &= |\pi_{W'}( \pi(y) - \pi(z) - (y-z) +  F(\pi(y)) -  F(\pi(z)) )| \\
&\leq |\pi(y) - \pi(z) - (y-z)| + |F(\pi(y)) -  F(\pi(z))| \\
&\leftstackrel{\eqref{e:pi-yz} \atop \eqref{e:Flip}}{\leq} (C+ C_4) \ve |y-z|.
\end{align}

By (<ref>), choosing $\ve$ small enough, we see that $f$ is bijection from $W \cap 39B_{i,k}$ onto $U = f(W \cap 39B_{i,k}) \subseteq W$. Thus, it has an inverse $f^{-1} : U \rightarrow W \cap 39B_{i,k}$. It is not difficult to show that (<ref>) and (<ref>) imply (<ref>) for $\ve$ small enough.

Now, we can define the function $A: U \rightarrow W^\perp$ by setting

\[ A(y) = \pi^\perp_W( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y))) ) - \pi_W^\perp(p). \]

See Figure <ref>.

Definition of $A$.

By the definition of $f$ (at the point $f^{-1}(y))$, we have

\[y = \pi_W ( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y)))).\]

Using this, and (<ref>), for any $y \in U$ we have

\begin{align}\label{e:AF}
\begin{split}
y + A(y) &= \pi_W ( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y)))) + \pi^\perp_W( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y))) ) - \pi_W^\perp(p) \\
&=  \pi_{W'} ( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y)))) + \pi^\perp_W( \pi(f^{-1}(y)) + F( \pi(f^{-1}(y))) ) \\
&=\pi(f^{-1}(y)) + F( \pi(f^{-1}(y))).
\end{split}
\end{align}

Now that we have defined $A$, we will establish (<ref>)-(<ref>). We start with (<ref>), which is the following.

We have

\[ \Gamma_A \cap D(x_{j,k+1},W,49r_{k+1}) = \Sigma_{k+1} \cap D(x_{j,k+1},W,49r_{k+1}). \]

Before we can show this, we need the following.
We have

\begin{align}\label{e:D}
D(x_{j,k+1},W,49r_{k+1}) \subseteq D(0,V,35r_k).
\end{align}

Let $y \in D(x_{j,k+1},W,49r_{k+1}).$ We want to show $|\pi(y)| \leq 35r_k$ and $|\pi(y) -y|  \leq 35 r_k$.  We start by showing $|\pi(y) - \pi_W(y)| \lesssim \ve r_k$ as follows. First,

\begin{align*}
|\pi(y) - \pi_W(y)| &\leq |\pi(y) - \pi_{W'}(y)| + |\pi_{W}^\perp(p)| \\
&\leftstackrel{\eqref{e:p}}{\leq} | \pi(y) - \pi (\pi_{W'}(y))| + |\pi(\pi_{W'}(y)) - \pi_{W'}(y)| + 20\ve r_k \\
&= |\pi(\pi_W^\perp(y))| + |\pi^\perp(\pi_{W'}(y))| + 20\ve r_k \\
&\leq C \ve(|\pi_W^\perp(y)| + |\pi_W(y)| + r_k),
\end{align*}

where the last inequality follows from (<ref>) and Lemma <ref>. Continuing from this, since $y,p \in W$ and $W^\perp$ is orthogonal to $W,$ we must have $\pi_W^\perp(y) = \pi_W^\perp(p).$ So,

\begin{align*}
|\pi(y) - \pi_W(y)| &\leq C \ve(|\pi_W^\perp(p)| + |\pi_W(y)| + r_k) \\
&\stackrel{\eqref{e:pi(y)}}{\leq} C\ve ( |p| + 49r_{k+1} + r_k ) \\
&\leq C\ve r_k.
\end{align*}

Our required bounds for $|\pi(y)|$ and $|\pi(y) - y|$ now follow for $\epsilon$ small enough since

\begin{align}\label{e:pi(y)}
\begin{split}
|\pi(y)| & \leq |\pi(y) - \pi_W(y)| + |\pi_W(y) - x_{j,k+1}| + |x_{j,k+1}| \\
&\leftstackrel{\eqref{e:2B_ik}}{\leq} |\pi(y) - \pi_W(y)| + 49r_{k+1} + 2r_k \\
&\leq  |\pi(y) - \pi_W(y)| + 7r_k \leq 8r_k \leq 35r_k
\end{split}
\end{align}

\begin{align}\label{e:pD2}
\begin{split}
|\pi(y) -y | &\leq |\pi(y) - \pi_W(y)| + |\pi_W(y) - y| \\
&\leq C\ve r_k + 49r_{k+1} \leq 5 r_k \leq 35r_k.
\end{split}
\end{align}

Since $\Gamma_A \subseteq \Sigma_{k+1}$ by (<ref>), the forward inclusion is clear. Our goal now is to show the backward inclusion. Let $p \in \Sigma_{k+1} \cap D(x_{j,k+1},W,49r_{k+1}),$ we must show $p \in \Gamma_A.$ By (<ref>) we have $p \in D(0,V,35r_k),$ which, since $\Sigma_{k+1}$ is the graph of $F$ over $D(0,V,38r_k)$, means we can find a point $y \in V \cap 35B_{i,k}$ such that

\[ p = y + F(y). \]

For $\ve$ small enough we have $V \cap 35B_{i,k} \subseteq \pi(W \cap 38B_{i,k})$ by Lemma <ref>, so we can find a point $z \in W \cap 38B_{i,k}$ such that $y = \pi(z).$ Then,

\[ p = \pi(z) + F(\pi(z)) . \]

Let $w = f(z) \in U.$ Since $f$ is bijective we can write

\[ p = \pi(f^{-1}(w)) + F( \pi(f^{-1}(w))) \]

and so (<ref>) implies

\[ p = w + A(w) \in \Gamma_A \]

as required.

To finish the proof of Lemma <ref> and hence Stage 1, we are left to show the following.

We have $|A(x_{j,k+1})| \leq C_1\ve r_k$ and $|DA(z)| \leq C_2 \ve$ for all $z \in W \cap 49B_{j,k+1}.$

We start with by estimating $|A(x_{j,k+1})|.$ To simplify notation we will write $x = x_{j,k+1}.$ Notice, in the definition of $A$, we can replace $p$ with any point in $W.$ Choosing the point $f^{-1}(x) \in W,$ we have

\begin{align*}
|A(x)| &= |  \pi^\perp_W( \pi(f^{-1}(x)) + F( \pi(f^{-1}(x))) ) - \pi_W^\perp(f^{-1}(x)) | \\
&=|\pi^\perp_W( \pi(f^{-1}(x)) - f^{-1}(x) + F( \pi(f^{-1}(x)))  )| \\
&\leq |\pi(f^{-1}(x)) - f^{-1}(x)| + |F( \pi(f^{-1}(x)))| \\
&\leftstackrel{\eqref{e:pi-y} \atop \eqref{e:Fnorm}}{\leq} (C + C_3)\ve |f^{-1}(x)| \leq C(1 +C_3) \ve |x| \\
&\leftstackrel{\eqref{e:2B_ik}}{\leq} C(1+C_3)\ve r_{k+1}.
\end{align*}

The required estimate holds as long as we choose

\[ C(1+C_3) \leq C_1. \]

Now we estimate $|DA(z)|,$ recall Remark <ref>. Let $y,z \in W \cap 49B_{j,k+1}.$ By (<ref>), we know that $f^{-1}(y),f^{-1}(z) \in W \cap 49.5B_{j,k+1}$, which by (<ref>), implies

\[\pi(f^{-1}(y)),\pi(f^{-1}(z)) \in V \cap 50B_{j,k+1} \stackrel{\eqref{e:2B_ik}}{\subseteq} V \cap 7B_{i,k}.\]

\begin{align*}
|A(y) - A(z)| &\leq |\pi( f^{-1}(y) - f^{-1}(z))| + |F( \pi(  f^{-1}(y)) - F( \pi(f^{-1}(z)))| \\
&\leftstackrel{\eqref{e:Flip1}}{\leq} (1 + C_5) |\pi(f^{-1}(y)) - \pi(f^{-1}(z))| \\
&\leq C(1 + C_5)\ve |f^{-1}(y)-f^{-1}(z)| \\
&\leq C(1+C_5)\ve |y-z|,
\end{align*}

where the third inequality follows from (<ref>) since $f^{-1}(y) - f^{-1}(z) \in W'.$ So, if

\[ C(1+C_5) \leq C_2 \]

then (<ref>) holds.

