Non-homogeneous Square Functions on General Sets: Suppression and Big Pieces Methods

Abstract

We aim to showcase the wide applicability and power of the big pieces and suppression methods in the theory of local Tb theorems. The setting is new: we consider conical square functions with cones $\big \{ x \in \mathbb {R}^n \setminus E: |x-y| < 2 {\text {dist}}(x,E)\big \}$, $y \in E$, defined on general closed subsets $E \subset \mathbb {R}^n$ supporting a non-homogeneous measure $\mu$. We obtain boundedness criteria in this generality in terms of weak type testing of measures on regular balls $B \subset E$, which are doubling and of small boundary. Due to the general set E we use metric space methods. Therefore, we also demonstrate the recent techniques from the metric space point of view, and show that they yield the most general known local Tb theorems even with assumptions formulated using balls rather than the abstract dyadic metric cubes.

Appendices

Appendix 1: A Sketch of the Proof of Lemma 2.6

Here we sketch the proof of Lemma 2.6 following the arguments in [2], Theorem 2.11. The constants M and L in the statement of Lemma 7.1 are related to properties of E as a geometrically doubling space and are used in the construction of random dyadic cubes.

Lemma 7.1

There exist two constants $C=C(M,L)>0$ and $\eta \in (0,1]$ so that the following holds. Fix some big enough (depending on $\gamma$) goodness parameter r. Suppose $B \subset E$ is a ball in E and construct the dyadic lattices $\mathcal {D}(\omega ), \omega \in \Omega ,$ in E using the center of B as the fixed reference dyadic point, see Sect. 2. For some fixed $\omega _0 \in \Omega$ write $\mathcal {D}_0=\mathcal {D}(\omega _0)$ and recall the lattices $\mathcal {D}_B(\omega ) \subset \mathcal {D}(\omega )$.

Assume $k_0 \in \mathbb {Z}$ such that $\ell (Q_B(\omega ))=\delta ^{k_0}$ for some, and hence for every, $\omega \in \Omega$. Let $k_1 \in \mathbb {Z}$ be any number such that $k_1 \ge k_0+r$. With the fixed $\omega _0$ define the probability space

$$\begin{aligned} \Omega _{k_0}^{k_1} := \{ \omega \in \Omega :\omega (m)= \omega _0(m) \hbox { if } m<k_0 \hbox { or } m > k_1\} \end{aligned}$$

equipped with the natural probability measure such that the coordinates $m \mapsto \omega (m), k_0 \le m \le k_1,$ are independent and uniformly distributed over $\{0, \dots , L\} \times \{1, \dots , M\}$.

Then, for every cube $R \in \mathcal {D}_0$ with $\ell (R) \ge \delta ^{k_1}$ it holds that

$$\begin{aligned} \mathbb {P}\big (\{ \omega \in \Omega ^{k_1}_{k_0} : R \text { is } \mathcal {D}_B(\omega ) \text {-bad}\}\big ) \le C \delta ^{\gamma r \eta }. \end{aligned}$$

(7.1)

Before the proof we define for $\varepsilon >0$ the $\varepsilon$-boundary $\partial _\varepsilon Q$ of a cube $Q \in \mathcal {D}(\omega )$ by

$$\begin{aligned} \partial _\varepsilon Q:=\{ y \in Q :d(y, E \setminus Q) <\varepsilon \ell (Q)\}. \end{aligned}$$

Proof of Lemma 7.1

Let $R \in \mathcal {D}_{0}$ be such that $\ell (R)= \delta ^m$, where $k_0 +r \le m \le k_1$ (if $m < k_0 + r$, then R is automatically good by definition). Fix some $\omega \in \Omega _{k_0}^{k_1}$ for the moment. First we show that if R is $\mathcal {D}_B(\omega )$-bad, then there exists $l \in \mathbb {Z}, k_0 \le l \le m-r$, and $Q \in \mathcal {D}_l(\omega )$ so that $c_R \in \partial _{7\ell (R)^\gamma /\ell (Q)^\gamma }Q$. Indeed, let $k_0 \le l \le m-r$ and suppose $Q \in \mathcal {D}_l(\omega )$ such that $c_R \in Q$. If $c_R \not \in \partial _{7\ell (R)^\gamma /\ell (Q)^\gamma }Q$, then because $R \subset B(c_R, 6\ell (R))$ by (2.2), it is seen that

$$\begin{aligned} d(R,E\setminus Q) \ge \ell (R)^\gamma \ell (Q)^{1-\gamma }. \end{aligned}$$

