square function estimate implies Carleson measure estimate Suppose that $E$ is a regular, $d$-Hausdorff dimensional set in $\mathbb{R}^n$, $d$ an integer.  By regular we mean that for any $x \in E$ and $R>0$ $R^d \approx H^d(B(x,R)\cap E).$  Suppose that $\psi$ is an odd, smooth, compactly supported function on $\mathbb{R}^n$.  I want to show that the square function estimate:
$$\sum_{k=-\infty}^\infty \int_E|\int_E \psi_k(x-y)f(y)dy|^2dx\leq C(\psi)\int_E|f(x)|^2dx$$
imples the Carleson measure estimate:
$$\sum_{2^k\in [0,R]} \int_{B(z,R)}|\int_E\psi_k(x-y)dy|^2dx \leq CR^d \space \forall z\in E, R>0$$
So far I have tried replacing $f$ above by the charcteristic function of a ball.  This gives me:
$$\sum_{k=-\infty}^\infty \int_E|\int_{B(z,R)}\psi_k(x-y)dy|^2dx \leq CR^d$$
This seems very close to what I want, but I can't go further.  I have tried to interchange the order of integration above by expanding the integrals into triple integrals using Fubini's theorem, and then introducing characteristic functions, but with no success.  Any hint would be greatly appreciated.
 A: A friend of mine outlined a solution for me yesterday.  Here it is with the details worked out:
Fix $z \in E$, $R>0$.  Set $B_1=B(0,R)$ and for $k>1$ $B_k=B(0,2^kR)\setminus B(0,2^{k-1}R)$.  Suppose that $spt(\psi)\subset B(0,L)$.  Then $2^{-k}(x-y)\in B(0,L) \implies x-y\in B(0,2^kL)$.  If $x \in B(z,R)$ and $y \in B_j$ then $|x-y|\geq (1+2^{j-1})R$.  So, for $x-y$ to lie outside of the support of $\psi_k$ it is sufficient for $(1+2^{j-1})R>2^kL$, which implies $j>\log_2L/R +R$.  Let $M$ be the smallest integer greater than $\log_2L/R +R$.  This fact will allow us to apply Minkowski's inequality below:
$
\sum_{2^k\in (0,R]}\int_{B(z,R)}\left|\int_E \psi_k(x-y)dy\right|^2dx\leq \sum_{2^k\in (0,R]}\int_{B(z,R)}\left|\sum_{j=1}^\infty\int_{B_j} \psi_k(x-y)dy\right|^2dx\\
\leq C\sum_{j=1}^M\sum_{2^k\in (0,R]}\int_{B(z,R)}\left|\int_{B_j}\psi_k(x-y)dy\right|^2dx\leq C\sum_{j=1}^M\sum_{k=-\infty}^\infty\int_E\left|\int_{B_j}\psi_k(x-y)dy\right|^2dx\\
\leq C'\sum_{j=1}^M 2^{(j-1)d}R^d=C''R^d$
