H|\lessapprox_{\Delta}\frac{1}{m_{Q}}\cdot C\cdot M\cdot r^{s}=C\cdot
M_{Q}\cdot r^{s}.$

Thus, $\sigma(\mathcal{T}_{Q})$ is a $(\overline{\delta},s,C_{Q}^{1},0)$-set
down from scale $r_{0}$ with $C_{Q}^{1}\approx_{\Delta}C_{1}$. Similarly,
$\sigma(\mathcal{T}_{Q})$ is a $(\overline{\delta},\kappa,C_{Q}^{2},k)$-set
with $C_{Q}^{2}\approx_{\Delta}C_{2}$. This shows item v and thus the proof of
the Proposition.

∎

###  6.2 Good multiscale decomposition

The idea is to apply Proposition 6.2, then apply Theorem 1.8 to bound
$|\mathcal{T}_{\Delta}|$ and Corollary 6.5 to bound $|\mathcal{T}_{Q}|$.
Unfortunately, while we use pigeonholing to ensure that
$\mathcal{D}_{\Delta}(S_{H}(\mathcal{P}))$ is a $(\Delta,t)$-set, we don’t
know that $S_{H}^{-1}\circ S_{Q}(S_{H}(\mathcal{P})\cap Q)$ is a
$(\frac{\delta}{\Delta},t)$-set. In fact, we won’t show this statement, but
rather a slightly weaker statement that is good enough. For this, a good
choice of $\Delta$ based on the branching structure of $\mathcal{P}$ is
needed.

First, we explain the pigeonholing preliminaries.

######  Lemma 6.3.

Given $P\subset H_{r}$, a $(r_{0},k)$-plane, there is a subset
$P^{\prime}\subset P$ with
$|P^{\prime}|_{\delta}\geq(\log(\frac{r_{0}}{\delta}))^{-1}|P|_{\delta}$ such
that $|Q\cap S_{H}(P)|$ is constant for all
$Q\in\mathcal{D}_{\delta/r_{0}}(S_{H}(P))$.

###### Proof.

Let $f(N)=\sum\\{|P\cap
S_{H}^{-1}(Q)|_{\delta}:Q\in\mathcal{D}_{\delta/r_{0}}([0,1]^{d}),|P\cap
S_{H}^{-1}(Q)|_{\delta}\in[N,2N]\\}$. Then $\sum_{N\text{
dyadic}}f(N)=|P|_{\delta}$. For each $N$, either $f(N)=0$ or $N\leq
f(N)\leq(r_{0}/\delta)^{d}\cdot N$. Hence, if $N_{0}$ is the largest $N$ for
which $f(N)>0$, we get $f(N_{0})\geq
N_{0}>\sum_{M<N_{0}(\delta/r_{0})^{d}/100\text{ dyadic}}f(M)$. Thus, we have

$\sum_{N_{0}(\delta/r_{0})^{d}/100<M<N_{0}\text{
dyadic}}f(M)>\frac{1}{2}|P|_{\delta}.$

Thus, by dyadic pigeonholing, there exists
$M\in(N_{0}(\delta/r_{0})^{d}/100,N_{0})$ such that
$f(M)\geq\frac{1}{20d}(\log(\frac{r_{0}}{\delta}))^{-1}|P|_{\delta}$. ∎

The next step is to make $S_{H}^{-1}\circ S_{Q}(S_{H}(\mathcal{P})\cap Q)$
satisfy a $t^{\prime}$-dimensional spacing condition with $t^{\prime}$ just
slightly less than $t$, for all $Q\in\mathcal{D}_{\Delta}(S_{H}(\mathcal{P}))$
at a certain scale $\Delta$. To do so, we need the following lemma.

######  Lemma 6.4.

Fix $C,\varepsilon>0$, and let $\frac{\delta}{r_{0}}=\Delta^{m}$ and $P\subset
H_{r}$ be a $(\delta,t,C,0)$-set in $H_{r}$, a $(r_{0},k)$-plane. Let
$L=\log(\frac{r_{0}}{\delta})\cdot(\log(1/\Delta))^{m}$. If
$t^{\prime}<\frac{t-d\varepsilon}{1-\varepsilon}$, then there exists
$m\varepsilon\leq k\leq m$ and a subset $P^{\prime}\subset P$ with
$|P^{\prime}|\geq L^{-1}|P|$ such that for any $k\leq j\leq m$,
$Q\in\mathcal{D}_{\Delta^{k}}(P^{\prime})$, and
$R\in\mathcal{D}_{\Delta^{j}}(P^{\prime})\cap Q$, we have

$|P^{\prime}\cap S_{H}^{-1}(R)|\leq|P^{\prime}\cap
S_{H}^{-1}(Q)|\cdot\Delta^{(j-k)t^{\prime}},$ (6.23)

and for $\delta\leq r\leq\frac{\delta}{r_{0}}$ and a ball $B_{r}$, we have

$|P^{\prime}\cap B_{r}|\leq C\cdot L\cdot|P^{\prime}\cap
S_{H}^{-1}(Q)|\left(\frac{r}{\Delta^{k}}\right)^{t^{\prime}}.$ (6.24)

###### Proof.

Throughout this proof we will not distinguish between $m\varepsilon$ and
$\lceil m\varepsilon\rceil$.

First, we will make $S_{H}(P)$ uniform at scales
$1,\Delta,\Delta^{2},\cdots,\Delta^{m}=\frac{\delta}{r_{0}}$. By Lemmas 6.3
and 2.17, we can find $|P^{\prime}|\geq L^{-1}|P|$ such that there is a
sequence $(N_{j})_{j=1}^{n}$ with $|S_{H}(P^{\prime})\cap
Q|_{\Delta^{k}}=N_{k}$ for all $1\leq k\leq n$ and
$Q\in\mathcal{D}_{\Delta^{k}}(P^{\prime})$.

Let $m\varepsilon\leq k\leq m$ be the largest index such that
$N_{k}\geq|P^{\prime}|\Delta^{m\varepsilon\cdot d+(k-m\varepsilon)t^{\prime}}$
for one (equivalently all) $Q\in\mathcal{D}_{\Delta^{k}}(P^{\prime})$.
Certainly $k=m\varepsilon$ is a valid index since
$|\mathcal{D}_{\Delta^{m\varepsilon}}|=\Delta^{-dm\varepsilon}$.

Now, we will check the given conditions. By maximality of $k$, we have for
$k\leq j\leq m$,

$N_{j}\leq|P^{\prime}|\Delta^{dm\varepsilon+(j-m\varepsilon)t^{\prime}}\leq
N_{k}\Delta^{(j-k)t^{\prime}}.$

Noticing that $|P^{\prime}\cap S_{H}^{-1}(Q)|=|S_{H}(P^{\prime})\cap Q|$ and
likewise for $R\in\mathcal{D}_{\Delta^{j}}(P^{\prime})\cap Q$, this proves
(6.23).

To check (6.24), we recall that $LN_{k}\geq
L\cdot|P^{\prime}|\Delta^{dm\varepsilon+(k-m\varepsilon)t^{\prime}}\geq|P|\Delta^{dm\varepsilon+(k-m\varepsilon)t^{\prime}}$.
Using $r\leq\frac{\delta}{r_{0}}=\Delta^{m}$,
$t^{\prime}\leq\frac{t-d\varepsilon}{1-\varepsilon}$, and that $P$ is a
$(\delta,t,C)$-set, we have

$|P^{\prime}\cap B_{r}|\leq|P\cap B_{r}|\leq C|P|r^{t}\leq C\cdot
N_{k}L\left(\frac{r}{\Delta^{k}}\right)^{t^{\prime}}.$

∎

Finally, we will need the following variant of Corollary 2.10.

######  Corollary 6.5.

Let $0\leq\max(s,k)<t\leq d-1$, $\delta\leq r\leq 1$, and let $C_{P}\geq
1,C_{T}\geq 0$. Let $\mathcal{P}\subset\mathcal{D}_{\delta}$ be a set
contained in an $(r_{0},k+1)$-plate $H$ satisfying the following conditions:

  * •

For all $\frac{\delta}{r_{0}}\leq r\leq 1$ and balls $B_{r}$, we have

$|\mathcal{P}\cap S_{H}^{-1}(B_{r})|\leq C_{P}\cdot|\mathcal{P}|\cdot r^{t}.$
(6.25)

  * •

For all $\delta\leq r\leq\frac{\delta}{r_{0}}$ and balls $B_{r}$, we have

$|\mathcal{P}\cap B_{r}|\leq C_{P}\cdot|\mathcal{P}|\cdot r^{t}.$ (6.26)

Assume that for every $p\in\mathcal{P}$ there exists a family
$\mathcal{T}(p)\subset\mathcal{T}^{\delta}$ of dyadic $\delta$-tubes
satisfying the following conditions:

  * •

$T\cap p\neq\emptyset$ for all $T\in\mathcal{T}(p)$;

  * •

$|\mathcal{T}(p)\cap\mathbf{T}|\leq C_{T}\cdot|\mathcal{T}(p)|\cdot
r_{0}^{k-s}x^{s}$ for all $x$-tubes $\mathbf{T}$ with $\delta\leq x\leq
r_{0}$.

Further assume that $|\mathcal{T}(p)|=M$ for some $M\geq 1$. If
$\mathcal{T}=\cup_{p\in\mathcal{P}}\mathcal{T}(p)$, then

$|\mathcal{T}|\gtrsim(C_{P}C_{T})^{-1}\cdot Mr_{0}^{s-k}\delta^{-s}.$

###### Proof.

Let

$j_{P}(\mathcal{P},\mathcal{T})=\\{(q,t)\in\mathcal{P}\times\mathcal{T}(p):t\in\mathcal{T}(q)\\}$

We have the following:

######  Lemma 6.6.

For all $p\in\mathcal{P}$, we have
$j_{p}(\mathcal{P},\mathcal{T})\lesssim_{s,t,k}C_{P}C_{T}|\mathcal{P}|\cdot
Mr_{0}^{k-s}\delta^{s}$.

###### Proof.

We count $j_{p}(\mathcal{P},\mathcal{T})$ by first choosing a dyadic
$\delta<r<1$, then counting the number of $q\in\mathcal{P}$ with $|p-q|\sim
r$, then finally counting the number of $t\in\mathcal{T}$ that pass through
$p,q$.

If $r>\frac{\delta}{r_{0}}$, we claim that if $|x-y|\in[r,2r]$ and some tube
through $x,y$ lies in $H$, then $|S_{H}(x)-S_{H}(y)|\leq 100r$, so $y\in
S_{H}^{-1}(B_{100r}(S_{H}(x)))$.