§.§ Stage 2
In this section we show the following.

If (<ref>)-(<ref>) hold for $k$ then (<ref>)-(<ref>) also hold for $k$.

Let $j \in J_k$. Our ultimate goal is to define a map $F = F_{j,k}$ so that (<ref>)-(<ref>) hold. We start with some preliminaries. By translation, we may assume $x_{j,k} = 0.$ Let $P_{j,k},B_{j,k}$ and $A_{j,k}$ satisfy (<ref>)-(<ref>). For ease of notation we denote

\[P = P_{j,k},  \ B = B_{j,k} \ \text{and} \  A =A_{j,k}.\]

Also, we define $\pi$ to be the orthogonal projection onto the plane $P$. Before we can define $F$, we need to prove some preliminary estimates and define an auxiliary function. The first of these estimates is Theorem <ref> (10).

Let $y \in \Sigma_k.$ Then

\begin{align}\label{e:sig-y}
|\sigma_k(y) - y| \leq C(C_1 + 1) \ve r_k.
\end{align}

If $y \not\in V_k^{10}$ then by definition $\sigma_k(y) = y$ and (<ref>) is trivial. Assume instead that $y \in V_k^{10}.$ Let $j \in J_k$ such that $y \in 10B_{j,k}$ Let $A_{j,k}$ be the map which satisfies (<ref>)-(<ref>). If $i \in J_k$ also satisfies $y \in 10B_{i,k}$ then by the Definition <ref> (3) we have

\[ d_{x_{j,k},100r_k}(P_{j,k},P_{i,k}) \leq \ve \]

and so

\begin{align}\label{e:before}
|\pi_{i,k}(y) - y| &= \dist(y,P_{i,k})  \leq  | \pi_{j,k}(y)- y|  + \dist( \pi_{j,k}(y),P_{i,k}) \\
&\leq  |A_{j,k}(\pi_{j,k}(y))| + 100\ve r_k \stackrel{\eqref{e:Anorm}}{\leq} (C_1 + C )\ve r_k.
\end{align}

Since the functions $\theta_{j,k}$ are supported on $10B_{j,k}$, which have bounded overlap (by the arguments in the proof of Lemma <ref>), we can use this and the above to get

\begin{align}
|\sigma_k(y) -y| \leq \sum_{i \in J_k} |\theta_{i,k}(y)||\pi_{i,k}(y) - y| \leq C(C_1 +1) \ve r_k,
\end{align}

which finishes the proof.

Let $y,z \in \Sigma_k \cap 45B.$ Then,

\begin{align}\label{e:siglip}
| (\sigma_k(y) - y) - (\sigma_k(z) - z)| \leq C(C_1 +C_2 +1)\ve|y-z|.
\end{align}

First, we write

\begin{align}
|(\sigma_k(y)-y) - (\sigma_k(z) -z)| &\leq \sum_{i \in J_k}|\theta_{i,k}(y) - \theta_{i,k}(z)||\pi_{i,k}(y) - y| \\
&\quad+ \sum_{i \in J_k} |\theta_{i,k}(y)| |\pi_{i,k}^\perp(y - z)| \\
&=S_1 + S_2.
\end{align}

Since $y,z \in 45B,$ if for some $i \in J_k$ we have either $\theta_{i,k}(y) \not=0$ or $\theta_{i,k}(z) \not=0$ then $|x_{i,k}| \leq 100r_k.$ By the Definition <ref> (3),  we get

\begin{align}\label{e:PPik}
d_{0,100r_k}(P,P_{i,k}) \leq \ve.
\end{align}

By exactly the same estimate as in (<ref>), this implies

\begin{align}
|\pi_{i,k}(y) - y|  \leq (C_1 + C)\ve r_k.
\end{align}

Then, since the balls $10B_{i,k}$ have bounded overlap and $|\theta_{i,k}(y) - \theta_{i,k}(z)| \lesssim r_k^{-1}$ by Lemma <ref>(3), it follows that

\begin{align}
S_1 \leq C(C_1 + 1)\ve|y-z|.
\end{align}

We now control $S_2.$ Recall that $\Sigma_k \cap 45B$ is contained in the graph of $A$ over $P \cap 49B$. In particular, $y = \pi(y) + A ( \pi(y) ),$ with a similar expression holding for $z$. Again, if $\theta_{i,k}(y)\not=0$ then (<ref>) holds, in particular, this implies $d_{0,100r_k}(P,P_{i,k}') \lesssim \ve$ by Lemma <ref>, where $P_{i,k}'$ is the linear space orthogonal to $P_{i,k}$. We use these facts, with Lemma <ref> and the fact that the balls $10B_{i,k}$ have bounded overlap, to control $S_2$ as follows,

\begin{align}
S_2 &=  \sum_{i \in J_k} | \theta_{i,k}(y) |\left| \pi_{i,k}^\perp\left[ \pi(y) + A(\pi(y)) - (\pi(z) + A(\pi(z)) \right] \right| \\
&\leq \sum_{i \in J_k} |\theta_{i,k}(y)|\left( | \pi_{i,k}^\perp (\pi(y)  - \pi(z) ) | + \left| A(\pi(y)) -  A(\pi(z))\right| \right) \\
&\leftstackrel{\substack{\eqref{e:Anorm} \\ \eqref{e:proj}}}{\leq} \sum_{i \in J_k} |\theta_{i,k}(y)| ( C+ C_2 )\ve |\pi(y)  - \pi(z)| \\
&\leq C(C_2+1)\ve|y-z|.
\end{align}

This completes the proof.

We are almost ready to define $F_{j,k}.$ Let us start with an auxiliary function. For $x \in P \cap 40B,$ write

\[ h(x) = x+ \pi( \sigma_k(x+A(x)) - (x + A(x))). \]

See Figure <ref>.

Definition of $h(x)$.

We have

\begin{align}\label{e:h-y}
|h(y) - y| \leq |\sigma_k(y+A(y)) - (y +A(y))| \stackrel{\eqref{e:sig-y}}{\leq} C\ve r_k
\end{align}

\begin{align}\label{e:liph}
|(h(y) - h(z)) - (y-z)| &\leq |\sigma_k(y+A(y)) -\sigma_k(z+A(z)) - \left( y+A(y) - (z+A(z)) \right)| \\
&\stackrel{\eqref{e:siglip}}{\leq} C \ve |y +A(y) - (z+A(z))| \\
&\leq C\ve|y-z|.
\end{align}

As in Stage 1 this implies that $h : P \cap 41B \rightarrow U$ is bijective, where $U = h(P \cap 41B).$ Denote its inverse by $h^{-1}: U \rightarrow P \cap 41B.$ By (<ref>) and (<ref>), $h^{-1}$ satisfies

\begin{align}\label{e:h}
|h^{-1}(x) - x| \lesssim \ve r_k, \quad \text{Lip}(h^{-1}) \leq 2.
\end{align}

We don't include the details here, but a degree theory argument similar to that in the proof of Lemma <ref> (see also [12]) implies

\[U \supset P \cap 40B. \]

\[ F(y) = \pi^\perp ( \sigma_k ( h^{-1}(y) + A(h^{-1}(y)))). \]

See Figure <ref>. We will now verify (<ref>)-(<ref>) for $F$. Let us start with (<ref>), which is the following.

Definition of $F_{j,k}$.

We have

\[\Sigma_{k+1} \cap D(0,P,40B) \subseteq \Gamma_F \cap D(0,P,40B). \]

Since $\Gamma_F \subseteq \Sigma_{k+1}$ by definition, the only non-trivial direction is to show

\[ \Sigma_{k+1} \cap D(0,P,40B) \subseteq \Gamma_F \cap D(0,P,40B). \]

Let $p \in \Sigma_{k+1} \cap D(0,P,40B)$ and let $z \in \Sigma_k$ such that $p = \sigma_{k}(z).$ Since $|z - p| \lesssim \ve r_k$ by (<ref>), we see that $\pi(z) \in P \cap 41B.$ This implies that $z$ is contained in the graph of the function $A$ over $P \cap D(0,P,49B)$, hence

\[ p = \sigma_k(z) = \sigma_k( \pi(z) + A(\pi(z))). \]

Since $h^{-1} : U \rightarrow 41B$ is bijective there exists $y \in U$ such that $h^{-1}(y) = \pi(z).$ This gives

\begin{align} p = \sigma_k( h^{-1}(y) + A(h^{-1}(y))) &= \pi( \sigma_k( h^{-1}(y) + A(h^{-1}(y)))  ) \\
&\hspace{4em} + \pi^\perp ( \sigma_k( h^{-1}(y) + A(h^{-1}(y)))  ) \\
&= y + \pi^\perp( \sigma_k( h^{-1}(y) + A(h^{-1}(y)))  \\
& = y + F(y) ,
\end{align}

where the penultimate equality follows from the definition of $h$. This finishes the proof.