In particular, we have

$$\begin{aligned} \max \big (d(R,Q'), d(R,E\setminus Q')\big ) \ge \ell (R)^\gamma \ell (Q')^{1-\gamma } \end{aligned}$$

for all $Q' \in \mathcal {D}_B(\omega )$ with $\ell (Q')=\delta ^l$. If this happens for all l such that $k_0 \le l \le m-r$, then the cube R is $\mathcal {D}_B(\omega )$-good. We have shown that

$$\begin{aligned} \mathbb {P}\big (\{ \omega \in \Omega ^{k_1}_{k_0} : R \text { is } \mathcal {D}_B(\omega ) \text {-bad}\}\big ) \le \sum _{l=k_0}^{m-r} \mathbb {P}\big (\{ \omega \in \Omega ^{k_1}_{k_0} : c_R \in \bigcup _{Q \in \mathcal {D}_l(\omega )} \partial _{7\ell (R)^\gamma /\ell (Q)^\gamma }Q\}\big ). \end{aligned}$$

(7.2)

Next we fix some $l \in \{k_0, \dots , m-r\}$ and estimate the corresponding term in the right-hand side of (7.2). Following the argument in [2], if $\omega (p)$ has a certain value for some $p \ge l$ such that $\delta ^p \ge 49 \delta ^{m\gamma } \delta ^{l(1-\gamma )}$, then

$$\begin{aligned} c_R \not \in \bigcup _{ Q \in \mathcal {D}_l(\omega )} \partial _{7\ell (R)^\gamma /\ell (Q)^\gamma } Q. \end{aligned}$$

(7.3)

This part of the argument is short, but to state it we would need to introduce more of the construction of the random dyadic cubes. We refer the reader to [2].

The requirement $\delta ^l \ge \delta ^p \ge 49 \delta ^{m\gamma } \delta ^{l(1-\gamma )}$ amounts to

$$\begin{aligned} l \le p \le \log _\delta 49 +m\gamma +l(1-\gamma ). \end{aligned}$$

(7.4)

Note that in particular every such p satisfies $k_0 \le p \le k_1$. For (7.4) to make sense we demand r to be so big that $r\gamma >2$, say, whence

$$\begin{aligned} \log _\delta 49 +m\gamma +l(1-\gamma )-l\ge (m-l)\gamma -1>1. \end{aligned}$$

(7.5)

Denote by $\lfloor \log _\delta 49 +(m-l)\gamma \rfloor$ the smallest integer less than or equal to $\log _\delta 49 +(m-l)\gamma$.

For every $p \in \mathbb {Z}$, the variable $\omega (p)$ has the probability $\tau := \frac{1}{M(L+1)}$ of getting a given value. Hence, by (7.3), we have

$$\begin{aligned} \mathbb {P}\big (\{ \omega \in \Omega ^{k_1}_{k_0} : c_R \in \bigcup _{Q \in \mathcal {D}_l(\omega )} \partial _{7\delta ^{(m-l)\gamma }} Q\}\big )\le & {} (1-\tau )^{\lfloor \log _\delta 49 +(m-l)\gamma \rfloor } \\\le & {} C (1-\tau )^{(m-l)\gamma }. \end{aligned}$$

Combining this with (7.2) we get

$$\begin{aligned} \mathbb {P}\big (\{ \omega \in \Omega ^{k_1}_{k_0} : R \text { is } \mathcal {D}_B(\omega ) \text {-bad}\}\big ) \le C\sum _{l=k_0}^{m-r} (1-\tau )^{ (m-l)\gamma } \sim (1-\tau )^{r\gamma }, \end{aligned}$$

and we can rewrite the bound as

$$\begin{aligned} (1-\tau )^{r\gamma }= \delta ^{r\gamma \log _\delta (1-\tau )} . \end{aligned}$$

This gives the required conclusion because $\log _\delta (1-\tau ) \in (0,1)$. $\square$

Appendix 2: $L^2$ Boundedness Implies Weak (1, 1) Boundedness

We verify here that if $\mu$ is a measure of order m in E and $\mathcal {C}_\mu$ is bounded in $L^2(\mu )$, then

$$\begin{aligned} \mathcal {C}:\mathcal {M}(E) \rightarrow L^{1, \infty }(\mu ) \end{aligned}$$

boundedly. The proof of this follows the standard steps using the Calderón–Zygmund decomposition, but we check the details because of our unusual set-up.