To prove this, we may assume $r_{0}\leq\frac{1}{50}$, as otherwise we can use
the simple fact $|S_{H}(x)-S_{H}(y)|\leq r_{0}^{-1}|x-y|\leq 100r$. Now choose
a coordinate system such that the first $k+1$ axes correspond to the long
sides of $H$, and the remaining axes correspond to the short sides of $H$. Let
$x-y=(\vec{a},\vec{b})\in\R^{k+1}\times\R^{d-k-1}$. Then
$|\vec{a}|\leq|x-y|\leq r$. Furthermore, we have $|\vec{b}|\leq
50r_{0}|\vec{a}|$, otherwise any tube through $x,y$ would be roughly
orthogonal to $H$ and intersect $H$ in a subtube with length $2r_{0}\leq 1$,
contradiction. Thus, we have
$|S_{H}(x)-S_{H}(y)|\leq|\vec{a}|+r_{0}^{-1}|\vec{b}|\leq 100r$.

Using the claim and condition (6.25), we see that there are $\lesssim
C_{P}|\mathcal{P}|\cdot r^{t}$ many choices for $q$. For each $q$, the set of
tubes $t\in\mathcal{T}(p)$ passing through $q$ lies in a
$\frac{\delta}{r}$-tube, so by the tube non-concentration condition (and
noting that $\frac{\delta}{r}<r_{0}$), we have $C_{T}\cdot
Mr_{0}^{k-s}\left(\frac{\delta}{r}\right)^{s}$ choices for $t$.

Thus, the contribution to $j_{p}(\mathcal{P},\mathcal{T})$ for a given dyadic
$r>\frac{\delta}{r_{0}}$ is $C_{P}C_{T}|\mathcal{P}|\cdot
Mr_{0}^{k-s}\delta^{s}\cdot r^{s-t}$, and summing over dyadic $r$ gives
$C_{P}C_{T}|\mathcal{P}|\cdot Mr_{0}^{k-s}\delta^{s}\cdot O_{s-t}(1)$.

If $r<\frac{\delta}{r_{0}}$, then by condition 6.26 we see that there are
$\lesssim C_{P}|\mathcal{P}|\cdot r^{t}$ many choices for $q$. For each $q$,
the set of tubes $t\in\mathcal{T}(p)$ passing through $q$ lies in a
$\frac{\delta}{r}$-tube $\mathbf{T}_{\delta/r}$. We note that
$\frac{\delta}{r}>r_{0}$, so the tube non-concentration doesn’t apply
directly, but luckily we note that $\mathbf{T}_{\delta/r}\cap H$ can be
covered by $(\frac{\delta}{rr_{0}})^{k}$ many $r_{0}$-tubes. Thus, by using
tube non-concentration at scale $r_{0}$, we have $C_{T}\cdot
Mr_{0}^{k}\cdot(\frac{\delta}{rr_{0}})^{k}$ choices for $t$.

Thus, the contribution to $j_{p}(\mathcal{P},\mathcal{T})$ for a given dyadic
$r<\frac{\delta}{r_{0}}$ is (after some manipulation)

$C_{P}C_{T}|P|Mr_{0}^{k}\left(\frac{\delta}{r_{0}}\right)^{t}\cdot\left(\frac{rr_{0}}{\delta}\right)^{t-k}.$

Since $t>\max(k,s)$, the sum is $\lesssim
C_{P}C_{T}|P|Mr_{0}^{k}\left(\frac{\delta}{r_{0}}\right)^{s}$.

Adding up both $r>\frac{\delta}{r_{0}}$ and $r<\frac{\delta}{r_{0}}$
contributions, we prove the Lemma. ∎

For $t\in\mathcal{T}$, let
$\mathcal{P}(t)=\\{p\in\mathcal{P}:t\in\mathcal{T}(p)\\}$. By Cauchy-Schwarz,
we have

$(M|\mathcal{P}|)^{2}=\left(\sum_{t\in\mathcal{T}}|\mathcal{P}(t)|\right)^{2}\leq|\mathcal{T}|\sum_{t\in\mathcal{T}}|\mathcal{P}(t)|^{2}=|\mathcal{T}|\sum_{p\in\mathcal{P}}j_{p}(\mathcal{P},\mathcal{T}).$

By Lemma 6.6, we get

$|\mathcal{T}|\geq\frac{M^{2}|\mathcal{P}|^{2}}{C_{P}C_{T}|\mathcal{P}|^{2}Mr_{0}^{k-s}\delta^{s}}=(C_{P}C_{T})^{-1}Mr_{0}^{s-k}\delta^{-s}.$

∎

###### Proof of Theorem 6.1.

A small reduction: we would like to assume $|\mathcal{T}(p)|\sim M$ for all
$p\in\mathcal{P}$. To assume this, we first observe that $|\mathcal{T}(p)|\geq
M_{0}=(\delta/r_{0})^{-\varepsilon}r_{0}^{k-s}\delta^{-s}$ for all
$p\in\mathcal{P}$. On the other hand, if for at least half of the
$p\in\mathcal{P}$ (call them $\mathcal{P}^{\prime}$) we have
$|\mathcal{T}(p)|\geq M_{0}(\delta/r_{0})^{-1}$, then we are immediately done
by Corollary 6.5 applied to $\mathcal{P}^{\prime}$ and $\mathcal{T}(p)$. Thus,
by reducing $\mathcal{P}$ if necessary, we may assume
$|\mathcal{T}(p)|\in(M_{0},M_{0}(\delta/r_{0})^{-1})$. Then by reducing
$\mathcal{P}$ further by a $\lesssim\log(\delta/r_{0})^{-1}$ factor, we may
assume $|\mathcal{T}(p)|\in(M,2M)$ for some
$M\in(M_{0},M_{0}(\delta/r_{0})^{-1})$. Finally, we may remove some tubes from
each $\mathcal{T}(p)$ to make $|\mathcal{T}(p)|=M$. Then
$(\mathcal{P}_{0},\mathcal{T}_{0})$ is a
$(\delta,s,C_{1}r_{0}^{k-s},\kappa,C_{2},M)$-nice configuration.

Pick $\beta(s,t,k)>0$ such that $\frac{t-d\beta}{1-\beta}>\max(s,k)$, and let
$t^{\prime}=\frac{1}{2}(\frac{t-d\beta}{1-\beta}+\max(s,k))$. Pick $\Delta>0$
such that $\log(1/\Delta)<\Delta^{-\varepsilon}$. Find
$\Delta^{\prime}=\Delta^{k}\in(\delta/r_{0},(\delta/r_{0})^{\beta})$ such that
the conclusion of Lemma 6.4 holds. Now by Proposition 6.2, we have

$\frac{|\mathcal{T}|}{M}\geq\frac{|\mathcal{T}_{Q}|}{M_{Q}}\cdot\frac{|\mathcal{T}^{\Delta^{\prime}}(\mathcal{T})|}{M_{\Delta^{\prime}}}.$

If $\varepsilon<\beta\eta^{2}$, where $\eta(s,t,\kappa,k,d)$ is the parameter
in Theorem 5.1, we have
$\frac{|\mathcal{T}^{\Delta^{\prime}}(\mathcal{T})|}{M_{\Delta^{\prime}}}\geq(\Delta^{\prime})^{-s-\sqrt{\varepsilon}}$.

Pick $Q$. Then $S_{H}^{-1}\circ S_{Q}(S_{H}(\mathcal{P})\cap Q)$ satisfies the
conditions of Corollary 6.5 with
$C_{P}=\left(\frac{\delta}{r_{0}}\right)^{-\varepsilon}\cdot
L\cdot\Delta^{-d}$. Thus, we have
$\frac{|\mathcal{T}_{Q}|}{M_{Q}}\geq\left(\frac{\delta}{r_{0}}\right)^{-\varepsilon}\Delta^{d}L^{-1}\cdot
r_{0}^{s-k}\left(\frac{\delta}{\Delta^{\prime}}\right)^{-s}$. Using these two
bounds and $M\geq(\delta/r_{0})^{\varepsilon}r_{0}^{s-k}\delta^{-s}$, we get

$|\mathcal{T}|\geq\left(\frac{\delta}{r_{0}}\right)^{2\varepsilon-\sqrt{\varepsilon}\beta}\Delta^{-d}L^{-1}r_{0}^{2(s-k)}\cdot\delta^{-2s}.$

It remains to choose $\varepsilon<\beta^{2}/100$ and also for
$\frac{\delta}{r_{0}}$ small enough, we have
$\Delta^{-d}<\left(\frac{\delta}{r_{0}}\right)^{-\varepsilon}$ and
$L\leq\left(\frac{\delta}{r_{0}}\right)^{-\varepsilon}\Delta^{-\varepsilon
m}\leq\left(\frac{\delta}{r_{0}}\right)^{-2\varepsilon}$. Thus,
$|\mathcal{T}|\geq\left(\frac{\delta}{r_{0}}\right)^{-\varepsilon}r_{0}^{2(s-k)}\cdot\delta^{-2s}$
and we are done. ∎

##  7 Power decay around $k$-planes

In this section, we will roughly deal with the following situation:

  * •

$\mu,\nu$ are $s$-Frostman measures with $k-1<s\leq k$;

  * •

$\nu$ gives mass $\leq\varepsilon$ to any $(r_{0},k)$-plate.

In other words, $\nu$ does not concentrate around $(r_{0},k)$-plates. We would
like to understand the $\nu$-mass of $(r,k)$-plates for $r$ much smaller than
$r_{0}$. A result of Shmerkin [25, Proposition B.1] says that there exist
$r_{1}(r_{0},s,k),\kappa(s,k)>0$, a subset $X\subset\mathrm{spt}\mu$ with
$\mu(X)>1-O(\varepsilon)$, and for each $x\in X$, a subset
$Y_{x}\subset\mathrm{spt}\nu$ with $\mu(Y_{x})>1-O(\varepsilon)$ such that
$\nu(H\cap Y_{x})\leq r^{\eta}$ for all $r\leq r_{1}$ and $(r,k)$-plates $H$
through $x$. Thus, we do obtain a power decay for sufficiently small $r$. But
what is the optimal starting point of the power decay? Can we hope for a power
decay $\nu(H\cap Y_{x})\lesssim K(\frac{r}{r_{0}})^{\eta}$ for all
$(r,k)$-plates through $x$? The answer is yes, and indeed we shall prove it by
making small but meaningful tweaks to Shmerkin’s argument. But before stating
our result, we shall introduce some convenient notation. We define thin
$k$-plates, a generalization of thin tubes, as follows.