We now focus on proving (<ref>)-(<ref>) for $F.$ Let us start with (<ref>).

For $y,z \in P \cap 40B$ we have $|F(y) - F(z)| \leq C_4 \ve |y-z|.$

Let $y,z \in P \cap 40B.$ To simplify notation we will write

\[ \tilde{y} = h^{-1}(y) + A(h^{-1}(y)), \]

with $\tilde{z}$ defined analogously. Then

\begin{align*}
|F(y) - F(z)| &= |\pi^\perp(\sigma_k(\tilde{y})) -\pi^\perp(\sigma_k(\tilde{z}))| \\
&\leq|\pi^\perp ( (\sigma_k(\tilde{y}) - \tilde{y}) - (\sigma_k(\tilde{z}) - \tilde{z}) ) + |\pi^\perp ( \tilde{y} - \tilde{z})| \\
&=|\pi^\perp ( (\sigma_k(\tilde{y}) - \tilde{y}) - (\sigma_k(\tilde{z}) - \tilde{z}) ) +| A(h^{-1}(y)) - A(h^{-1}(z)| \\
&\leftstackrel{\substack{\eqref{e:Alip} \\ \eqref{e:siglip}}}{\leq} C(C_1 + C_2 +1) \ve | \tilde{y} - \tilde{z}| + C_2 \ve|y-z| \\
&\leq C(C_1 + C_2 +1)\ve |y-z|.
\end{align*}

If $C_4$ is chosen so that

\[ C_4 \geq C(C_1 + C_2 +1) \]

then (<ref>) follows.

Finally, we prove (<ref>) and (<ref>). Both of the constants $C_3$ and $C_5$ will be chosen independent of all other constants and this closes the induction.

We have $|F(x_{j,k})| \leq C_3 \ve r_k$ and $|F(y) - F(z)| \leq C_5 \ve |y-z|$ for all $y,z \in P \cap 7B.$

Let us start with the first statement, it will be useful to prove the same bound for any point in $P \cap 7B.$ Let $y \in P \cap 7B$ and let $w = h^{-1}(y) +A(h^{-1}(y)).$ By (<ref>) and (<ref>) we have

\begin{align}\label{e:above}
|\sigma_k(w) - y| &\leq |\sigma_k(w) - w| + |w -y| \\
&\leq C\ve r_k + |h^{-1}(y) - y| + |A(h^{-1}(y))| \\
&\leq C(1+C_1) \ve r_k
\end{align}

so that $\sigma_k(w) \in 8B$ for $\ve$ small enough. By Lemma <ref>, this implies

\begin{align}
|F(y)| = \dist(\sigma_k(w),P) \leq  | \sigma_k(w) - \pi(w) | \leq C \ve r_k.
\end{align}

Taking $y = x_{j,k}$ proves (<ref>) if we take $C_3$ large enough.

We prove the second statement. Let $y,z \in 7B$ and let $w_y = h^{-1}(y) + A(h^{-1}(y))$ and $w_z = h^{-1}(z) + A(h^{-1}(z)).$ Examining (<ref>), we see $w_y,w_z \in 8B$ for $\ve$ small enough. By Lemma <ref>, (<ref>), and (<ref>), we have

\begin{align*}
|F(y) - F(z)| &= |\pi^\perp ( \sigma_k(w_y) - \sigma_k(w_z) ) | \\
&=|\pi^\perp ( \sigma_k(w_y) - \sigma_k(w_z) - \pi(w_y) - \pi(w_z) ) | \\
&\leq C\ve |w_y - w_z| \\
&\leq C\ve ( |h^{-1}(y) - h^{-1}(z)| + |A(y) - A(z)| )\\
&\leq C\ve ( |y-z| + C_2\ve|y-z| ) \\
&\leq C \ve |y-z|.
\end{align*}

This completes the proof of (<ref>) if we take $C_5$ large enough.

Finally, we show Theorem <ref>(13), which is the following.

§.§ Remaining results
First, let us proof Theorem <ref>(13), which is the following.

Let $\Sigma$ be the surface produced by Theorem <ref>. Then, for $x \in \Sigma$ and $r > 0$,

\[ \mathscr{H}^d(\Sigma \cap B(x,r)) \gtrsim r^d. \]

Consider the map $A:H \rightarrow H$ given by $A(y) =\tfrac{y-x}{r},$ and the set $\tilde{\Sigma} = A(\Sigma).$ Then, $0 \in \tilde{\Sigma}$, and since $\Sigma$ is $C\ve$-Reifenberg flat, the same is true of $\tilde{\Sigma}$. Let $P$ be a $d$-plane through $0$ so that

\[ d_{B(0,1)}(\tilde{\Sigma},P) \leq  C\ve. \]

By <cit.>, for $\ve > 0$ small enough, there exists a bi-Hölder map $g: P \rightarrow H$ which is the identity outside of $B(0,3/2)$, satisfies $|g(y) -y| \lesssim \ve$ for $y \in P$, and so that $\tilde{\Sigma} \cap B(0,1) = g(P) \cap B(0,1)$. Let $\pi$ denote the orthogonal projection onto $P$, and define $h = \pi \circ g.$ By definition, $h$ is continuous and is the identity outside $B(0,3/2),$ so, it is surjective. Additionally, for $y \in P,$

\[ |h(y) -y| = | \pi ( g(y) ) - \pi(y) | \leq |g(y) - y| \lesssim \ve. \]

Let $z \in P \cap B(0,1/2)$. Since $h$ is surjective, we can find $w \in P$ so that $z = h(w).$ For $\ve$ small enough, it follows from the inequality above that $w \in B(0,3/4)$ and so

\[ |g(w)| \leq |g(w) - w| + |w| \leq C\ve +\frac{3}{4} \leq 1.  \]

In particular, $g(w) \in \tilde{\Sigma} \cap B(0,1)$ and $z = \pi(g(w)).$ We have just show that $P \cap B(0,1/2) \subseteq \pi(\tilde{\Sigma} \cap B(0,1)).$ This gives,

\[ \mathscr{H}^d(\tilde{\Sigma} \cap B(0,1)) \geq  \mathscr{H}^d(\pi(\tilde{\Sigma} \cap B(0,1))) \geq \mathscr{H}^d(P \cap B(0,\tfrac{1}{2}) )\gtrsim 1. \]

The lemma follow from this since

\[ \mathscr{H}^d(\Sigma \cap B(x,r)) = \mathscr{H}^d( A^{-1}(\tilde{\Sigma} \cap B(0,1))) = r^d \mathscr{H}^d(\tilde{\Sigma} \cap B(0,1)) \gtrsim r^d. \]

As mentioned, we direct the reader to the proofs of the remaining results in Theorem <ref>. For (1),(3),(8) and (9), see <cit.>, respectively. For (12), see <cit.>. This proof relies on <cit.> (which we do not prove), but the same estimate can be obtained from (<ref>).