First we record a few lemmas, and begin with the non-homogeneous Calderón–Zygmund decomposition whose proof can be found, for example, in [21]. We say that a collection $\{B_i\}_i$ of balls in $\mathbb {R}^n$ has bounded overlap if there exists a constant C such that

$$\begin{aligned} \sum _i 1_{B_i}(x) \le C \end{aligned}$$

for every $x \in \mathbb {R}^n$.

Lemma 7.2

Let $\mu$ be a Radon measure in $\mathbb {R}^n$ and suppose $\nu$ is a complex measure in $\mathbb {R}^n$ with compact support. Let $\lambda > 2^{n+1}\frac{|\nu |(\mathbb {R}^n)}{\mu (\mathbb {R}^n)}$.

There exists a countable family $\{B_i\}_{i \in \mathcal {I}}$ of closed balls with bounded overlap and with centers in ${\text {spt}}\nu$, and a function $f \in L^1(\mu )$ with $\Vert f \Vert _{L^\infty (\mu )} \le \lambda$ so that

$$\begin{aligned}&|\nu |(B_i) > 2^{-n-1} \lambda \mu (2B_i) \ \text { for all }i, \end{aligned}$$

(7.6)

$$\begin{aligned}&|\nu |(\eta B_i) \le 2^{-n-1} \lambda \mu (2\eta B_i) \ \text { for all }i \text { and }\eta >2, \end{aligned}$$

(7.7)

$$\begin{aligned}&1_{\mathbb {R}^n \setminus \bigcup _i B_i}\nu = f \mu . \end{aligned}$$

(7.8)

For every $i \in \mathcal {I}$ suppose $R_i$ is $(6,6^{m+1})$-doubling ball (with respect to $\mu$) concentric with $B_i$ and $r(R_i) > 4r(B_i)$. Define the functions $w_i:= \frac{1_{B_i}}{\sum _{j} 1_{B_j}}$. Then there exists a family $\{\varphi _i\}_{i \in \mathcal {I}}$ of functions such that each function is of the form $\varphi _i=\alpha _i h_i$, where $\alpha _i \in \mathbb {C}$ and $h_i$ is a non-negative function, and this family satisfies the properties

$$\begin{aligned}&{\text {spt}}\varphi _i \subset R_i, \end{aligned}$$

(7.9)

$$\begin{aligned}&\int \varphi _i \,\mathrm {d}\mu = \int w_i \,\mathrm {d}\nu , \end{aligned}$$

(7.10)

$$\begin{aligned}&\sum _{i \in \mathcal {I}} |\varphi _i| \le C_1 \lambda , \end{aligned}$$

(7.11)

$$\begin{aligned}&\Vert \varphi _i\Vert _{L^{\infty }(\mu )} \mu (R_i) \le c |\nu |(B_i). \end{aligned}$$

(7.12)

Here $C_1$ is a constant depending on m and n, and c is an absolute constant.

The next two simple lemmas can also be found for example in [21].

Lemma 7.3

Suppose $\mu$ is a measure of order m in $\mathbb {R}^n$. Let $b > a^m$. If B is a ball in $\mathbb {R}^n$, then there exists $s>1$ such that the ball sB is (a, b)-doubling.

Lemma 7.4

Suppose $\mu$ is a Radon measure in $\mathbb {R}^n$. Let $b> a^m, a>1$. Suppose $B_1$ and $B_2$ are two balls in $\mathbb {R}^n$ with center x and $B_1 \subset B_2$. Assume none of the balls $a^kB_1$ is (a, b)-doubling for those $k \in \mathbb {Z}$ such that $B_1 \subsetneq a^k B_1 \subset B_2$. Then

$$\begin{aligned} \int _{B_2 \setminus B_1} \frac{1}{|y-x|^m} \,\mathrm {d}\mu (y) \lesssim _{a, b,m} \frac{\mu (B_2)}{r(B_2)^m}. \end{aligned}$$

Theorem 7.5

Let $\mu$ be a measure of order m in E. Suppose that $\mathcal {C}_\mu$ is bounded in $L^2(\mu )$. Then

$$\begin{aligned} \mathcal {C}:\mathcal {M}(E) \rightarrow L^{1,\infty }(\mu ) \end{aligned}$$

is bounded with a constant depending on the kernel parameters, the dimension n, and the $L^2(\mu )$ norm of $\mathcal {C}_\mu$.

Proof

Let $\nu \in \mathcal {M}(E)$ and $\lambda >0$. We want to show that

$$\begin{aligned} \mu \big (\{ y \in E :\mathcal {C}\nu (y) > \lambda \}\big ) \lesssim \frac{1}{\lambda } |\nu |(E). \end{aligned}$$

We may assume that $\lambda >2^{n+1} \frac{|\nu |(E)}{\mu (E)}$, since otherwise we have nothing to prove.