######  Definition 7.1.

Let $K,t\geq 0$, $1\leq k\leq d-1$, and $c\in(0,1]$. Let
$\mu,\nu\in\mathbb{P}(\R^{d})$ supported on $X,Y$. Fix $G\subset X\times Y$.
We say $(\mu,\nu)$ has $(t,K,c)$-thin $k$-plates on $G$ down from scale
$r_{0}$ if

$\nu(H\cap G|_{x})\leq K\cdot r^{t}\quad\text{ for all }r\in(0,r_{0})\text{
and all }(r,k)\text{-plates }H\text{ containing }x.$ (7.27)

######  Remark 7.2.

In this paper, we will choose $G=(A\cup B)^{c}$ where $\mu\times\nu(B)$ is
small. (The complement is taken with respect to $\R^{d}\times\R^{d}$.) In this
case, the equation (7.27) becomes

$\nu(H\setminus(A|_{x}\cup B|_{x}))\leq K\cdot r^{t}\quad\text{ for all
}r\in(0,r_{0})\text{ and all }(r,k)\text{-plates }H\text{ containing }x.$

Now, we can state the main proposition, which generalizes and extends
Proposition B.1 of [25]. It may be of independent interest.

######  Proposition 7.3.

Let $1\leq k\leq d-1$ and $k-1<s\leq k$. There exist $\eta(\kappa,k,d)>0$ and
$K_{0}(\kappa,k,d)>0$ with the following property. Fix $r_{0}\leq 1$ and
$K\geq K_{0}$. Suppose that $\mu,\nu$ are positive measures with
$|\mu|,|\nu|\geq 1$ and for any $(r,k-1)$-plate $H$, we have

$\displaystyle\mu(H)\leq C_{\mu}r^{\kappa},$ $\displaystyle\nu(H)\leq
C_{\nu}r^{\kappa}.$

Let $A\subset X\times Y$ be the pairs of points that lie in some
$K^{-1}$-concentrated $(r_{0},k)$-plate. Then there exists $B$ with
$\mu\times\nu(B)\leq K_{0}K^{-1}$ such that $(\mu,\nu)$ have
$(\eta,Kr_{0}^{-\eta})$-thin $k$-plates on $(A\cup B)^{c}$. (The complement is
taken with respect to $\R^{d}\times\R^{d}$.)

######  Remark 7.4.

(a) We can apply Proposition 7.3 in case $\mu,\nu$ are $s$-dimensional with
$s>k-1$.

(b) In Proposition B.1 of [25], the exponents for $\mu,\nu$ are allowed to
differ. The proof of Proposition 7.3 is easily modified to include this
detail.

To prove Proposition 7.3, we need the following two lemmas. Fix $r\leq r_{0}$.
The first says that there are few dense $(r,k)$-plates, and the second says
that for most $x\in X$, the dense $(r,k)$-plates through $x$ lie in some
$(r_{0},k)$-plate.

######  Lemma 7.5.

There is $N=N(\kappa,k,d)$ such that the following holds: let $\nu$ be a
measure with mass $\leq 1$ such that $\nu(W)\leq C_{\nu}\rho^{\kappa}$ for all
$(\rho,k-1)$-plates $W$, $1>\rho>r$. Let $\mathcal{E}_{r,k}$ be a set of
$(r,k)$-plates such that every $(s,k)$-plate contains
$\lesssim\left(\frac{s}{r}\right)^{(k+1)(d-k)}$ many $r$-plates of
$\mathcal{E}_{r,k}$ (as in Section 2.2). Let
$\mathcal{H}=\\{H\in\mathcal{E}_{r,k}:\nu(H)\geq a\\}$. Then
$|\mathcal{H}|\lesssim(\frac{C_{\nu}}{a})^{N}$.

Lemma 7.5 follows from the condition on $\mathcal{E}_{r,k}$ and the following
generalization of [25, Lemma B.3]. In the case $a=\delta^{\eta}$, the
resulting bound is stronger but the assumption is also stronger.

######  Lemma 7.6.

Suppose $\nu(W)\leq C_{\nu}\rho^{\kappa}$ for all $(\rho,k-1)$-plates $W$,
$1>\rho>r$. Then there exists a family of $\lesssim a^{-1}$ many
$(r(C_{\nu}/a^{2})^{1/\kappa},k)$-plates $\\{T_{j}\\}$ such that every
$(r,k)$-plate $H$ with $\nu(H)\geq a$ is contained in some plate $T_{j}$.

###### Proof.

Choose a maximal set of $(r,k)$-plates $\\{Y_{j}\\}_{j=1}^{m}$ such that

  1. 1.

$\nu(Y_{i})\geq a$,

  2. 2.

$\nu(Y_{i}\cap Y_{j})\leq a^{2}/2$ for $1\leq i<j\leq m$.

We claim $m\leq 2a^{-1}$. Indeed, if $S=\sum_{i=1}^{m}\nu(Y_{i})$ and
$f=\sum_{i=1}^{m}\mathbbm{1}_{Y_{i}}$, then, then

$S^{2}=\left(\int f\,d\nu\right)^{2}\leq\int f^{2}\,d\nu=S+\sum_{1\leq i<j\leq
m}\nu(Y_{i}\cap Y_{j})\leq S+m^{2}a^{2}/2.$ (7.28)

Now, $S\geq ma>2$, so $S^{2}-S>\frac{S^{2}}{2}$. Combining with (7.28) gives
$S^{2}<m^{2}a^{2}$, a contradiction.

Let $\\{T_{j}\\}_{j=1}^{m}$ be the $(r(C_{\nu}/a^{2})^{1/\kappa},k)$-plates
with same central $k$-plane as $Y_{j}$. We show the problem condition. Given
an $(r,k)$-plate $H$ with $\nu(H)\geq a$, by maximality there exists $Y_{j}$
such that $\nu(H\cap Y_{j})\geq a^{2}/2$. Thus, if $\angle(H,Y_{j})$ is the
largest principal angle between the central planes of $H$ and $Y_{j}$, then
$H\cap Y_{j}$ is contained in a box of dimensions

$\underbrace{1\times\cdots\times 1}_{(k-1)\text{ times}}\times
r/\angle(H,Y_{j})\times\underbrace{r\times\cdots\times r}_{d-k\text{ times}}.$

Thus, $H\cap Y_{j}$ is contained in a $(r/\angle(H,Y_{j}),k-1)$-plate, so
$\nu(H\cap Y_{j})\leq C_{\nu}(r/\angle(H,Y_{j}))^{\kappa}$. Thus,
$\angle(H,Y_{j})\lesssim r(C_{\nu}/a^{2})^{1/\kappa}$, so $H$ is contained in
$T_{j}$. ∎

######  Remark 7.7.

We would like to present an alternative proof of Lemma 7.5, which was the
original one found by the author. It gives slightly worse bounds but we
believe it is slightly more motivated.

If $a>1$ then $\mathcal{H}=\emptyset$, so assume $a\leq 1$. Let
$\xi=\left(\frac{a}{2C_{\nu}}\right)^{1/\kappa}\leq 1$. By induction, for each
$0\leq i\leq k$, there exist $x_{0},\cdots,x_{i}$ such that $|x_{0}\wedge
x_{1}\wedge\cdots\wedge x_{i}|\geq\xi^{i}$ and that lie in at least
$|\mathcal{H}|(\frac{a}{2})^{i+1}$ many elements of $\mathcal{H}$.

The base case $i=1$ is trivial. For the inductive step, suppose
$x_{0},\cdots,x_{i}$ are found. Let $\Omega$ be the $\tilde{r}$-neighborhood
of the span of $x_{1},\cdots,x_{i}$. Then since $\mu(\Omega)\leq
C_{\nu}\xi^{\kappa}<\frac{1}{2}a$ for every $H\in\mathcal{H}$, we have
$\mu(H\setminus\Omega)\geq\frac{1}{2}a$. Thus, there is
$x_{i+1}\in\R^{d}\setminus\Omega$ such that $x_{0},\cdots,x_{i+1}$ lie in at
least $|\mathcal{H}|(\frac{a}{2})^{(i+2)}$ many elements of $\mathcal{H}$, and
by construction, $|x_{0}\wedge x_{1}\wedge\cdots\wedge x_{i+1}|\geq\xi^{i+1}$.
This completes the inductive step and thus the proof of the claim.

Finally, the set of $(r,k)$-plates through $x_{1},\cdots,x_{k}$ must lie in a
$(r\xi^{-k},k)$-plate, so at most $\xi^{-k(k+1)(d-k)}$ many $(r,k)$-plates of
$\mathcal{E}_{r,k}$ can lie in it. Thus,
$|\mathcal{H}|\leq(\frac{a}{2})^{-(k+1)}\xi^{-k(k+1)(d-k)}\lesssim(\frac{C_{\nu}}{a})^{N}$.

The following lemma is in the same spirit as [25, Proposition B.2].

######  Lemma 7.8.

Let $\mathcal{H}$ be a collection of $(r,k)$-plates, and suppose $\mu(W)\leq
C_{\mu}\rho^{\kappa}$ for all $(\rho,k-1)$-plates $W$, $1>\rho>r$. Then for
all $x\in X$ except a set of $\mu$-measure $\leq
C_{\mu}\left(\frac{r}{r_{0}}\right)^{\kappa}|\mathcal{H}|^{2}$, there exists
an $(r_{0},k)$-plate that contains every $(r,k)$-plate in $\mathcal{H}$ that
passes through $x$.

###### Proof.

The exceptional set is contained in the set of $x\in X$ that lies in two
plates of $\mathcal{H}$ with “angle” $\geq\frac{1}{r_{0}}$. The intersection
of two such plates is contained in a box with dimensions
$\underbrace{r\times\cdots\times r}_{d-k\text{
times}}\times\frac{r}{r_{0}}\times\underbrace{1\times\cdots\times
1}_{k-1\text{ times}}$, which in turn is contained in a
$(\frac{r}{r_{0}},k-1)$-plate (since $r_{0}\leq 1$). Thus, by assumption on
$\mu$, this box has mass $\lesssim
C_{\mu}\left(\frac{r}{r_{0}}\right)^{s-(k-1)}$. Finally, there are
$|\mathcal{H}|^{2}$ pairs of plates in $\mathcal{H}$. ∎

###### Proof of Proposition 7.3.

Fix $r\leq r_{0}$, and let $\eta=\frac{\kappa}{4N}$, where $N$ is the constant
in Lemma 7.5. We may assume $N\geq 2$. By Lemmas 7.5 and 7.8, we can find a
set $E_{r}$ with $\mu(E_{r})\leq K^{-2}\left(\frac{r}{r_{0}}\right)^{\eta}$
and, for each $x\notin E_{r}$, a set $P_{r}(x)\subset Y$ that is either empty
or a $(r_{0}^{1/2}r^{1/2},k)$-plate through $x$ such that $\nu(W)\leq
K\left(\frac{r}{r_{0}}\right)^{\eta}$ for every $W$ intersecting $Y\setminus
P_{r}(x)$.