§ CONSTANTS

Suppose that for any $\tilde{C}_0 > 1$ and $\tilde{A} > \max\{6\tilde{C}_0,10^5\}$ there exists $\ve > 0$ so that (<ref>) holds with constant $\tilde{C}_0,\tilde{A}$ and $\tilde{\ve}.$ Then, for any $C_0 > 1$ and $A > 10^5$ there exists $\ve > 0$ so that (<ref>) holds with constant $C_0,A$ and $\ve.$

Let $C_0 > 1$, $A > 10^5$, and $E \subseteq H$ be $(c,d)$-lower content $d$-regular. If $A > 6C_0$ then there is nothing to do, so let us assume $A \leq 6C_0.$ Choose $\kappa \sim_{A,C_0} 1$ so that $A > 6 \kappa C_0 > A/2$ and choose $K = K(A,C_0)$ to be the smallest integer so that $C_0 \rho^K + 1 \leq \kappa C_0.$ Choose $\ve > 0$ so that (<ref>) holds with constants $A,\kappa C_0$ and $\ve.$ First, we separate

\begin{align}
\sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d &= \sum_{k=0}^{K-1} \sum_{\substack{Q \in \mathscr{D}_k \\ Q \subseteq Q_0}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d + \sum_{k=K}^\infty \sum_{\substack{Q \in \mathscr{D}_k \\ Q \subseteq Q_0 \\ Q \in \BWGL(A,\ve)}}  \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d \\
&\hspace{4em} + \sum_{k=K}^\infty \sum_{\substack{Q \in \mathscr{D}_k \\ Q \subseteq Q_0 \\ Q \not\in \BWGL(A,\ve)}}  \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d \\
&=I_1 + I_2 + I_3.
\end{align}

The term $I_2$ is bounded by $\BWGL(Q_0,A,\ve)$ be definition. Additionally, since $E$ is lower regular and the balls $c_0B_Q, \ Q \in \mathscr{D}_k$, are disjoint, we have

\[ I_1 \lesssim \sum_{k=0}^{K-1} \sum_{\substack{Q \in \mathscr{D}_k \\ Q \subseteq Q_0}} \mathscr{H}^d(E \cap c_0B_Q) \leq K \mathscr{H}^d(Q_0). \]

We are left to bound $I_3.$ Let $\mathscr{C}$ to be the collection of cubes $Q \in \bigcup_{k=K}^\infty \mathscr{D}_k$ so that $Q \subseteq Q_0$ and $Q \not\in \BWGL(A,\ve).$ Let $\mathscr{C}_1$ be the set of cubes $Q \in \mathscr{C}$ so that $Q^{(K)} \not\in \BWGL(A,\ve)$.  Also, set $\mathscr{C}_2$ to be the set of cubes $Q \in \mathscr{C}$ so that $Q^{(K)} \in \BWGL(A,\ve)$. Notice that $Q^{(K)} \subseteq Q_0$ in either case.  Then

\[ I_3 = \sum_{Q \in \mathscr{C}_1} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d + \sum_{Q \in \mathscr{C}_2} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d = I_{3,1} + I_{3,2}.  \]

We start by estimating $I_{3,1}.$ For $Q \in \mathscr{C}_1,$ set $Q^* = Q^{(K)}.$ We begin by showing that the map $Q \mapsto Q^*$ is at most $C$-to-$1$ for some universal constant $C.$ Indeed, let $R \in \mathscr{D}$ and consider $\mathscr{C}_1(R) = \{Q \in \mathscr{C}_1 : Q^* = R\}.$ Since $R \not\in \BWGL(A,\ve),$ there exists a $d$-plane $P$ so that $d_{AB_R}(E,P) \leq \ve.$ So, for $\ve$ small enough, $\dist(x_Q,P) \leq \ve A \ell(R) \leq \tfrac{c_0}{2} \ell(Q)$ and $c_0B_Q \subseteq B_{R} = B(x_{R},\rho^{-K}\ell(Q))$ for all $Q \in \mathscr{C}_1(R).$ By Lemma <ref>, this implies $\# \mathscr{C}_1(R) \lesssim 1$ as required. Now, by our choice of $K$, $C_0B_Q \subseteq \kappa C_0 B_{Q^*}$. So, by our assumption in the statement of Lemma <ref>, Lemma <ref>, and the above, we have

\begin{align}
I_{3,1} \lesssim \sum_{Q \in \mathscr{C}_1 } \beta_{E}^{d,p}(\kappa C_0B_{Q*})^2 \ell(Q^*)^d \lesssim \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}}  \beta^{d,p}_E(\kappa C_0B_Q)^2\ell(Q)^d \lesssim \mathscr{H}^d(Q_0) + \BWGL(Q_0,A,\ve).
\end{align}

Now for $I_{3,2}$. For $Q \in \mathscr{C}_2$, define $Q^*$ to be the smallest cube so that $Q \subseteq Q^*$ and $Q^* \in \BWGL(A,\ve).$ Such a cube exists since $Q^{(K)} \in \BWGL(A,\ve)$ by definition. Again, we show the map $Q \mapsto Q^*$ is at most $C$-to-$1$ for some universal constant $C.$ Let $R \in \mathscr{D}$ for which the set $\mathscr{C}_2(R) = \{Q \in \mathscr{C}_2 : Q^* = R \}$ is non-empty. Note, there are at most $K$ scales for which $\#\mathscr{C}_2(R) \cap \mathscr{D}_k \not=\emptyset.$ Choose $k$ to be one of these scales and let $Q \in \mathscr{C}_2(R) \cap \mathscr{D}_k.$ Since $R$ is the smallest cube containing $Q$ so that $R \in \BWGL(A,\ve),$ there exists a child $R'$ of $R$ so that $R' \not\in \BWGL(A,\ve).$ As above, since $B_R \subseteq AB_{R'}$ (because $A > 10^5$), we can find a $d$-plane $P$ so that $\dist(x_Q,P) \leq \frac{c_0}{2}\ell(Q)$ and $c_0B_Q \subseteq B_R \subseteq B(x_R, \rho^{-K}\ell(Q)).$ Then, by Lemma <ref>, $\#\mathscr{C}_2(R) \cap \mathscr{D}_k \lesssim 1$, and so $\mathscr{C}_2(R) \lesssim 1.$ Hence,

\begin{align}
I_{3,2} \lesssim \sum_{Q \in \mathscr{C}_2} \ell(Q^*)^d \lesssim \sum_{\substack{  Q \subseteq Q_0 \\ Q \in \BWGL(A,\ve)}} \ell(Q)^d = \BWGL(Q_0,A,\ve).
\end{align}

This finishes the proof.

Let $\ve > 0$, $A \geq 1$ and $M \geq C_0$, and suppose

\begin{align}\label{e:Section6a} \mathscr{H}^d(Q_0) + \BWGL(Q_0,A,\ve) \lesssim \ell(Q_0)^d + \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,1}(MB_Q)^2\ell(Q)^d.
\end{align}

Then, (<ref>) holds with constant $\ve,A$ and $C_0$.

The proof will be similar to the proof of Lemma <ref>. First, by Lemma <ref>,

\begin{align}\label{e:consts1}
\sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,1}(MB_Q)^2\ell(Q)^d \lesssim \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,p}(MB_Q)^2\ell(Q)^d.
\end{align}

The remainder of the proof will be focussed controlling the right-hand side of (<ref>) in terms of the same quantity but with $C_0$ in place of $M$. Let $K = K(C_0,M)$ be the smallest integer so that $1+M\rho^K \leq C_0$ and let $\delta > 0$ be a constant to be chosen small. Let $\mathscr{C}$ be the set of cubes $Q \in \mathscr{D}$ so that $Q \subseteq Q_0$ and $\beta_E^{d,p}(C_0B_Q) \leq \delta.$ Notice that

\begin{align}
\sum_{\substack{Q \in \mathscr{D} \setminus \mathscr{C}\\ Q \subseteq Q_0}}  \beta_E^{d,p}(MB_Q)^2\ell(Q)^d \lesssim_\delta \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,p}(C_0B_Q)^2\ell(Q)^d,
\end{align}

so we only need to control the sum over cubes in $\mathscr{C}.$ Let $\mathscr{C}_1$ be the set of cubes $Q \in \mathscr{C} \cap \bigcup_{k=K}^\infty \mathscr{D}_k$ so that $Q^{(K)}$ satisfies $\beta_E^{d,p}(C_0B_{Q^{(K)}}) \leq \delta$ and let $\mathscr{C}_2$ be the set of cubes $Q \in\mathscr{C} \cap  \bigcup_{k=K}^\infty \mathscr{D}_k$ so that $Q^{(K)}$ satisfies $\beta_E^{d,p}(C_0B_{Q^{(K)}}) > \delta$. Note that $Q^{(K)} \subseteq Q_0$ in either case. Then,

\begin{align}
\sum_{Q \in \mathscr{C}} \beta_E^{d,p}(MB_Q)^2\ell(Q)^d  &\lesssim \sum_{k=0}^K \sum_{Q \in \mathscr{C} \cap \mathscr{D}_k} \beta_E^{d,p}(MB_Q)^2\ell(Q)^d + \sum_{Q \in \mathscr{C}_1} \beta_E^{d,p}(MB_Q)^2\ell(Q)^d \\
&\hspace{4em} + \sum_{Q \in \mathscr{C}_2} \beta_E^{d,p}(MB_Q)^2\ell(Q)^d  \\
&= I_0 + I_1 + I_2.
\end{align}