Suppose first that $\nu$ is compactly supported. We apply Lemma 7.2 with $\lambda$ to the measures $\nu$ and $\mu$ to get a function f and an almost disjoint collection $\{B_i\}$ of closed balls with centers in E such that (7.6), (7.7), and (7.8) hold. For each i, let $R_i \supsetneq 4B_i$ be the smallest $(8,8^{m+1})$-doubling closed ball concentric with $B_i$, which exists by Lemma 7.3. We apply Lemma 7.2 with the balls $R_i$ to get a collection $\{\varphi _i\}$ of functions such that (7.9), (7.10), (7.11), and (7.12) hold.

Using the balls $B_i$ and the functions $\varphi _i$ we can write the measure $\nu$ as

$$\begin{aligned} \begin{aligned} \nu&=1_{\mathbb {R}^n \setminus \bigcup _iB_i}\nu + 1_{\bigcup _i B_i} \nu = f \mu +\sum _i w_i \nu \\&= \left( f+\sum _i \varphi _i \right) \mu + \sum _i (w_i\nu - \varphi _i\mu ). \end{aligned} \end{aligned}$$

Write $g= f+\sum _i\varphi _i$ and $b= \sum _i b_i=\sum _i( w_i\nu -\varphi _i\mu )$. Then we have

$$\begin{aligned} \mu \big (\{y \in E :\mathcal {C}\nu (y)> \lambda \}\big ) \le \mu \left( \left\{ y \in E :\mathcal {C}_\mu g> \frac{\lambda }{2}\right\} \right) + \mu \left( \left\{ y \in E :\mathcal {C}b > \frac{\lambda }{2}\right\} \right) . \end{aligned}$$

(7.13)

The $L^2(\mu )$-boundedness of $\mathcal {C}_\mu$ and the fact that $\Vert f +\sum _i \varphi _i\Vert _{L^{\infty }(\mu )} \lesssim \lambda$ by (7.11) give

$$\begin{aligned} \mu \big (\{y \in E :\mathcal {C}_\mu g > \frac{\lambda }{2}\}\big ) \lesssim \frac{1}{\lambda ^2} \Vert f +\sum _i \varphi _i \Vert _{L^2(\mu )}^2 \lesssim \frac{1}{\lambda } \Vert f +\sum _i \varphi _i \Vert _{L^1(\mu )}. \end{aligned}$$

Using Eqs. (7.8), (7.9), (7.12), and the bounded overlapping property of $\{B_i\}$ we get

$$\begin{aligned} \begin{aligned} \Vert f +\sum _i \varphi _i \Vert _{L^1(\mu )}&\le \int |f| \,\mathrm {d}\mu + \sum _i \int |\varphi _i| \,\mathrm {d}\mu \le |\nu |(\mathbb {R}^n) +\sum _i \Vert \varphi _i\Vert _{L^\infty (\mu )} \mu (R_i) \\&\lesssim |\nu |(\mathbb {R}^n)+ \sum _i |\nu |(B_i) \\&\lesssim |\nu |(\mathbb {R}^n). \end{aligned} \end{aligned}$$

Next, we consider the second term on the right-hand side of (7.13). Note that

$$\begin{aligned} \mu \big ( \bigcup _i 2B_i\big ) \le \sum _i \mu (2B_i) \le \frac{2^{n+1}}{\lambda }\sum _i |\nu |(B_i) \lesssim \frac{1}{\lambda } |\nu |(\mathbb {R}^n). \end{aligned}$$

Hence we need to show that

$$\begin{aligned} \mu \big (\{y \in E \setminus \bigcup _i 2B_i :\mathcal {C}b > \frac{\lambda }{2}\}\big ) \lesssim \frac{1}{\lambda } |\nu |(\mathbb {R}^n). \end{aligned}$$

First estimate as

$$\begin{aligned} \begin{aligned} \mu \big (\{y \in E \setminus \bigcup _i 2B_i :\mathcal {C}b > \frac{\lambda }{2}\}\big )&\le \frac{2}{\lambda } \int _{E \setminus \bigcup _i 2B_i} \mathcal {C}b \,\mathrm {d}\mu \\&\le \frac{2}{\lambda } \sum _i \int _{E \setminus 2B_i} \mathcal {C}b_i \,\mathrm {d}\mu . \end{aligned} \end{aligned}$$

We will prove that

$$\begin{aligned} \int _{E \setminus 2B_i} \mathcal {C}b_i \,\mathrm {d}\mu \lesssim |\nu | (B_i) \end{aligned}$$

(7.14)

holds for every i, which then concludes the proof because $\sum _i |\nu | (B_i) \lesssim |\nu |( \mathbb {R}^n)$.