Now, let $E=\cup_{n\geq 0}E_{r_{0}K^{-2^{n}}}$ and $P(x)=\cup_{n\geq
0}P_{r_{0}K^{-2^{n}}}(x)$. We claim that $\mu(E)\leq K^{-1}$ and if $x\notin
E$, then $\nu(P(x)\setminus A|_{x})\lesssim K^{-1}$. Then if $r\geq
r_{0}K^{-1}$, then $\nu(W)\leq 1\leq K\left(\frac{r}{r_{0}}\right)^{\eta}$ for
$\eta<1$; for any $r_{0}K^{-2^{n}}\leq r<r_{0}K^{-2^{n-1}}$, we have for any
$(r,k)$-plate $W$,

$\nu(W\setminus P_{r_{0}K^{-2^{n}}}(x))\leq
K\left(\frac{r_{0}K^{-2^{n-1}}}{r_{0}}\right)^{\eta}\leq
K\left(\frac{r}{r_{0}}\right)^{\eta/2}$

Then $(\mu,\nu)$ have $(\eta/2,K^{2}r_{0}^{-\eta/2},1-K^{-1})$-thin $k$-plates
relative to $A$.

To prove the first claim, we observe that $\mu(E)\leq
K^{-2}\sum_{n=0}^{\infty}K^{-\eta 2^{n}}\leq K^{-1}$ if $K_{0}$ is
sufficiently large in terms of $\eta$.

Next, by definition of $P_{r_{0}K^{-2^{n-1}}}$, we have
$\nu(P_{r_{0}K^{-2^{n}}}(x)\setminus P_{r_{0}K^{-2^{n-1}}}(x))\leq
K\left(\frac{r_{0}K^{-2^{n-1}}}{r_{0}}\right)^{\eta/2}\leq K^{1-2^{n-2}\eta}$.
We also have the bound $\nu(P_{r_{0}2^{-n}}(x)\setminus A|_{x})\leq K^{-1}$
from the given condition (note that $P_{r_{0}2^{-n}}(x)$ is a
$(r_{0}2^{-(n-1)},k)$-plate). Thus,

$\displaystyle\nu(P(x)\setminus A|_{x})$
$\displaystyle\leq\sum_{n=0}^{\log\eta^{-1}}\nu(P_{r_{0}K^{-2^{n}}}(x)\setminus
A|_{x})+\sum_{n=\log\eta^{-1}}^{\infty}\nu(P_{r_{0}K^{-2^{n}}}(x)\setminus
P_{r_{0}K^{-2^{n-1}}}(x))$ $\displaystyle\leq\log\eta^{-1}\cdot
K^{-1}+\sum_{n=\log\eta^{-1}}^{\infty}K^{1-2^{n-2}\eta}$
$\displaystyle\lesssim K^{-1},$

if $K_{0}$ is chosen large enough. ∎

##  8 Radial projection estimates

In this section, we will first prove a key special case, and then the general
case of Theorem 1.13.

###  8.1 Maximal plate concentration case

This subsection is based on ideas from [19].

######  Theorem 8.1.

Let $k\in\\{1,2,\cdots,d-1\\}$, $k-1<\sigma<s\leq k$, and fix $K\geq 1$. There
exists $N\in\mathbb{N}$ and $K_{0}$ depending on $\sigma,s,k$ such that the
following holds. Fix $r_{0}\leq 1$ and $K_{1},K_{2}\geq K_{0}$. Let $\mu,\nu$
be $\sim 1$-separated $s$-dimensional measures with constant $C_{\mu},C_{\nu}$
supported on $E_{1},E_{2}$, which lie in an $(r_{0},k)$-plate $H_{r}$. Assume
that $|\mu|,|\nu|\leq 1$. Let $A$ be the pairs of $(x,y)\in E_{1}\times E_{2}$
that lie in some $K_{1}^{-1}$-concentrated $(\frac{r_{0}}{K_{2}},k)$-plate.
Then there exists a set $B\subset E_{1}\times E_{2}$ with
$\mu\times\nu(B)\lesssim K_{1}^{-1}$ such that for every $x\in E_{1}$ and
$r$-tube $T$ through $x$, we have

$\mu(T\setminus(A|_{x}\cup
B|_{x}))\lesssim\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)}}(K_{1}K_{2})^{N}.$

The implicit constant may depend on $s,k$.

Theorem 8.1 is the special case of Theorem 1.13 where $(\mu,\nu)$ are
concentrated in a $(r_{0}K,k)$-plate for some small $K\ll r_{0}$ (we call this
the maximal plate concentration case). For this, we closely follow the
bootstrapping approach of [19]. There are three ingredients.

  * •

The next Proposition 8.2 will be the base case for the bootstrapping argument
($\sigma=0$).

  * •

Proposition 7.3 will ensure power decay for $\mu,\nu$ around $k$-planes.

  * •

Theorem 1.11 will be used in the bootstrapping step to upgrade $\sigma$ to
$\sigma+\eta$.

######  Proposition 8.2.

Let $1\leq k\leq d-1$ and $k-1<s\leq k$, then there exists $N=N(s,k)$ such
that the following holds. Fix $K\geq K_{0}$. Then for any $s$-dimensional
measures $\mu,\nu$ with constant $\sim 1$ contained in the
$r_{0}$-neighborhood of a $k$-plane and $d(\mu,\nu)\gtrsim 1$, there exists
$B\subset X\times Y$ with $\mu\times\nu(B)\leq K^{-1}$ such that $(\mu,\nu)$
has $(0,K^{N}r_{0}^{k-1})$-thin tubes on $B^{c}$ down from scale $r_{0}$.

###### Proof.

Let $\tilde{\mu},\tilde{\nu}$ be the projected measures on the $k$-plane. Then
$\tilde{\mu},\tilde{\nu}$ satisfy $s$-dimensional Frostman conditions for
$r_{0}\leq r\leq 1$. Let

$B=\\{(x,y):x,y\in T\text{ for some }r_{0}\text{-tube }T\text{ with
}\nu(T)\geq K^{N}r_{0}^{k-1}.\\}$

The rest is a standard argument following [8, Proof of Lemma 3.6]. Define the
radial projection $P_{y}(x)=\frac{x-y}{|x-y|}$. Orponen’s radial projection
theorem [17, Equation (3.5)] can be written in the form (where $p=p(s,k)>1$):

$\int\|P_{x}\tilde{\mu}\|_{L^{p}}^{p}\,d\tilde{\mu}(x)\lesssim 1.$ (8.29)

To effectively use (8.29), we will show that $|P_{x}(B|_{x})|$ is small for
$x\in X$. Indeed, let $\mathcal{T}_{x}$ be a minimal set of finitely
overlapping $2r_{0}$-tubes through $x$ such that any $r_{0}$-tube through $x$
with $\nu(T)\geq K^{N}r_{0}^{k-1}$ lies in a $2r_{0}$-tube in
$\mathcal{T}_{x}$. Then each $2r_{0}$-tube in $\mathcal{T}_{x}$ has
$\nu$-measure $\geq K^{N}r_{0}^{k-1}$. Since $d(x,\nu)\gtrsim 1$, we conclude
that $|\mathcal{T}_{x}|\lesssim K^{-N}r_{0}^{1-k}$. Therefore, since the
Lebesgue measure $|P_{x}(T)|\lesssim r_{0}^{k-1}$ for a $2r_{0}$-tube $T$
through $x$, we obtain $|P_{x}(B|_{x})|\lesssim K^{-N}$. Finally, we can use
Holder’s inequality and (8.29) to upper bound $\mu\times\nu(B)$:

$\displaystyle\mu\times\nu(B)$ $\displaystyle=\int\nu(B|_{x})d\mu(x)$
$\displaystyle=\int\left(\int_{P_{x}(B|_{x})}P_{x}(\nu)\right)d\mu(x)$
$\displaystyle\leq\sup_{x}|P_{x}(B|_{x})|^{1-1/p}\int\|P_{x}\nu\|_{L^{p}}d\mu(x)$
$\displaystyle\lesssim K^{-N(1-1/p)}.$

Choose $N=1+(1-1/p)^{-1}$ to finish (the implicit constant is dominated by
$K\geq K_{0}$ if $K_{0}$ is large enough). ∎

The bootstrapping step is as follows:

######  Proposition 8.3.

Let $k\in\\{1,\cdots,d-1\\}$, $0\leq\sigma\leq k$, $\max(\sigma,k-1)<s\leq k$,
$\kappa>0$. There exist $\eta(\sigma,s,\kappa,k,d)$ and $K_{0}(\eta,k)>0$ such
that the following holds. Fix $r_{0}\leq 1$ and $K\geq K_{0}$. Let $\mu,\nu$
be $\sim K^{-1}$-separated $s$-dimensional measures with constant $K$
supported on $X,Y$, which lie in an $(r_{0},k)$-plate $H$. Let $G\subset
X\times Y$. Suppose that $(\mu,\nu)$ and $(\nu,\mu)$ have
$(\sigma,Kr_{0}^{-(\sigma-(k-1))})$-thin tubes and
$(\kappa,Kr_{0}^{-\kappa})$-thin $k$-plates on $G$ down from scale $r_{0}$.
Then there exists a set $B\subset X\times Y$ with $\mu\times\nu(B)\leq K^{-1}$
such that $(\mu,\nu)$ and $(\nu,\mu)$ have
$(\sigma+\eta,K^{d+1}r_{0}^{-(\sigma+\eta-(k-1))})$-thin tubes on $G\setminus
B$ down from scale $r_{0}$. Furthermore, $\eta(\sigma,s,\kappa,k,d)$ is
bounded away from zero on any compact subset of
$\\{(\sigma,s,\kappa,k):\max(\sigma,k-1)<s\leq k\leq d-1\\}$.

######  Remark 8.4.

The reader is advised to set $r_{0}=1$ in the following argument, in which
case it is a straightforward modification of [19, Lemma 2.8], with one small
technical exception in the proof of the concentrated case, where we improve
upon the dyadic pigeonholing step. Also if $r_{0}=1$, then the simpler Theorem
1.8 can be used instead of Theorem 1.11 in the proof.