Let us estimate $I_1.$ For $Q \in \mathscr{C}_1$ let $Q^* = Q^{(K)}$. Let $R \in \mathscr{D}$ be so that $\mathscr{C}_1(R) = \{Q \in \mathscr{C}_1 : Q^* = R \}$ is non-empty. By Lemma <ref>, $\beta^d_{E,\infty}(\tfrac{C_0}{2}B_R) \lesssim \delta^\frac{1}{d+1}$ so there exists a $d$-plane $P$ so that $\dist(y,P) \lesssim \ve \ell(R)$ for all $y \in E \cap \tfrac{C_0}{2}B_R.$ Then, for $\ve > 0$ small enough, a similar argument to Lemma <ref> shows the map $Q \mapsto Q^*$ is at most $C$-to-$1$. Now, since $MB_Q \subseteq C_0B_{Q^*},$ Lemma <ref> implies

\begin{align}\label{e:consts2}
I_1 \lesssim \sum_{Q \in \mathscr{C}_1} \beta^{d,p}_E(C_0B_{Q^*})^2\ell(Q^*)^d \lesssim \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}}  \beta^{d,p}_E(C_0B_{Q})^2\ell(Q)^d
\end{align}

Now for $I_2.$ For $Q \in \mathscr{C}_2$, define $Q^*$ to be the smallest cube so that $Q \subseteq Q^*$ and $\beta_E^{d,p}(C_0B_{Q^*}) > \delta.$ Let $R \in \mathscr{D}$ for which the set $\mathscr{C}_2(R) = \{Q \in \mathscr{C}_2 : Q^* = R \}$ is non-empty. We will show

\[ \#\mathscr{C}_2(R) \lesssim 1. \]

Note, there are at most $K$ scales for which $\#\mathscr{C}_2(R) \cap \mathscr{D}_k \not=\emptyset.$ Choose $k$ to be one of these scales and let $Q \in \mathscr{C}_2(R) \cap \mathscr{D}_k.$ Since $R$ is the smallest cube containing $Q$ with $\beta_E^{d,p}(C_0B_R) > \delta$, there exists a child $R'$ of $R$ so that $\beta_E^{d,p}(C_0B_{R'}) \leq \delta.$ As above, since $B_R \subseteq C_0B_{R'}$ (because we assumed $C_0 \geq 2\rho^{-1}$), we can find a $d$-plane $P$ so that $\dist(x_Q,P) \leq \frac{c_0}{2}\ell(Q)$ and $c_0B_Q \subseteq B_R \subseteq B(x_R, \rho^{-K}\ell(Q)).$ Then, by Lemma <ref>, $\#\mathscr{C}_2(R) \cap \mathscr{D}_k \lesssim 1$ and so $\#\mathscr{C}_2(R) \lesssim 1.$ Hence,

\begin{align}
I_2 \lesssim_\delta \sum_{Q \in \mathscr{C}_2} \beta_E^{d,p}(C_0B_{Q^*})^2\ell(Q^*)^d \lesssim \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}}  \beta^{d,p}_E(C_0B_{Q})^2\ell(Q)^d.
\end{align}

Finally, we deal with $I_0.$ Let $\delta > 0$ be as above. If $\beta_E^{d,p}(C_0B_{Q_0}) \leq \delta$ then the arguments above (<ref>) show that $\#\{Q \in \mathscr{D}_k , 0 \leq k \leq K-1\} \lesssim 1.$ So, in this case,

\[ I_0 \lesssim \ell(Q_0)^d. \]

Suppose instead that $\beta_E^{d,p}(C_0B_{Q_0}) > \delta.$ For each $Q \in \bigcup_{k = 0}^{K-1} \mathscr{D}_k$ let $Q^*$ be the smallest cube so that $Q \subseteq Q^*$ and $\beta_E^{d,p}(C_0B_{Q^*}) > \delta.$ In this case, by exactly the same arguments as for $I_2,$ we get

\[ I_0 \lesssim_\delta \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,p}(C_0B_Q)^2 \ell(Q)^d. \]

Combining the above gives

\[\sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,1}(MB_Q)^2\ell(Q)^d \lesssim \sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta_E^{d,p}(C_0B_Q)^2\ell(Q)^d \]

as required.

§ REDUCTION TO EUCLIDEAN SPACE

Here we prove Lemma <ref> and Lemma <ref>. However, since Lemma <ref> is a special case of Lemma <ref>, it suffices to only prove Lemma <ref>, which we restate below for convenience.

Suppose for any $n \geq 2$, $1 \leq d < n,$ $1 \leq p < p(d),$ $C_0 > 1$, and $A > \max\{6C_0,10^5\}$, that there exists $\ve > 0$ so that (<ref>) holds with constants $d,p,C_0,A,$ and $\ve$, for any lower content $d$-regular set $E \subseteq \R^n$, with constants independent of $n$. Then, there exists $C \geq 1$ so that (<ref>) holds with constant $d,p,C_0,2A$, and $C\ve,$ for all $E \subseteq H$ lower content $d$-regular sets $E \subseteq H.$

Let $1 \leq d < \infty, \ 1 \leq  p < p(d)$, $C_0 > 1$, $A > \max\{6C_0,10^5\}$, $\ve > 0$, and $E \subseteq H$ be a lower content $d$-regular set. Let $X_m, \ m \in \Z,$ be a sequence of maximally $\rho^m$-separated nets and let $\mathscr{D}$ be the Christ-David cubes for $E$ from Theorem <ref>. Fix $Q_0 \in \mathscr{D}$ and assume without loss of generality that $x_{Q_0} = 0$ and $\ell(Q_0) = 1.$ We may also assume $\mathscr{H}^d(Q_0) < \infty$ since otherwise (<ref>) is trivial. For $k \geq 0$ let

\[ \mathscr{D}^k(Q_0) = \bigcup_{m = 0}^k \{Q \in \mathscr{D}_m : Q \subseteq Q_0 \}. \]

Notice that

\begin{align}\label{e:red1}
\sum_{\substack{Q \in \mathscr{D} \\ Q \subseteq Q_0}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d = \lim_{k \rightarrow \infty} \sum_{Q \in \mathscr{D}^k(Q_0)}  \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d,
\end{align}

so it suffices to show

\begin{align}\label{e:red1.5}
\sum_{Q \in \mathscr{D}^k(Q_0)}  \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d \lesssim \mathscr{H}^d(Q_0) +\BWGL(Q_0,A,\ve)
\end{align}

with constant independent $k$. This is our goal for the rest of the section. Let $k \in \Z$ and let $k^*$ be the smallest integer so that $\rho^{k^*} \leq 5\ve \rho^k.$ Since $E$ is lower content $d$-regular, $\# \mathscr{D}^{k^*}(Q_0) < \infty$, because for each $0 \leq m \leq k^*$,

\begin{align}
\#\{Q \in \mathscr{D}_m : Q \subseteq Q_0\} \rho^{md} \lesssim \sum_{\substack{Q \in \mathscr{D}_m \\ Q \subseteq Q_0}} \mathscr{H}^d(E \cap c_0B_Q) \leq \mathscr{H}^d(Q_0) < \infty.
\end{align}

Let $V$ be the linear space spanned by the centres of the cubes in $\mathscr{D}^{k^*}(Q_0)$ (recall $x_{Q_0} = 0$). By the above, we can identify $V$ with $\R^n$ for some $n \leq \# \mathscr{D}^{k^*}(Q_0) < \infty.$ We now define a subset $\tilde{E}$ of $\R^n$ which well approximates $E$. Let $T$ be a $d$-dimensional simplex centred at the origin, with vertices in $S^{n-1} \subseteq \R^n$ (the unit sphere). For $Q \in \mathscr{D}_{k^*},$ define $T_Q = x_Q + c_0\ell(Q)T.$ Then each $T_Q$ is Ahlfors $d$-regular and contained in the ball $B_n(x_Q,c_0\ell(Q)) = B_n(x_Q,5c_0\rho^{k^*}).$ Define $\tilde{E}$ by setting

\[ \tilde{E} = \bigcup_{Q \in \mathscr{D}_{k^*}} T_Q \subseteq \R^n. \]

