# Density argument

A ${\gamma}$-density argument is useful in enlarging some sets a little bit while keeping the measure roughly the same.
Given a closed set ${A\subset \mathbb{R}^n}$ and ${0<\gamma<1}$, a point ${x\in\mathbb{R}^n}$ is said to be ${\gamma}$-dense with respect to ${A}$ if $\displaystyle \inf_{r>0} \frac{|A\cap B(x,r)|}{|B(x,r)|}>\gamma,$

and the set of ${\gamma}$-dense points is denoted as ${A^\ast}$. ${A}$ being close we see that ${A^\ast \subset A}$ and is closed as well. Even if ${\left(A^\ast\right)^{c}\supset A^{c}}$ we can somewhat reverse this: the measures ${|\left(A^\ast\right)^{c}|}$ and ${|A^{c}|}$ are in fact comparable, when finite.

Lemma 1 There exists a constant ${C=C_{n,\gamma}}$ such that $\displaystyle \left|\left(A^\ast\right)^c\right|\leq C_{n,\gamma} \left|A^c\right|.$

Proof: One just has to write ${\left(A^\ast\right)^c}$ explicitly, $\displaystyle \left(A^\ast\right)^c=\left\{x\,:\, \exists B, \frac{|A\cap B|}{|B|}\leq \gamma\right\},$

where ${B}$ is a ball centered in ${x}$ and the condition then can be reformulated as $\displaystyle \frac{|A^c \cap B|}{|B|}>1-\gamma$

for some ball with center ${x}$. But $\displaystyle \frac{|A^c \cap B|}{|B|}\leq M\left(\chi_{A^c}\right),$

where ${M}$ is the Hardy-Littlewood maximal function, thus the measure of ${\left(A^\ast\right)^c}$ is dominated by ${|\{M\left(\chi_{A^c}\right)>1-\gamma\}|}$, which is in turn dominated by ${\frac{C_n}{1-\gamma}\|\chi_{A^c}\|_{L^1}=\frac{C_n}{1-\gamma}|A^c|}$. $\Box$

Notice the only point where we use that the measure is Lebesgue’s is when using the maximal function properties. Thus the theorem is true in a more general setting, in particular for doubling measures on metric spaces (see).
The lemma has the consequence that non-tangential maximal functions don’t actually depend on the aperture of the cone where you are taking the supremum. Say ${F(x,t)}$ is a function ${F\,:\, \mathbb{R}^n\times [0,+\infty[\rightarrow \mathbb{R}}$. Then the non-tangential maximal function of parameter ${a}$ is $\displaystyle F^\ast_{a}(x):=\sup_{|x-y|

where the supremum is taken in both ${t}$ and ${y}$, and ${a}$ therefore determines the aperture of this cone. It’s obvious that if ${a>b}$ then pointwise $\displaystyle F^\ast_{a}(x)\geq F^\ast_{b}(x),$

but actually we have some inverse control on ${F^\ast_{a}(x)}$:

Lemma 2 If ${a>b}$ then $\displaystyle \int_{\mathbb{R}^n}{F^\ast_{a}(x)}\,dx\lesssim_{a,b} \int_{\mathbb{R}^n}{F^\ast_{b}(x)}\,dx.$

Actually the control is directly over upper level sets. A quick remark: if instead of ${F}$ we consider ${|F|^p}$ then the above inequality becomes an ${L^p}$ inequality, provided the functions belong to ${L^p}$ for that exponent.
Proof: In this case one has to take $\displaystyle A:=\{x\,:\, F^\ast_{b}(x)>\alpha\}^c$

for a fixed ${\alpha}$. Notice that if ${x\in \{ F^\ast_{a}>\alpha\}}$ then by definition there exist ${\overline{y},\overline{t}}$ s.t. ${|x-\overline{y}| and ${|F\left(\overline{y},\overline{t}\right)|>\alpha}$. Then, since ${F^\ast_b}$ takes a supremum on a ball of radius ${bt}$, we have that the entire ball ${B\left(\overline{y},b\overline{t}\right)}$ is contained inside ${A^c}$ (all points near ${\overline{y}}$ have ${\overline{y}}$ as neighbour!). This ball is further contained inside ${B\left(x,(a+b)\overline{t}\right)}$ for obvious geometric reasons, and thus $\displaystyle \frac{|A^c\cap B\left(x,(a+b)\overline{t}\right)|}{|B\left(x,(a+b)\overline{t}\right)|}\geq \frac{|B\left(\overline{y},b\overline{t}\right)|}{|B\left(x,(a+b)\overline{t}\right)|}=\left(\frac{b}{a+b}\right)^n.$

Intersecting with ${A}$ instead of ${A^c}$, if ${\gamma}$ is chosen such that ${\gamma> 1-\left(\frac{b}{a+b}\right)^n}$, we see that our ${x}$ (which was taken in ${\{ F^\ast_{a}>\alpha\}}$) is contained in ${\left(A^\ast\right)^c}$, and thus with a constant depending on ${n}$ and ${\gamma}$ (i.e. on ${a, b}$) we have $\displaystyle |\{ F^\ast_{a}>\alpha\}| \lesssim_{n,a,b} |A^c|=|\{ F^\ast_{b}>\alpha\}|.$

Since ${\int{|\{|f|>\alpha\}|}\,d\alpha=\|f\|_{L^1}}$, it suffices to integrate on both sides. $\Box$
To summarize the argument above, we proved that all points of $\{ F^\ast_{a}>\alpha\}$ have a small density in the complement of $\{ F^\ast_{b}>\alpha\}$.
A useful observation is that the constant is given by ${c_n \left(\frac{a+b}{b}\right)^n}$.