###### Proof.

We are given that for all $r\in(0,r_{0}]$,

$\displaystyle\nu(T\cap G|_{x})\leq
K\cdot\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)}}\text{ for all }r\text{-tubes
}T\text{ containing }x\in X,$ (8.30) $\displaystyle\nu(W\cap G|_{x})\leq
K\cdot\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)}}\text{ for all
}(r,k)\text{-plates }W\text{ containing }x\in X,$ (8.31)
$\displaystyle\mu(T\cap G|^{y})\leq
K\cdot\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)}}\text{ for all }r\text{-tubes
}T\text{ containing }y\in Y,$ (8.32) $\displaystyle\mu(W\cap G|^{y})\leq
K\cdot\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)}}\text{ for all
}(r,k)\text{-plates }W\text{ containing }y\in Y.$ (8.33)

For $x\in X$ and $r\leq r_{0}$, let $\mathcal{T}^{\prime\prime}_{x,r}$ denote
the $r$-tubes through $x$ such that

$\nu(T\cap G|_{x})\geq
K^{d+1}\cdot\frac{r^{\sigma+\eta}}{r_{0}^{\sigma+\eta-(k-1)}}.$ (8.34)

Now, let $\mathcal{T}^{\prime}_{x,r}$ denote a covering of
$\mathcal{T}^{\prime\prime}_{x,r}$ by essentially distinct $2r$-tubes. Then
for $x\in X$, since $d(x,Y)\geq K^{-1}$, we have that the tubes in
$\mathcal{T}^{\prime}_{x,r}$ have $\lesssim K^{d-1}$-overlap on $\nu$, so
$|\mathcal{T}^{\prime}_{x,r}|\lesssim\frac{r^{-(\sigma+\eta)}}{r_{0}^{-(\sigma+\eta-(k-1))}}$.
For a dyadic $r\in(0,r_{0}]$, let $H_{r}=\\{(x,y)\in
G:y\in\cup\mathcal{T}^{\prime}_{x,r}\\}$, where
$\cup\mathcal{T}^{\prime}_{x,r}$ denotes the union of the tubes in
$\mathcal{T}^{\prime}_{x,r}$.

Claim. There are $\eta(\sigma,s,\kappa,k,d)>0$ and $K_{0}(\eta)>0$ such that
the following holds for $K\geq K_{0}$. If $\frac{r}{r_{0}}<K^{-1/\eta}$, then
$\mu\times\nu(H_{r})\leq 2\left(\frac{r}{r_{0}}\right)^{\eta}$. Furthermore,
$\eta(\sigma,s,\kappa,k,d)$ is bounded away from zero on any compact subset of
$\\{(\sigma,s,\kappa,k,d):\max(\sigma,k-1)<s\leq k\leq d-1\\}$.

We will be done if we show the claim. Indeed, let $B_{1}=\cup_{r\leq
r_{0}\text{ dyadic }}H_{r}$; then for any dyadic $r\leq r_{0}$ and any
$r$-tube $T$ through some $x\in X$, we either have
$T\in\mathcal{T}^{\prime}_{x,r}$, which means $T\cap G|_{x}\setminus
B_{1}|_{x}=\emptyset$, or the negation of (8.34) holds. In either case, we get

$\nu(T\cap G|_{x}\setminus B_{1}|_{x})\leq
K^{d+1}\cdot\frac{r^{\sigma+\eta}}{r_{0}^{\sigma+\eta-(k-1)}}.$ (8.35)

We have (8.35) for dyadic $r\leq r_{0}$, but it also holds for all $r\leq
r_{0}$ at the cost of introducing a multiplicative factor of
$2^{\sigma+\eta}\leq 2^{k+1}$ on the RHS of (8.35). Thus, $(\mu,\nu)$ have
$(\sigma+\eta,2^{k+1}\cdot K^{d}r_{0}^{-(\sigma+\eta-(k-1))})$-thin tubes on
$G\setminus B_{1}$ down from scale $r_{0}$. Now we move to upper-bounding
$\mu\times\nu(B_{1})$. By (8.30) and (8.34), we have $H_{r}\neq\emptyset$ for
all $r>r_{0}K^{-d/\eta}$, and so if $K\geq K_{0}$ from Claim, then

$\mu\times\nu(B_{1})\leq\sum_{r\leq r_{0}K^{-d/\eta}\text{ dyadic
}}\mu\times\nu(H_{r})\leq\sum_{r\leq r_{0}K^{-d/\eta}\text{ dyadic
}}2\left(\frac{r}{r_{0}}\right)^{\eta}\leq C_{\eta}K^{-d}.$

Let $K_{0}$ be the maximum of the value of $K_{0}$ from Claim, $2C_{\eta}$,
and $2^{k+1}$. Since $d\geq 2$, we get
$\mu\times\nu(B_{1})\leq\frac{1}{2}K^{-1}$ and $(\mu,\nu)$ have
$(\sigma+\eta,K^{d+1}r_{0}^{-(\sigma+\eta-(k-1))})$-thin tubes on $G\setminus
B_{1}$ down from scale $r_{0}$. We can analogously find $B_{2}\subset X\times
Y$ with $\mu\times\nu(B_{2})\leq\frac{1}{2}K^{-1}$ such that $(\nu,\mu)$ have
$(\sigma+\eta,K^{d+1}r_{0}^{-(\sigma+\eta-(k-1))})$-thin tubes on $G\setminus
B_{2}$ down from scale $r_{0}$, and so $B=B_{1}\cup B_{2}$ would be a good
choice. Now we turn to proving the Claim.

Proof of Claim. We will choose
$\eta=\min\\{\frac{1}{2}(6+\frac{15(d-1)}{s-\max(\sigma,k-1)})^{2},\frac{1}{5}\varepsilon^{2}\\}$,
where $\varepsilon$ is obtained from Theorem 1.11. From Remark 1.12 and the
continuity of the function $(s,\sigma,k)\mapsto(s-\max(\sigma,k-1))^{-1}$, we
see that $\eta(\sigma,s,\kappa,k,d)$ is bounded away from zero on any compact
subset of $\\{(\sigma,s,\kappa,k,d):\max(\sigma,k-1)<s\leq k\leq d-1\\}$.

Suppose that Claim is false. Let $\mathbf{X}=\\{x\in
X:\nu(H_{r})\geq\left(\frac{r}{r_{0}}\right)^{\eta}\\}$. Then
$\mu(\mathbf{X})\geq\left(\frac{r}{r_{0}}\right)^{\eta}$.

Recall that for $x\in X$, the fiber $H_{r}|_{x}$ is covered by
$\mathcal{T}^{\prime}_{x,r}$, which is a set of cardinality
$\lesssim\frac{r^{-(\sigma+\eta)}}{r_{0}^{-(\sigma+\eta-(k-1))}}$. Let

$\mathcal{T}_{x}=\\{T\in\mathcal{T}^{\prime}_{x,r}:\nu(T\cap
H_{r}|_{x})\geq\frac{r^{\sigma+3\eta}}{r_{0}^{\sigma+3\eta-(k-1)}}\\},\qquad
Y_{x}=(H_{r}|_{x})\cap\bigcup\mathcal{T}_{x}.$

Then
$\nu(Y_{x})\geq\left(\frac{r}{r_{0}}\right)^{\eta}-\left(\frac{r}{r_{0}}\right)^{2\eta}\geq\left(\frac{r}{r_{0}}\right)^{2\eta}$
for all $x\in\mathbf{X}$. Furthermore, for every $T\in\mathcal{T}_{x}$, we
have

$\frac{r^{\sigma+3\eta}}{r_{0}^{\sigma+3\eta-(k-1)}}\leq\nu(T\cap
Y_{x})\leq\frac{r^{\sigma-\eta}}{r_{0}^{\sigma-\eta-(k-1)}}.$ (8.36)

The upper bound follows from $Y_{x}\subset H_{r}|_{x}\subset G|_{x}$, (8.30),
and $K\leq\left(\frac{r}{r_{0}}\right)^{-\eta}$. In fact, we have in general,

$\nu(T^{(\rho)}\cap
Y_{x})\leq\left(\frac{r}{r_{0}}\right)^{-\eta}\rho^{\eta},\qquad\rho\in[r,1],T\in\mathcal{T}_{x}.$

We also take the time to state the thin plates assumption:

$\nu(W^{(\rho)}\cap
Y_{x})\leq\left(\frac{r}{r_{0}}\right)^{\kappa-\eta}\qquad\rho\in[r,1],W\text{
is }(\rho,k)\text{-plate}.$

Since $\cup\mathcal{T}_{x}$ covers $Y_{x}$, we get by the upper bound in
(8.36),
$|\mathcal{T}_{x}|\gtrsim\frac{r^{-\sigma+\eta}}{r_{0}^{-\sigma+\eta+(k-1)}}\nu(Y_{x})\geq\frac{r^{-\sigma+3\eta}}{r_{0}^{-\sigma+3\eta+(k-1)}}$.
Hence, $\mathcal{T}_{x}$ is a
$(r,\sigma,r_{0}^{-(\sigma-(k-1))}\left(\frac{r}{r_{0}}\right)^{-5\eta})$-set
and $(r,\kappa,r_{0}^{-\kappa}\left(\frac{r}{r_{0}}\right)^{-5\eta},k-1)$-set
for each $x\in\mathbf{X}$.

Let $\gamma=\frac{15\eta}{s-\max(\sigma,k-1)}$. Call a tube
$T\in\mathcal{T}_{x}$ concentrated if there is a ball $B_{T}$ with radius
$\left(\frac{r}{r_{0}}\right)^{\gamma}$ such that

$\nu(T\cap B_{T}\cap Y_{x})\geq\frac{1}{3}\cdot\nu(T\cap Y_{x}).$ (8.37)

Suppose that there is $\mathbf{X}^{\prime}\subset\mathbf{X}$ with
$\mu(\mathbf{X}^{\prime})\geq\mu(\mathbf{X})/2$ such that for each
$x\in\mathbf{X}^{\prime}$, at least half the tubes of $\mathcal{T}_{x}$ are
non-concentrated. Since
$\mu(\mathbf{X}^{\prime})\geq\frac{1}{2}\mu(\mathbf{X})/2\geq\frac{1}{2}\left(\frac{r}{r_{0}}\right)^{2\eta}$
and $\mu$ is Frostman with constant
$K\leq\left(\frac{r}{r_{0}}\right)^{-\eta}$, we can find a
$(r,\sigma,\left(\frac{r}{r_{0}}\right)^{-3\eta})$-set
$P\subset\mathbf{X}^{\prime}$. For each $x\in\mathbf{X}^{\prime}$, the set of
non-concentrated tubes $\mathcal{T}^{\prime}_{x}\subset\mathcal{T}_{x}$ is a
$(r,\sigma,2r_{0}^{-(\sigma-(k-1))}\left(\frac{r}{r_{0}}\right)^{-5\eta})$-set
and
$(r,\kappa,2r_{0}^{-\kappa}\left(\frac{r}{r_{0}}\right)^{-5\eta},k-1)$-set.
Let $\mathcal{T}=\cup_{x\in P}\mathcal{T}^{\prime}_{x}$. By Lemma 2.8, since
$d(X,Y)\geq K^{-1}$, we have that $\mathcal{T}$ is contained in the $O(K)\cdot
r_{0}$-neighborhood of $H$. Now, we apply Theorem 1.11 with
$\overline{r}_{0}:=\min(O(K)\cdot r_{0},1)$. Since
$K\leq\left(\frac{r}{r_{0}}\right)^{-\eta}$ and $\sigma\leq k$, we still have
that for each $x\in\mathbf{X}^{\prime}$, the set of non-concentrated tubes
$\mathcal{T}_{x}^{\prime}$ is a
$(r,\sigma,2\overline{r}_{0}^{-(\sigma-(k-1))}\left(\frac{r}{\overline{r}_{0}}\right)^{-7\eta})$-set
and
$(r,\kappa,2\overline{r}_{0}^{-\kappa}\left(\frac{r}{\overline{r}_{0}}\right)^{-7\eta},k-1)$-set.
At this point, let us remark that implicit constants are dominated by
$\left(\frac{r}{\overline{r}_{0}}\right)^{-\eta}\geq K^{\eta}$ if $K\geq
K_{0}(\eta)$ is chosen large enough.