Fix some i, and recall the ball $R_i$ related to the ball $B_i$. We begin the proof of (7.14) by writing

$$\begin{aligned} \int _{E \setminus 2B_i} \mathcal {C}b_i \,\mathrm {d}\mu = \int _{E \setminus 8R_i} \mathcal {C}b_i \,\mathrm {d}\mu + \int _{8R_i \setminus 2B_i} \mathcal {C}b_i \,\mathrm {d}\mu =: I + II. \end{aligned}$$

We consider the term I first. Let $y \in E \setminus 8R_i$ and $x \in \Gamma (y)$. Let $c_{R_i}$ be the center of $R_i$, whence it follows that $|x-c_{R_i}| \ge 2r(R_i)$. Since ${\text {spt}}b_i \subset R_i$ and $b_i(R_i)=0$, we may apply the y-continuity (1.2) of the square function kernel to get

$$\begin{aligned} \begin{aligned} |Tb_i(x)| \lesssim \frac{r(R_i)^\beta }{|x-c_{R_i}|^{m+\alpha +\beta }} |b_i|(R_i) \sim \frac{r(R_i)^\beta }{(|x-y|+ |y-c_{R_i}|)^{m+\alpha +\beta } } |b_i| (R_i). \end{aligned} \end{aligned}$$

Also

$$\begin{aligned} \begin{aligned} |b_i | (R_i)&= |w_i \nu - \varphi _i\mu | (R_i) \le \int _{R_i} w_i \,\mathrm {d}|\nu | + \int _{R_i} |\varphi _i| \,\mathrm {d}\mu \\&\le |\nu |(B_i) +\Vert \varphi _i \Vert _{L^\infty (\mu )} \mu (R_i) \lesssim |\nu |(B_i). \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} \mathcal {C}b_i(y)^2&\lesssim \int _{\Gamma (y)} \frac{r(R_i)^{2\beta } d(x,E)^{2\alpha }}{(|x-y|+|y-c_{R_i}|)^{2(m+\alpha +\beta )}} \,\mathrm {d}\sigma (x) |b_i |(R_i)^2 \\&\lesssim \frac{r(R_i)^{2\beta }}{|y-c_{R_i}|^{2(m+\beta )}} |\nu |(B_i)^2, \end{aligned} \end{aligned}$$

where we applied Lemma 2.4 in the second step. Because $\mu$ is of order m we get

$$\begin{aligned} \begin{aligned} I&=\int _{E \setminus 8R_i} \mathcal {C}b_i \,\mathrm {d}\mu \lesssim \int _{E \setminus 8R_i} \frac{r(B_i)^{\beta }}{|y-c_{R_i}|^{m+\beta }} \,\mathrm {d}\mu (y) |\nu |(B_i) \\&\lesssim |\nu |(B_i)|. \end{aligned} \end{aligned}$$

It remains to consider the term II. Since $b_i= w_i \nu - \varphi _i \mu$, we have

$$\begin{aligned} II=\int _{8R_i \setminus 2B_i} \mathcal {C}b_i \,\mathrm {d}\mu \le \int _{8R_i \setminus 2B_i} \mathcal {C}(w_i\nu ) \,\mathrm {d}\mu +\int _{8R_i \setminus 2B_i} \mathcal {C}_\mu \varphi _i \,\mathrm {d}\mu =: II_1+II_2. \end{aligned}$$

The $L^2(\mu )$-boundedness of $\mathcal {C}_\mu$ gives

$$\begin{aligned} \begin{aligned} II_2 \le \mu (8R_i) ^\frac{1}{2} \Vert \mathcal {C}_\mu \varphi _i \Vert _{L^2(\mu )}&\lesssim \mu (R_i)^\frac{1}{2} \Vert \varphi _i \Vert _{L^2(\mu )} \le \Vert \varphi _i \Vert _{L^\infty (\mu )} \mu (R_i) \lesssim |\nu | (B_i), \end{aligned} \end{aligned}$$

where we used the fact that $R_i$ is $(8,8^{m+1})$-doubling and the properties (7.9) and (7.12) of the function $\varphi _i$.