The set $\tilde{E}$ is lower content $d$-regular

Let $x \in \tilde{E}$ and $0 < r < \diam(\tilde{E})$. Assume first of all that $r < 20c_0\rho^{k^*}.$ Let $Q \in \mathscr{D}_{k^*}$ be the cube so that $x \in T_Q.$ Since $T_Q$ is lower content $d$-regular (by virtue of being Ahlfors $d$-regular) and $\diam(T_Q) = 10c_0\rho^{k^*} \gtrsim r$, this implies

\begin{align}\label{e:EtildeLR}
\mathscr{H}^d_\infty(\tilde{E} \cap B(x,r)) \geq \mathscr{H}^d_\infty(T_Q \cap B(x,r)) \gtrsim r^d.
\end{align}

Assume instead that $r \geq 20c_0\rho^{k^*}.$ For $A \subseteq \R^n$ and $0 < \delta < \infty$, set

\begin{align}
\mathscr{H}^d_{\delta,\infty}(A) = \inf \left\{ \sum_i \diam(A_i)^d : A \subseteq \bigcup_i A_i \mbox{ and } \diam A_i \geq \delta \right\}.
\end{align}

This is like the Hausdorff content except we include a lower bound on the diameters of the covering sets. The lemma will be proved once we show the following,

\begin{align}\label{e:lowerboundcontent}
\mathscr{H}^d_\infty(\tilde{E} \cap B(x,r)) \gtrsim \mathscr{H}^d_{5c_0\rho^{k^*},\infty}(\tilde{E} \cap B(x,r/2)) \gtrsim r^d.
\end{align}

We start with the left-hand side. First note that $\mathscr{H}^d_\infty(\tilde{E} \cap B(x,r)) > 0$ by (<ref>). Thus, we can find a collection of sets $\{U_i\}_{i \in I}$ so that $\tilde{E} \cap B(x,r) \subseteq \bigcup_{i \in I} U_i$ and

\[ \sum_{i \in I } \diam(U_i)^d \leq 2\mathscr{H}^d_\infty(\tilde{E} \cap B(x,r)). \]

We will modify $\{U_i\}$ to replace those sets with $\diam(U_i) < 5c_0\rho^{k^*}.$ Let $\mathscr{G}$ be the set of cubes in $\mathscr{D}_{k^*}$ so that $T_Q \cap B(x,r/2) \neq\emptyset$. For each $Q \in \mathscr{G},$ let $I_Q$ be the set of $i \in I$ so that $T_Q \cap U_i \not=\emptyset$ and $\diam(U_i) < 5c_0\rho^{k^*}.$ Let $\alpha > 0$ (to be chosen shortly) and further sub-divide $\mathscr{G}$ into

\begin{align}
\mathscr{G}_1 &=  \left\{ Q \in \mathscr{G} : \sum_{i \in I_Q} \diam(U_i)^d > \alpha (10c_0)^d \rho^{dk^*} \right\} ; \\
\mathscr{G}_2 &=  \left\{Q \in \mathscr{G} : \sum_{i \in I_Q} \diam(U_i)^d \leq \alpha (10c_0)^d \rho^{dk^*} \right\}.
\end{align}

If $Q \in \mathscr{G}_1$ then $T_Q \subseteq B(x_Q,5c_0\rho^{k^*})$ and

\[ \diam(B(x_Q,5c_0\rho^{k^*}))^d = (10c_0)^d \rho^{dk^*}  < \frac{1}{\alpha} \sum_{i \in I_Q} \diam(U_i)^d. \]

For $Q \in \mathscr{G}_2$ we claim there exists $i_Q \in I$ so that $T_Q \cap U_{i_Q} \not=\emptyset$ and $\diam(U_{i_Q}) \geq 5c_0\rho^{k^*}.$ in particular $i_Q \not\in I_Q.$ Assuming this to be true for the moment, we let $\tilde{I} \subseteq I$ be the set of indices for which there exists $Q \in \mathscr{G}_2$ so that $i_Q = i$. For each $i \in \tilde{I}$ we pick a point $x_i \in U_i$ and set $B_i = B(x_i,2\diam(U_i)).$ In this way

\[ \diam(B_i) \geq 20c_0\rho^{k^*} \mbox{ and } \bigcup_{\substack{Q \in \mathscr{G}_2 \\ i_Q = i}} T_Q \subseteq B_i. \]

Putting everything together we have

\[ \tilde{E} \cap B(x,r/2) \subseteq \bigcup_{Q \in \mathscr{G}} T_Q \subseteq \bigcup_{Q \in \mathscr{G}_1} B(x_Q,5c_0\rho^{k^*}) \cup \bigcup_{i \in \tilde{I}} B_{i}, \]

\begin{align}
\mathscr{H}^d_{5c_0\rho^{k^*},\infty}(\tilde{E} \cap B(x,r/2)) &\leq \sum_{Q \in \mathscr{G}_1} \diam(B(x_Q,5c_0\rho^{k^*}))^d + \sum_{i \in \tilde{I}} \diam(B_i)^d \\
& \leq \left(\frac{1}{\alpha} + 2 \right) \sum_{i \in I} \diam(U_i)^d \lesssim \mathscr{H}^d(\tilde{E} \cap B(x,r)).
\end{align}

This completes the left-hand side of (<ref>) modulo our claim. Let us see why the claim is true. Since $T_Q \cap B(x,r/2) \not=\emptyset, \ \diam(T_Q) =10c_0\rho^{k^*} \leq r/2,$ we have $T_Q \subseteq \tilde{E} \cap B(x,r).$ In particular, the collection of set $\{U_i : i \in I, \ T_Q \cap U_i\neq \emptyset\}$ form a cover for $T_Q.$ Then, since $T_Q$ is lower content $d$-regular, we have

\begin{align}
\rho^{dk^*} \lesssim \sum_{\substack{i \in I \\ T_Q \cap U_i \not=\emptyset}} \diam(U_i)^d = \sum_{\substack{i \in I\setminus I_Q \\ T_Q \cap U_i \not=\emptyset}} \diam(U_i)^d + \sum_{\substack{i \in I_Q}} \diam(U_i)^d \lesssim \sum_{\substack{i \in I\setminus I_Q \\ T_Q \cap U_i \not=\emptyset}} \diam(U_i)^d + \alpha \rho^{dk^*}.
\end{align}

For $\alpha > 0$ small enough, the penultimate term must be non-zero which implies the existence of the required index $i \in I$ and proves the claim.

Now for the right-hand side of (<ref>). Let $\{V_j\}_{j \in J}$ be a collection of subsets of $\R^n$ so that $5c_0\rho^{k^*} \leq \diam(V_j) < \infty$, $\tilde{E} \cap B(x,r/2) \subseteq \bigcup_{j \in J} V_j$ and

\[ \mathscr{H}^d_{5c_0\rho^{k^*},\infty}(\tilde{E} \cap B(x,r/2)) \sim \sum_{j \in J} \diam(V_j)^d. \]

For each $j \in J,$ let $y_j \in V_j$ and $\tilde{B}_j = B(y_j,c_0^{-1}\diam(V_j)) \subseteq H.$ For each $Q \in \mathscr{D}_{k^*}$ so that $x_Q \in B(x,r/2)$, there exists $j \in J$ so that $x_Q \in V_j,$ which by the triangle inequality and the lower bound on $\diam(V_j)$, implies $B(x_Q,2\rho^{k^*}) \subseteq \tilde{B}_j.$ In particular,

\[ \bigcup_{\substack{Q \in \mathscr{D}_{k^*} \\ x_Q \in B(x,r/2) }} B(x_Q,2\rho^{k^*}) \subseteq \bigcup_{j \in J} \tilde{B}_j. \]

By maximality, the collection $\{B(x_Q,2\rho^{k^*}) : Q \in \mathscr{D}_{k^*}, \  x_Q \in B(x,r/2) \}$ forms a cover of $E \cap B(x,r/4)$. By the above, the same is true of the collection of balls $\{\tilde{B}_j\}_{j \in J}.$ Since $E$ is lower content $d$-regular, this implies

\begin{align}
r^d \lesssim \mathscr{H}^d_\infty(E \cap B(x,r/4)) \leq \sum_{j \in J} \diam(\tilde{B}_j)^d \sim \sum_{j \in J} \diam(V_j)^d \sim \mathscr{H}^d_{5c_0\rho^{k^*},\infty}(\tilde{E} \cap B(x,r/2))
\end{align}

as required.