Then if $\eta\leq\varepsilon^{2}/4$, where $\varepsilon$ is obtained from
Theorem 1.11, then

$|\mathcal{T}|\geq\frac{r^{-2\sigma-2\sqrt{\eta}}}{\overline{r}_{0}^{-2(\sigma-(k-1))-2\sqrt{\eta}}}\geq\frac{r^{-2\sigma-\sqrt{\eta}}}{r_{0}^{-2(\sigma-(k-1))-\sqrt{\eta}}}.$

In other words, we get a gain of
$\left(\frac{r}{r_{0}}\right)^{-\sqrt{\eta}}$, which means a two-ends argument
gives an immediate contradiction. Specifically, by (8.36) and (8.37), we have
for each non-concentrated $T\in\mathcal{T}$, $\nu\times\nu(\\{(x,y):x,y\in
T,d(x,y)\geq\left(\frac{r}{r_{0}}\right)^{\gamma}\\})\geq\frac{2}{3}\nu(T\cap
Y_{x})^{2}\geq\frac{r^{2\sigma+6\eta}}{r_{0}^{2\sigma+6\eta-2(k-1)}}$. Thus,
by Fubini, there exists a pair $(x,y)$ with
$d(x,y)\geq\left(\frac{r}{r_{0}}\right)^{\gamma}$ such that $x,y\in T$ for
$\gtrsim\frac{r^{2\sigma+6\eta}}{r_{0}^{2\sigma+6\eta-2(k-1)}}|\mathcal{T}|\geq\left(\frac{r}{r_{0}}\right)^{-\sqrt{\eta}+6\eta}$
many tubes $T\in\mathcal{T}$. However, since
$d(x,y)\geq\left(\frac{r}{r_{0}}\right)^{\gamma}$, we have that $x,y$ can only
lie in $\lesssim\left(\frac{r}{r_{0}}\right)^{-(d-1)\gamma}$ many essentially
distinct $2r$-tubes. Since $\sqrt{\eta}-6\eta\geq(d-1)\gamma$, we get a
contradiction.

Now we focus on the concentrated case: assume there is a subset
$\mathbf{X}^{\prime}\subset\mathbf{X}$ with
$\mu(\mathbf{X}^{\prime})\geq\mu(\mathbf{X})/2$ such that at least half of the
tubes in $\mathcal{T}_{x}$ are concentrated for all $x\in\mathbf{X}^{\prime}$.
This case is where we use the fact that $\nu$ is a $s$-dimensional measure.
Let $\mathcal{T}_{x}^{\prime}$ denote the concentrated tubes and
$\\{B_{T}:T\in\mathcal{T}_{x}^{\prime}\\}$ denote the corresponding heavy
$\left(\frac{r}{r_{0}}\right)^{\gamma}$-balls. Because the family
$\mathcal{T}_{x}$ has $K$-overlap on $\mathrm{spt}(\nu)$, the set

$H^{\prime}=\\{(x,y):x\in\mathbf{X}^{\prime},y\in T\cap B_{T}\cap Y_{x}\text{
for some }T\in\mathcal{T}_{x}^{\prime}\\}$

has measure

$(\mu\times\nu)(H^{\prime})\gtrsim
K^{-1}\cdot\mu(\mathbf{X}^{\prime})\cdot\inf_{x\in\mathbf{X}^{\prime}}|\mathcal{T}_{x}^{\prime}|\cdot\inf_{x\in\mathbf{X}^{\prime},T\in\mathcal{T}_{x}^{\prime}}\nu(T\cap
B_{T}\cap Y_{x})\\\
\gtrsim\left(\frac{r}{r_{0}}\right)^{2\eta}\cdot\frac{r^{-\sigma+3\eta}}{r_{0}^{-(\sigma-3\eta-(k-1))}}\cdot\frac{r^{\sigma+3\eta}}{r_{0}^{\sigma+3\eta-(k-1)}}=\left(\frac{r}{r_{0}}\right)^{8\eta}.$

Notice that if $(x,y)\in H^{\prime}$, then there is a tube
$T(x,y)\in\mathcal{T}^{r}$ containing $x,y$ such that

$\nu(B(y,2(r/r_{0})^{\gamma})\cap
T(x,y))\gtrsim\frac{r^{\sigma+3\eta}}{r_{0}^{\sigma+3\eta-(k-1)}}.$

Thus, $\nu$ can’t be too concentrated near $y$:

$\nu(B(y,r))\leq K\cdot
r^{s}\leq\frac{1}{2}\nu(B(y,2\left(\frac{r}{r_{0}}\right)^{\gamma})\cap
T(x,y)),$

assuming $4\eta<s-\sigma$ and $k-1<s$. (The relevant inequalities are
$K\leq\left(\frac{r}{r_{0}}\right)^{-\eta}$ and $r^{s-\sigma-3\eta}\leq
r_{0}^{s-\sigma-3\eta}\leq r_{0}^{k-1-\sigma-3\eta}$.)

Therefore, for each $(x,y)\in H^{\prime}$, we can choose a dyadic number
$r\leq\xi(x,y)\leq(r/r_{0})^{\gamma}$ such that

$\nu(A(y,\xi(x,y),2\xi(x,y))\cap
T(x,y))\geq\left(\frac{r}{r_{0}}\right)^{\sigma+4\eta}\left(\frac{\xi(x,y)}{(r/r_{0})^{\gamma}}\right)^{\eta}r_{0}^{k},$

where the annulus $A(y,\xi,2\xi):=B(y,2\xi)\setminus B(y,\xi)$. (One remark:
[19] used dyadic pigeonholing at this step, but we can’t do this because then
we would introduce a $\log r_{0}^{-1}$ factor. Fortunately, we are allowed to
introduce the decaying tail
$\left(\frac{\xi(x,y)}{(r/r_{0})^{\gamma}}\right)^{\eta}$, which is summable
in $\xi(x,y)$.)

Then, recalling that
$(\mu\times\nu)(H^{\prime})\gtrsim\left(\frac{r}{r_{0}}\right)^{7\eta}$, we
can further find $r\leq\xi\leq\left(\frac{r}{r_{0}}\right)^{\gamma}$ such that

$(\mu\times\nu)(H^{\prime\prime})\geq\left(\frac{r}{r_{0}}\right)^{8\eta}\left(\frac{\xi(x,y)}{(r/r_{0})^{\gamma}}\right)^{\eta},\text{
where }H^{\prime\prime}=\\{(x,y)\in H^{\prime}:\xi(x,y)=\xi\\}\subset G.$

By Fubini, we can find $y\in Y$ such that
$\mu(H^{\prime\prime}|^{y})\geq\left(\frac{r}{r_{0}}\right)^{8\eta}\left(\frac{\xi(x,y)}{(r/r_{0})^{\gamma}}\right)^{\eta}$.
Then by construction, $H^{\prime\prime}|^{y}$ can be covered by a collection
of tubes $\mathcal{T}_{y}\subset\mathcal{T}^{r}$ containing $y$ that satisfy

$\nu(A(y,\xi,2\xi)\cap T)\geq\nu(A(y,\xi(x,y),2\xi(x,y))\cap
T(x,y))\geq\left(\frac{r}{r_{0}}\right)^{\sigma+4\eta}\left(\frac{\xi(x,y)}{(r/r_{0})^{\gamma}}\right)^{\eta}r_{0}^{k}.$

Finally, we claim that $\mathcal{T}_{y}$ contains a subset
$\mathcal{T}_{y}^{\prime}$ whose directions are separated by $\geq(r/\xi)$,
such that $|\mathcal{T}_{y}^{\prime}|\gtrsim\mu(H^{\prime\prime}|^{y})\cdot
r^{\eta}\cdot\left(\frac{\xi r_{0}}{r}\right)^{\sigma}r_{0}^{-k}$ if
$\xi>\frac{r}{r_{0}}$ and
$|\mathcal{T}_{y}^{\prime}|\gtrsim\mu(H^{\prime\prime}|^{y})\cdot
r^{\eta}\cdot\left(\frac{\xi r_{0}}{r}\right)^{\sigma}r_{0}^{-k}$ if
$\xi>\frac{r}{r_{0}}$ if $r<\xi<\frac{r}{r_{0}}$. Indeed, if
$\xi>\frac{r}{r_{0}}$, then any $r/\xi$-tube $\mathbf{T}$ containing $y$ has

$\mu(\mathbf{T}\cap H^{\prime\prime}|^{y})\leq\mu(\mathbf{T}\cap A|^{y})\leq
K\cdot\left(\frac{r}{\xi
r_{0}}\right)^{\sigma}r_{0}^{k}\leq\left(\frac{r}{r_{0}}\right)^{\eta}\cdot\left(\frac{r}{\xi
r_{0}}\right)^{\sigma}r_{0}^{k}.$

If $\xi<\frac{r}{r_{0}}$, then any $r/\xi$-tube $\mathbf{T}$ containing $y$
lies in the union of $(\frac{r}{\xi r_{0}})^{k}$ many $r_{0}$-tubes, and so

$\mu(\mathbf{T}\cap H^{\prime\prime}|^{y})\leq\mu(\mathbf{T}\cap A|^{y})\leq
K\cdot\left(\frac{r}{\xi
r_{0}}\right)^{-k}r_{0}^{k}\leq\left(\frac{r}{r_{0}}\right)^{\eta}\cdot\left(\frac{r}{\xi}\right)^{k}.$

Thus, if $\xi>\frac{r}{r_{0}}$, then it takes
$\gtrsim\mu(H^{\prime\prime}|^{y})\cdot r^{\eta}\cdot\left(\frac{\xi
r_{0}}{r}\right)^{\sigma}r_{0}^{-k}$ many $(r/\xi)$-tubes to cover
$H^{\prime\prime}|^{y}$, and perhaps even more to cover $\cup\mathcal{T}_{y}$.
We may now choose $\mathcal{T}_{y}^{\prime}\subset\mathcal{T}_{y}$ to be a
maximal subset with $(r/\xi)$-separated directions to prove the claim for
$\xi>\frac{r}{r_{0}}$. A similar argument holds for $\xi<\frac{r}{r_{0}}$.