Finally, we consider the term $II_1$. Suppose $y \in 8R_i \setminus 2B_i$ and $x \in \Gamma (y)$. Then, by Lemma 2.2,

$$\begin{aligned} |T (w_i\nu )(x)| \lesssim \int _{B_i} \frac{\,\mathrm {d}|\nu |(z)}{|x-z|^{m+\alpha }} \sim \frac{|\nu |(B_i)}{(|x-y|+|y-c_{B_i}|)^{m+\alpha }}, \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} \mathcal {C}(w_i\nu )(y)&\lesssim \Big ( \int _{\Gamma (y)} \frac{d(x,E)^{2\alpha }}{(|x-y|+|y-c_{B_i}|)^{2(m+\alpha )}} \,\mathrm {d}\sigma (x) \Big )^{\frac{1}{2}} |\nu |(B_i) \\&\lesssim \frac{1}{|y-c_{B_i}|^{m} }|\nu |(B_i). \end{aligned} \end{aligned}$$

Integrating this over $8R_i \setminus 2B_i$ gives

$$\begin{aligned} \begin{aligned} II_1&=\int _{8R_i \setminus 2B_i} \mathcal {C}(w_i\nu ) \,\mathrm {d}\mu \lesssim \int _{8R_i \setminus R_i} \frac{1}{|y-c_{B_i}|^{m}} \,\mathrm {d}\mu (y)|\nu |(B_i) \\&\quad + \int _{R_i \setminus 2B_i} \frac{1}{|y-c_{B_i}|^{m}} \,\mathrm {d}\mu (y)|\nu |(B_i) \\&\lesssim |\nu |(B_i), \end{aligned} \end{aligned}$$

where we used Lemma 7.4 to estimate the integral over $R_i \setminus 2B_i$. This finishes the proof (7.14), and hence also the proof Theorem 7.5 in the case when $\nu$ is compactly supported.

Suppose then $\nu$ is not compactly supported but $\mu$ is compactly supported. Suppose $M>0$ is such that ${\text {spt}}\mu \subset B(0,M/2)$. Write $\tilde{\nu }:= \nu \lfloor ( E \setminus B(0,M))$. Then for any $y \in {\text {spt}}\mu$ and $x \in \Gamma (y)$ we have

$$\begin{aligned} \begin{aligned} \big |T\tilde{\nu }(x)\big |&\lesssim \int _{E \setminus B(0,M)} \frac{\,\mathrm {d}|\nu |(z)}{(|x-y|+|y-z|)^{m+\alpha }} \\&\le \frac{|\nu | (\mathbb {R}^n)}{\big (|x-y|+d(y,E \setminus B(0,M))\big )^{m+\alpha }}, \end{aligned} \end{aligned}$$

and this gives by Lemma 2.4 that

$$\begin{aligned} \begin{aligned} \mathcal {C}\tilde{\nu }(y)&\lesssim \Big (\int _{\Gamma (y)} \frac{d(x,E)^{2\alpha }}{\big (|x-y|+d(y,E \setminus B(0,M))\big )^{2(m+\alpha )}}\,\mathrm {d}\sigma (x) \Big )^\frac{1}{2}|\nu | (\mathbb {R}^n) \\&\lesssim \frac{1}{d(y,E \setminus B(0,M))^{m}} |\nu |(\mathbb {R}^n). \end{aligned} \end{aligned}$$

Hence, if M is big enough, we get

$$\begin{aligned} \begin{aligned} \mu \big (\{y \in E :\mathcal {C}\nu (y)> \lambda \}\big )&\le \mu \big (\{y \in E :\mathcal {C}\big (\nu \lfloor B(0,M)\big )(y) > \frac{\lambda }{2}\} \big )\\&\lesssim \frac{1}{\lambda } |\nu | \big (B(0,M)\big ) \le \frac{1}{\lambda } |\nu |(\mathbb {R}^n), \end{aligned} \end{aligned}$$

where the second inequality holds because $\nu \lfloor B(0,M)$ is compactly supported.

Suppose finally that neither $\nu$ nor $\mu$ is compactly supported. Then for every $M>0$ it holds that

$$\begin{aligned} \begin{aligned} \mu \lfloor B(0,M) \big (\{ y \in E :\mathcal {C}\nu (y) > \lambda \}\big ) \lesssim \frac{1}{\lambda } |\nu |(\mathbb {R}^n), \end{aligned} \end{aligned}$$

because $\mu \lfloor B(0,M)$ is compactly supported. Letting M tend to infinity concludes the proof. $\square$