Let $Y_m, \ m \in \Z$, be a sequence of nested maximally $\rho^m$-separated nets in $\tilde{E}.$ We can construct these so that $Y_m = X_m$ for each $m \leq k^*$ (recall that $X_m$ were the nets we used to construct the cubes for $E$). Let $\tilde{\mathscr{D}}$ be the Christ-David cubes for $\tilde{E}$ constructed from the $Y_m.$ In this way, for each $Q \in \mathscr{D}^{k^*}$ there exists a unique $\tilde{Q} \in \tilde{\mathscr{D}}^{k^*}$ so that $x_Q = x_{\tilde{Q}}$ and $\ell(Q) = \ell(\tilde{Q}),$ and vice-versa. These also satisfy

\begin{align}\label{e:Y_m}
Y_m \cap Q = Y_m \cap \tilde{Q}  \mbox{ for all } m \leq k^*.
\end{align}

We have $\mathscr{H}^d(\tilde{Q}_0) \lesssim \mathscr{H}^d(Q_0).$

First observe that

\begin{align}\label{e:tilde{Q}_0}
\tilde{Q}_0 \subseteq \bigcup_{\substack{Q \in \mathscr{D}_{k^*} \\ Q \subseteq Q_0}} T_Q.
\end{align}

Indeed, suppose the opposite was true. Then we could find a point $z \in \tilde{Q}_0$ so that $z \in T_Q$ for some $Q \in \mathscr{D}_{k^*}$ satisfying $Q \cap Q_0 =\emptyset.$ Let $\tilde{Q}$ be the corresponding cube in $\tilde{\mathscr{D}}.$ Then $x_{\tilde{Q}} \not\in Q_0$ and so $x_{\tilde{Q}} \not\in \tilde{Q}_0$ by (<ref>). Thus $\tilde{Q} \cap \tilde{Q}_0 = \emptyset.$ This, however, is a contradiction since

\[ z \in T_Q \subseteq \tilde{E} \cap c_0B_Q = \tilde{E} \cap c_0B_{\tilde{Q}} \subseteq \tilde{Q}. \]

Continuing from (<ref>), since each of the $T_Q, \ Q \in \mathscr{D}_{k^*},$ are Ahlfors $d$-regular, and the balls $E \cap c_0B_Q$ are disjoint and contained in $Q_0$, we have

\begin{align}
\mathscr{H}^d(\tilde{Q}_0) \leq \sum_{\substack{Q \in \mathscr{D}_{k^*} \\ Q \subseteq Q_0}} \mathscr{H}^d(T_Q) \lesssim \sum_{\substack{Q \in \mathscr{D}_{k^*} \\ Q \subseteq Q_0}} \ell(Q)^d \lesssim \sum_{\substack{Q \in \mathscr{D}_{k^*} \\ Q \subseteq Q_0}} \mathscr{H}^d(E \cap c_0B_Q) \leq \mathscr{H}^d(Q_0)
\end{align}

Let $\tilde{Q} \in \mathscr{D}^{k}(\tilde{Q}_0)$ and let $Q \in \mathscr{D}^{k}(Q_0)$ be the corresponding cube from above. If $\tilde{Q} \in \BWGL(\tfrac{1}{2}A,c\ve)$ then $Q \in \BWGL(A,\ve).$

We will prove the contrapositive. In particular, let us suppose that $Q \not\in \BWGL(A,\ve)$, we will show $\tilde{Q} \not\in \BWGL(\tfrac{1}{2}A,c\ve).$

For brevity, we write $B = AB_Q = AB_{\tilde{Q}}.$ Since $Q \not\in \BWGL(A,\ve),$ there exists a $d$-plane $P_Q$ through $x_Q$ so that $d_{B}(E,P_Q) \leq 2\ve.$ Choose a collection of point $z_i, \ i = 1,\dots,d$, in $P_Q \cap (B \setminus \tfrac{1}{2}B)$ so that the vectors $z_i - x_Q$ are mutually orthogonal. Since $d_{B}(E,P_Q) \leq 2\ve,$ there exists $y_i \in E \cap \tfrac{3}{4}B$ so that $|y_i - z_i| \leq 2\ve A \ell(Q)$. By maximality, there exists $Q_i \in \mathscr{D}_{k^*}$ so that $x_{Q_i} \in B$ and $|y_i - x_{Q_i}| \leq \ell(Q_i) = 5\rho^{k^*} \leq \ve \ell(Q).$ Let $P_{\tilde{Q}} \subseteq \R^n$ be the $d$-plane spanned by the vectors $x_{Q_i} - x_Q.$ Since, by the triangle inequality, $|z_i - x_{Q_i}| \lesssim \ve \ell(Q)$, it follows that

\begin{align}\label{e:nearend}
d_{B}(P_Q,P_{\tilde{Q}}) \lesssim \ve.
\end{align}

Let us now estimate $d_{B/2}(\tilde{E},P_{\tilde{Q}}).$ Let $y \in \tilde{E} \cap \tfrac{1}{2}B$, we will find $z \in P_{\tilde{Q}} \cap  \tfrac{1}{2}B$ so that $|y-z| \lesssim \ve \ell(Q).$ First, since the centres of cubes in $\mathscr{D}_{k^*}$ are contained in $E$ and form a maximally $\rho^{k^*}$ separated net for $\tilde{E},$ we can find $y_1 \in E \cap (\tfrac{1}{2} + \ve)B$ so that $|y-y_1| \leq \rho^{k^*} \leq \ve \ell(Q).$ Since $d_{B}(E,P_Q) < 2\ve$, there exists $y_2 \in P_Q \cap (\tfrac{1}{2} + 3\ve)B$ so that $|y_1 - y_2| \leq 2\ve A \ell(Q).$ By (<ref>), we can find $y_3 \in P_{\tilde{Q}} \cap (\tfrac{1}{2} + C \ve)B$ so that $|y_2 - y_3| \lesssim \ve\ell(Q).$ Finally, by considering the line segment joining $y_3$ and $x_{\tilde{Q}}$, we can find a point $z \in P_{\tilde{Q}} \cap \tfrac{1}{2}B$ so that $|y_3 - y_z| \leq CA\ve \ell(Q).$ Combining the above chain of inequalities shows $z \in P_{\tilde{Q}} \cap \tfrac{1}{2}B$ is the required point.

A similar argument (essentially working backwards) shows that for each $z \in P_{\tilde{Q}} \cap \tfrac{1}{2}B$ we can find $y \in \tilde{E} \cap \tfrac{1}{2}B$ so that $|y-z| \lesssim \ve \ell(Q).$ This concludes the argument.

We can now prove Lemma <ref>.

Recall that it suffices to show (<ref>). Let $\mathscr{D}_{\text{Bad}}$ denote the set of cubes $Q$ in $\mathscr{D}^k(Q_0)$ for which there exists a point $y \in E \cap 3C_0B_Q \setminus Q_0,$ and let $\mathscr{G} = \mathscr{D}^k(Q_0) \setminus \mathscr{D}_{\text{Bad}}.$ Then, the left-hand side of (<ref>) can be written as

\[ \sum_{Q \in \mathscr{D}_{\text{Bad}}}\beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d + \sum_{Q \in \mathscr{G}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d.\]

Let us estimate the sum over $\mathscr{D}_{\text{Bad}}$. First, we claim that the collection of balls $\tfrac{c_0}{2}B_Q,$ $Q \in \mathscr{D}_{\text{Bad}}$, have bounded overlap. To see this, suppose $Q,R \in \mathscr{D}_\text{Bad}$ are such that $\tfrac{c_0}{2}B_Q \cap \tfrac{c_0}{2}B_R \neq \emptyset.$ The claim will follow once we show $\ell(Q) \sim \ell(R)$ since the balls $c_0B_Q$ are disjoint for cubes in the same generation. Assume without loss of generality that $\ell(Q) \leq \ell(R).$ Assume towards a contradiction that $\ell(Q) \leq \tfrac{c_0}{12C_0} \ell(R).$ It follows that $E \cap 3C_0B_Q \subseteq E \cap c_0B_R$ since for any $y \in 3C_0B_Q,$

\begin{align}
|y - x_R| \leq |y-x_Q| + |x_Q - x_R| \leq 3C_0\ell(Q) + \frac{c_0}{2} \ell(Q) + \frac{c_0}{2}\ell(R) \leq c_0\ell(R).
\end{align}