Finally, let’s first assume $\xi>\frac{r}{r_{0}}$. Since $\mathcal{T}^{y}$ has
bounded overlap in $\R^{d}\setminus B(y,\xi)$, we obtain

$\left(\frac{r}{r_{0}}\right)^{\sigma+13\eta}\left(\frac{\xi}{(r/r_{0})^{\gamma}}\right)^{2\eta}\cdot\left(\frac{\xi
r_{0}}{r}\right)^{\sigma}r_{0}^{-k}\\\
\lesssim\inf_{T\in\mathcal{T}_{y}^{\prime}}\nu(A(y,\xi,2\xi)\cap
T)\cdot|\mathcal{T}_{y}^{\prime}|\lesssim\nu(B(y,2\xi))\leq C\cdot(2\xi)^{s}.$

We will obtain a contradiction if we show the opposite inequality holds, for
$\gamma=\frac{15\eta}{s-\max(k,\sigma)}$. Since $2\eta+\sigma<s$ and
$\xi\leq(\frac{r}{r_{0}})^{\gamma}$, it suffices to check
$\xi=(\frac{r}{r_{0}})^{\gamma}$.

If $\xi<\frac{r}{r_{0}}$, then we obtain

$\left(\frac{r}{r_{0}}\right)^{\sigma+13\eta}\left(\frac{\xi}{(r/r_{0})^{\gamma}}\right)^{2\eta}\cdot\left(\frac{\xi
r_{0}}{r}\right)^{k}r_{0}^{-k}\\\
\lesssim\inf_{T\in\mathcal{T}_{y}^{\prime}}\nu(A(y,\xi,2\xi)\cap
T)\cdot|\mathcal{T}_{y}^{\prime}|\lesssim\nu(B(y,2\xi))\leq C\cdot(2\xi)^{s}.$

Again, since $2\eta+k<s$, it suffices to check $\xi=\frac{r}{r_{0}}$.

This proves the result. ∎

###### Proof of Theorem 8.1.

By Propositions 7.3 (with $\frac{r_{0}}{K_{2}}$ for $r_{0}$) and 8.2, there
exists a set $B_{0}\subset X\times Y$ with $\mu\times\nu(B_{0})\lesssim
K_{1}^{-1}$ such that $(\mu,\nu)$ and $(\nu,\mu)$ have
$(0,K_{1}^{N}r_{0}^{k-1})$-thin tubes on $B_{0}^{c}$ down from scale $r_{0}$,
and $(\mu,\nu)$ and $(\nu,\mu)$ have
$(\kappa,K_{1}\left(\frac{r_{0}}{K_{2}}\right)^{-\kappa})$-thin $k$-plates on
$(A\cup B_{0})^{c}$. Then iterate Proposition 8.3 applied to a uniform
$\eta(\sigma,s,\kappa,k,d)$. So initially we have
$K=\max(K_{1}^{N}K_{2}^{\kappa},K_{0}(\eta,k))$, and after each iteration, $K$
becomes $K^{d+1}$. After iterating $\lesssim\eta^{-1}$ many times and letting
$B_{1}\subset X\times Y$ be the union of the $B$’s outputted from the
Proposition (so $\mu\times\nu(B_{1})\lesssim K^{-1}\leq K_{1}^{-1}$), we find
that $(\mu,\nu)$ and $(\nu,\mu)$ have
$(\sigma,K^{(d+1)\eta^{-1}}r_{0}^{k-1})$-thin tubes on $(A\cup B_{0}\cup
B_{1})^{c}$. Then we can take $B:=B_{0}\cup B_{1}$ to be our desired set. ∎

###  8.2 Proof of Theorem 1.13, general case

We will prove Theorem 1.13, which we restate here.

######  Theorem 8.5.

Let $k\in\\{1,2,\cdots,d-1\\}$, $k-1<\sigma<s\leq k$, and $\varepsilon>0$.
There exist $N,K_{0}$ depending on $\sigma,s,k$, and $\eta(\varepsilon)>0$
(with $\eta(1)=1$) such that the following holds. Fix $r_{0}\leq 1$, and
$K\geq K_{0}$. Let $\mu,\nu$ be $\sim 1$-separated $s$-dimensional measures
with constant $C_{\mu},C_{\nu}$ supported on $E_{1},E_{2}$, which lie in
$B(0,1)$. Assume that $|\mu|,|\nu|\leq 1$. Let $A$ be the pairs of $(x,y)\in
E_{1}\times E_{2}$ that lie in some $K^{-1}$-concentrated $(r_{0},k)$-plate.
Then there exists a set $B\subset E_{1}\times E_{2}$ with
$\mu\times\nu(B)\lesssim K^{-\eta}$ such that for every $x\in E_{1}$ and
$r$-tube $T$ through $x$, we have

$\nu(T\setminus(A|_{x}\cup
B|_{x}))\lesssim\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)+N\varepsilon}}K^{N}.$

The implicit constant may depend on $C_{\mu},C_{\nu},\sigma,s,k$.

######  Remark 8.6.

Note that in Theorem 8.1, we demand the stronger conclusion
$\mu\times\nu(B)\lesssim K^{-1}$.

The idea is to apply Theorem 8.1 at different scales. As a start, if
$\varepsilon=1$, then we can directly apply Theorem 8.1 with $K_{1}=K_{2}=K$
(and thus we may take $\eta(1)=1$).

We may assume $\varepsilon=\frac{1}{M}$ for some $M$. Let $N$ be the large
constant in Lemma 7.5, and let $\eta_{n}=(N+2)^{n-M}$. For $1\leq n\leq M$,
let $A_{n}$ be the pairs of $(x,y)\in E_{1}\times E_{2}$ that lie in some
$K^{-\eta_{n}}$-concentrated $(r_{0}^{n\varepsilon},k)$-plate. We remark that
$A_{M}=A$.

######  Lemma 8.7.

Fix $n\geq 1$. There exists a set $B_{n}\subset A_{n}$ with
$\mu\times\nu(B_{n})\lesssim K^{-\eta_{n}}$ such that for every $x\in E_{1}$
and $r$-tube through $x$ that intersects $A_{n}|_{x}$, we have

$\nu(T\setminus(A_{n+1}|_{x}\cup
B|_{x}))\lesssim\frac{r^{\sigma}}{r_{0}^{n\varepsilon(\sigma-(k-1))+N\varepsilon}}K^{N}.$

###### Proof.

By Lemma 2.8, there exists an absolute constant $C$ such that every $r$-tube
through some $(x,y)\in A_{n}$ lies in some $K^{-\eta_{n}}$-concentrated
$(CK^{-n},k)$-plate. We can find a collection $\mathcal{H}$ of essentially
distinct $K^{-\eta_{n}}$-concentrated $(2CK^{-n},k)$-plates such that each
$K^{-\eta_{n}}$-concentrated $(CK^{-n},k)$-plate is contained within some
element of $\mathcal{H}$. By Lemma 7.5, $|\mathcal{H}|\lesssim
K^{-N\eta_{n}}$. By construction, every $r$-tube through some $(x,y)\in A_{n}$
is contained in some member of $\mathcal{H}$. Apply Theorem 8.1 to each
$H\in\mathcal{H}$ with measures $\mu|_{H},\nu|_{H}$ and $K_{1}\rightarrow
K^{-\eta_{n+1}}$, $K_{2}\rightarrow 2Cr_{0}^{-\varepsilon}$, and
$r_{0}\rightarrow r_{0}^{n\varepsilon}$ to obtain a set $B_{H}$ with
$\mu\times\nu(B_{H})\lesssim K^{-\eta_{n+1}}$. Let
$B_{n}=\cup_{H\in\mathcal{H}}B_{H}$, and then $\mu\times\nu(B_{n})\leq
K^{N\eta_{n}}\cdot K^{-\eta_{n+1}}<K^{-\eta_{n}}$ since
$(N+1)\eta_{n}<\eta_{n+1}$. ∎

###### Proof of Theorem 8.5.

Let $B=\cup_{n=1}^{M}B_{n}$; then $\mu\times\nu(B)\leq K^{-\eta_{0}}$. Fix an
$r$-tube $T$ and $x\in E_{1}$. Let $n\leq M-1$ be the largest number such that
$T$ passes through points in $A_{n}|_{x}$. Then by Lemma 8.7, we have
$\nu(T\setminus(A_{n+1}|_{x}\cup
B|_{x}))\lesssim\frac{r^{\sigma}}{K^{-n(\sigma-(k-1)}}K^{N}$. If $m<M-1$, then
$T\cap A_{n+1}|_{x}=\emptyset$. In any case, we have
$\nu(T\setminus(A|_{x}\cup
B|_{x}))\lesssim\frac{r^{\sigma}}{r_{0}^{\sigma-(k-1)+N\varepsilon}}K^{N}$,
completing the proof of Theorem 8.5.

∎

##  9 Corollaries of Radial Projection Estimates

We prove a variant of Corollary 1.1.

######  Proposition 9.1.

Fix $s\in(k-1,k]$ and $\eta>0$. Let $\mu,\nu\in\mathcal{P}(\R^{d})$ be
measures with $\mathcal{E}_{s}(\mu),\mathcal{E}_{s}(\nu)<\infty$ and $\sim
1$-separated supports. Suppose that $\mu(H)=\nu(H)=0$ for each $k$-plane
$H\in\mathbb{A}(\R^{d},k)$. Then for $\mu$-almost all $x$, for all sets $Y$ of
positive $\nu$-measure,

$\dim_{H}(\pi_{x}Y)\geq s-\eta.$

###### Proof.

The proof is standard and follows [27, Proof of Proposition 6.9]. By Lemma
2.30, by passing to subsets of nearly full measure and replacing $s$ by an
arbitrary $s^{\prime}<s$, we may assume that $\mu(B_{r}),\nu(B_{r})\lesssim
r^{s}$ for all $r\in(0,1]$.

Fix $\varepsilon>0$. By a compactness argument, there exists $r_{0}>0$ such
that $\mu(H),\nu(H)<\varepsilon$ for all $(r_{0},k)$-plates $H$. In Theorem
8.1, we know that for $\varepsilon>0$ sufficiently small, the set
$A=\emptyset$. Thus, there exists $B\subset X\times Y$ with
$\mu\times\nu(B)\lesssim\varepsilon$ such that for every $x\in X$ and $r$-tube
through $x$, we have

$\nu(T\setminus B|_{x})\lesssim_{\eta,\varepsilon,s}r^{s-\eta}.$

Thus, there is a set $X$ with $\mu(X)>1-O(\varepsilon)$ such that if $x\in X$,
then

$\dim_{H}(\pi_{x}Y)\geq s-\eta\text{ for all }Y\text{ with }\nu(Y)\geq
O(\varepsilon).$

Taking $\varepsilon\to 0$ completes the proof. ∎

Using this, we prove Corollary 1.2.

######  Corollary 9.2.

Let $s\in(d-2,d]$, then there exists $\varepsilon(s,d)>0$ such that the
following holds. Let $\mu,\nu$ be Borel probability measures on $\R^{d}$ with
disjoint supports that satisfy
$\mathcal{E}_{s}(\mu),\mathcal{E}_{s}(\nu)<\infty$ and
$\dim_{H}(\mathrm{spt}(\nu))<s+\varepsilon(s,d)$. Further, assume that
$\mu,\nu$ don’t simultaneously give full measure to any affine $(d-1)$-plane
$H\subset\R^{d}$. Then there exist restrictions of $\mu,\nu$ to subsets of
positive measure (which we keep denoting $\mu,\nu$) such that the following
holds. For almost every affine 2-plane $W\subset\R^{d}$ (with respect to the
natural measure on the affine Grassmanian), if the sliced measures $\mu_{W}$,
$\nu_{W}$ on $W$ is non-trivial, then they don’t simultaneously give full
measure to any line. In other words,

$(\gamma_{d,2}\times\mu)\\{(V,x):\mu_{V,x}(\ell)\nu_{V,x}(\ell)=|\mu_{V,x}||\nu_{V,x}|>0\text{
for some }\ell\in\mathbb{A}(V+x,1)\\}=0$

where we parametrize affine 2-planes as $V+x$, for $x\in\R^{d}$ and $V$ in the
Grassmannian $\mathrm{Gr}(d,2)$ with the rotationally invariant Haar measure
$\gamma_{d,2}$.

###### Proof.

First, if $\mu(H)>0$ for some affine $(d-1)$-plane $H$, then $\nu(H^{c})>0$
where $H^{c}$ denotes the complement of $H$ in $\R^{d}$. By restricting $\mu$
to $H$ and $\nu$ to $H^{c}$ (and calling the results $\mu,\nu$), we see that
the sliced measures $\mu_{W}$ and $\nu_{W}$ can’t give full mass to any line
$\ell$ for any affine $(d-1)$-plane $W$, for the simple reason that
$\mu_{W}(\ell)>0$ forces $\ell\subset H$, and $\nu_{W}(\ell)>0$ forces
$\ell\subset H^{c}$. Likewise, we are done if $\nu(H)>0$ for some affine
$(d-1)$-plane $H\subset\R^{d}$. Thus, assume $\mu(H)=\nu(H)=0$ for all affine
$(d-1)$-planes $H$.

With this assumption, the remainder of the proof is nearly identical to the
proof of Proposition 6.8 in [27], except using Proposition 9.1 instead of [27,
Proposition 6.9]. One can take $\varepsilon(s,d)$ to be arbitrarily close to
$s-(d-2)$. ∎

Finally, we can deduce Theorem 1.1 from either Proposition 9.1 or Proposition
9.2, see [19, Section 4] for details. The only case not yet considered in this
paper is when either $\mu,\nu$ gives positive mass to a $k$-plane. But this
special case was considered in [19, Section 4] (briefly, if $X$ gives positive
mass to some $k$-plane, then radial projections become orthogonal projections
and then we apply Kaufman’s projection theorem; if $Y$ gives positive mass to
some $k$-plane $H$, then for $x\notin H$, we have
$\dim_{H}(\pi_{x}(Y))=\dim_{H}(Y)$.)

##  Appendix A Proof of Balog-Szemerédi-Gowers

By a standard covering argument (e.g. see Section 3 of [12]), Theorem 2.27
follows from the case $\delta=0$, which we prove below.

######  Theorem A.1 (refined Theorem 4.1 of [30]).

Let $K\geq 1$ be a parameter. Let $A,B$ be finite subsets of $\R^{n}$, and let
$P\subset A\times B$ satisfy $|P|\geq K^{-1}|A||B|$. Suppose that
$|A\overset{P}{+}B|\leq K(|A||B|)^{1/2}$, where
$A\overset{P}{+}B=\\{a+b:(a,b)\in P\\}$. Then one can find subsets
$A^{\prime}\subset A,B^{\prime}\subset B$ with
$|A^{\prime}|\geq\frac{1}{16K^{2}}|A|,|B^{\prime}|\geq\frac{1}{16K^{2}}|B|$
such that $|A^{\prime}+B^{\prime}|\leq 2^{12}K^{8}(|A||B|)^{1/2}$ and
$|P\cap(A^{\prime}\times B^{\prime})|\geq\frac{|A||B|}{16K^{2}}$.

###### Proof.

We follow the exposition in [24].

Claim. There exist subsets $A^{\prime}\subset A,B^{\prime}\subset B$ with
$|P\cap(A^{\prime}\times B^{\prime})|\geq\frac{|A||B|}{16K^{2}}$, such that
for each $a\in A^{\prime},b\in B^{\prime}$, there are
$\geq\frac{|A||B|}{2^{12}K^{5}}$ many pairs $(a^{\prime},b^{\prime})\in
A\times B$ such that $(a,b^{\prime})$, $(a^{\prime},b^{\prime})$, and
$(a^{\prime},b)\in P$.

Assuming the claim, we will see how the theorem follows. First, we get
$|A^{\prime}||B^{\prime}|\geq|P\cap(A^{\prime}\times
B^{\prime})|\geq\frac{|A||B|}{16K^{2}}$. Since $|A^{\prime}|\leq|A|$ and
$|B^{\prime}|\leq|B|$, we get $|A|\geq\frac{|A|}{16K^{2}}$ and
$|B|\geq\frac{|B|}{16K^{2}}$.

Next, for $a\in A^{\prime},b\in B^{\prime}$, we have

$a+b=(a+b^{\prime})-(a^{\prime}+b^{\prime})+(a^{\prime}+b).$

Thus, there are $\geq|A||B|2^{-12}K^{-5}$ many solutions to $a+b=x-y+z$ with
$x,y,z\in A\overset{P}{+}B$. Since $|A\overset{P}{+}B|\leq K(|A||B|)^{1/2}$,
we get
$|A^{\prime}+B^{\prime}|\lesssim\frac{K^{3}(|A||B|)^{3/2}}{|A||B|2^{-12}K^{-5}}=2^{12}K^{8}|A|^{1/2}|B|^{1/2}$.

Now, we prove the claim. For convenience, we can prune $P$ to satisfy
$|P|=K^{-1}|A||B|$ (this is not necessary but will make the proof look nicer).
Treat $(A\cup B,P)$ as a bipartite graph with an edge between $a\in A$ and
$b\in B$ if $(a,b)\in P$. Then we want to find $A^{\prime},B^{\prime}$ such
that there are many paths of length $3$ between any $a\in A^{\prime},b\in
B^{\prime}$.

The average degree of a vertex in $A$ is $K^{-1}|B|$. Thus, if we delete the
vertices in $A$ with degree $\leq\frac{1}{2}K^{-1}|B|$, then at least
$\frac{1}{2K}|A||B|$ many edges remain. Let $E$ be the set of edges. For $v\in
A\cup B$, let $N(v)$ be the set of neighbors of $v$.

Now pick a vertex $b\in B$. On average, it has
$\frac{|E|}{|B|}\geq\frac{1}{2K}|A|$ many neighbors.

Now, we say $(a,a^{\prime})\in A^{2}$ is bad if $|N(a)\cap
N(a^{\prime})|<\frac{1}{128K^{3}}|B|$. For $v\in B$, let $\mathrm{Bad}_{v}$ be
the set of bad pairs in $N(v)^{2}$. There are $\binom{|A|}{2}$ many pairs in
$A$, so (expectation is taken over uniformly chosen $v\in B$)

$\mathbb{E}[|\mathrm{Bad}_{v}|]<\binom{|A|}{2}\cdot\frac{1}{128K^{3}}<\frac{|A|^{2}}{256K^{3}}.$

If $A_{bad,v}$ is the set of vertices of $A$ that lie in at least
$\frac{|A|}{32K^{2}}$ many pairs of $B_{v}$, then

$\mathbb{E}[|A_{bad,v}|]\leq\frac{2\mathbb{E}[|B_{v}|]}{|A|/(32K^{2})}<\frac{|A|}{4K}.$

Finally, let $A_{v}=N(v)\setminus A_{bad,v}$. Then by linearity of
expectation,

$\mathbb{E}[|A_{v}|]=\mathbb{E}[|N(v)|]-\mathbb{E}[|A_{bad,v}|]>\frac{|A|}{2K}-\frac{|A|}{4K}=\frac{|A|}{4K}.$

Thus, there exists $v\in B$ such that $|A_{v}|>\frac{|A|}{4K}$. Then, let
$A^{\prime}=A_{v}$ and

$B^{\prime}=\\{w\in B:|N(w)\cap A^{\prime}|\geq\frac{|A|}{16K^{2}}.$

Let $E(X,Y)$ be the number of edges between $X$ and $Y$. We first check that
$E(A^{\prime},B^{\prime})\geq\frac{|A||B|}{16K^{2}}$. Indeed, since every
vertex of $A$ has degree $\geq\frac{|B|}{2K}$, we have

$|E(A^{\prime},B)|\geq\frac{|A^{\prime}||B|}{2K}\geq\frac{|A||B|^{2}}{8K^{2}}.$

On the other hand, every vertex in $B\setminus B^{\prime}$ corresponds to
fewer than $\frac{|A|}{16K^{2}}$ many edges of $A^{\prime}$, so
$|E(A^{\prime},B\setminus B^{\prime})|\leq\frac{|A||B|^{2}}{16K^{2}}$. Hence,
$|E(A^{\prime},B^{\prime})|\geq\frac{|A||B|^{2}}{16K^{2}}$.

Finally, for any $v\in A^{\prime}$, $w\in B^{\prime}$, we know that $w$ has at
least $\frac{|A|}{16K^{2}}$ many neighbors in $A^{\prime}$, and fewer than
$\frac{|A|}{32K^{2}}$ of those form a bad pair with $w$. For the remaining
$\geq\frac{|A|}{32K^{2}}$ vertices $v^{\prime}$ that do not form a bad pair
with $w$, there are $\geq\frac{|B|}{128K^{3}}$ many vertices $w^{\prime}\in B$
that are common neighbors of $v,v^{\prime}$. Thus, we get at least
$\frac{|A|}{32K}\cdot\frac{|B|}{128K^{3}}=\frac{|A||B|}{2^{12}K^{5}}$ many
paths $(v,w^{\prime},v^{\prime},w)$ between $v$ and $w$. ∎

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