But this means that $E \cap 3C_0B_Q \subseteq Q_0$ which contradicts the definition of $\mathscr{D}_\text{Bad}.$ Using this, we have

\begin{align}
\sum_{Q \in \mathscr{D}_\text{Bad}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d \lesssim \sum_{Q \in \mathscr{D}_\text{Bad}} \ell(Q)^d \lesssim \sum_{Q \in \mathscr{D}_\text{Bad}} \mathscr{H}^d(E \cap c_0B_Q) \lesssim \mathscr{H}^d(Q_0).
\end{align}

We turn our attention to the sum over $\mathscr{G}.$ It will be useful to note (for later on) that if $Q \in \mathscr{G}$ then $E \cap 3C_0B_Q \subseteq Q_0.$ We begin by applying Lemma <ref>,

\begin{align}
\sum_{Q \in \mathscr{G}} \beta^{d,p}_E(C_0B_Q)^2\ell(Q)^d &\lesssim \sum_{Q \in \mathscr{G}} \beta^{d,p}_{\tilde{E}}(2C_0B_Q)^2\ell(Q)^d \\
&\hspace{4em}+ \sum_{Q \in \mathscr{G}} \left( \frac{1}{\ell(Q)^d} \int_{E \cap 2C_0B_Q} \left(\frac{\dist(y,\tilde{E})}{\ell(Q)} \right)^p \, d\mathscr{H}^d_\infty \right)^\frac{2}{p} \ell(Q)^d \\
&\eqqcolon I_1 + I_2.
\end{align}

By construction, for each $Q \in \mathscr{G}$ there exists a unique $\tilde{Q} \in\tilde{\mathscr{D}}$ so that $2C_0B_Q = 2C_0B_{\tilde{Q}}.$ Using this along with Theorem <ref>, Lemma <ref>, and Lemma <ref>, we have

\begin{align}
I_1  \leq \sum_{m=0}^k\sum_{Q \in \tilde{\mathscr{D}}_m}\beta^{d,p}_{\tilde{E}}(2C_0B_Q)^2\ell(Q)^d &\lesssim \mathscr{H}^d(\tilde{Q}_0) + \sum_{\substack{Q \in \BWGL(\frac{1}{2}A,c\ve) \cap \tilde{\mathscr{D}}^k(\tilde{Q}_0)}} \ell(Q)^d \\
&\lesssim \mathscr{H}^d(Q_0) +\sum_{\substack{Q \in \BWGL(A,\ve) \cap \mathscr{D}^k(Q_0)}} \ell(Q)^d  \\
&\leq \mathscr{H}^d(Q_0) + \BWGL(Q_0,A,\ve).
\end{align}

We are left to estimate $I_2.$ We claim

\begin{align}\label{e:I_2last}
I_2 \lesssim \mathscr{H}^d(Q_0) + \BWGL(Q_0,A,\ve).
\end{align}

Indeed, notice that

\begin{align}
I_2 \leq \sum_{Q \in \mathscr{G}\setminus \BWGL(A,\ve)} \left( \frac{1}{\ell(Q)^d} \int_{E \cap 2C_0B_Q} \left(\frac{\dist(y,\tilde{E})}{\ell(Q)} \right)^p \, d\mathscr{H}^d_\infty \right)^\frac{2}{p} \ell(Q)^d  + \BWGL(Q_0,A,\ve) \\
\end{align}

Denoting the first term above by $I_3,$ the proof of (<ref>) will be complete once we show

\[ I_3 \lesssim \mathscr{H}^d(Q_0). \]

Let $\mathscr{G}' = \mathscr{G} \setminus \BWGL(A,\ve)$ and for $m \in \Z$ set $\mathscr{G}_m' = \mathscr{G}' \cap \mathscr{D}_m.$ First, we may assume $p \geq 2$ by Lemma <ref>.  Notice, if $y \in 2C_0B_R$ for some $R \in \mathscr{D}_k$ then $\dist(y,\tilde{E}) \lesssim \ell(R)$ since $x_R \in \tilde{E}.$ Hence,

\begin{align}\label{e:red2}
I_3 &\lesssim \sum_{m=0}^{k} \sum_{Q \in \mathscr{G}_m'} \left( \frac{1}{\ell(Q)^d} \sum_{\substack{R \in \mathscr{D}_k \\ B_R \cap 2C_0B_Q \not=\emptyset}}  \int_{E \cap B_R}  \left(\frac{\dist(y,\tilde{E})}{\ell(Q)} \right)^p \, d\mathscr{H}^d_\infty \right)^\frac{2}{p} \ell(Q)^d \\
&\lesssim \sum_{m=0}^k \sum_{Q \in \mathscr{G}_m'} \left( \sum_{\substack{R \in \mathscr{D}_k \\ B_R \cap 2C_0B_Q \not=\emptyset}} \frac{\ell(R)^{d+p}}{\ell(Q)^{d+p}}  \right)^\frac{2}{p} \ell(Q)^d \lesssim \sum_{m=0}^k \sum_{Q \in \mathscr{G}_m'} \sum_{\substack{R \in \mathscr{D}_k \\ B_R \cap 2C_0B_Q \not=\emptyset}} \frac{\ell(R)^{d\frac{2}{p} +2}}{\ell(Q)^{d(\frac{2}{p} -1) +2}}
\end{align}

Let $R \in \mathscr{D}_k$. For $m \leq k$, let $\mathscr{G}_{R,m}'$ be the set of cubes $Q$ in $\mathscr{G}_m'$ so that $B_R \cap 2C_0B_Q \not=\emptyset.$ We will show

\begin{align}\label{e:red4}
\# \mathscr{G}_{R,m}' \lesssim 1.
\end{align}

If $\mathscr{G}_{R,m}'$ is empty then there is nothing to show. Let us assume it is not empty, and fix some $Q^* \in \mathscr{G}_{R,m}'.$ For any other $Q \in \mathscr{G}_{R,m}'$, since $B_R \cap 2C_0B_Q \neq \emptyset$ and $\ell(R) \leq \ell(Q),$ it follows that $c_0B_Q \subseteq B \coloneqq 6C_0B_{Q^*}.$ Also, since $A > 6C_0$ and $Q \not\in \BWGL(A,\ve)$ (by definition of $\mathscr{G}'$) we can find a $d$-plane $P$ through $x_B$ so that $d_B(E,P) \lesssim \ve.$ Thus, for $\ve > 0$ small enough, the balls $c_0B_Q, \ Q \in \mathscr{G}_{R,m}',$ satisfy the conditions of Lemma <ref>, and we conclude (<ref>) since

\begin{align}
\# \mathscr{G}_{R,m}' \rho^{md} \lesssim \sum_{Q \in \mathscr{G}_{R,m}'} \ell(Q)^d \lesssim \rho^{md}.
\end{align}

Let us return to $I_3.$ Notice, if $R \in \mathscr{D}_k$, $m \leq k$, and $Q \in \mathscr{G}'_m$ is such that $B_R \cap 2C_0B_Q \not=\emptyset$, then $E \cap B_R \subseteq E \cap 3C_0B_Q \subseteq Q_0.$ Thus, continuing from (<ref>) by swapping the order of summation, using (<ref>), and summing over a geometric series, we have

\begin{align}
I_3 \lesssim  \sum_{\substack{R \in \mathscr{D}_k \\ R \subseteq Q_0}} \sum_{m=0}^k \sum_{\substack{Q \in \mathscr{G}_m' \\ B_R \cap 2C_0B_Q \not=\emptyset}} \frac{\ell(R)^{d\frac{2}{p} +2}}{\ell(Q)^{d(\frac{2}{p} -1) +2}} \lesssim \sum_{\substack{R \in \mathscr{D}_k \\ R \subseteq Q_0}} \ell(R)^d \lesssim  \sum_{\substack{R \in \mathscr{D}_k \\ R \subseteq Q_0}} \mathscr{H}^d(E \cap c_0B_R) \leq \mathscr{H}^d(Q_0).
\end{align}

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