Compact Reductions for Maximal Superlevel Sets

From a Superlevel Set to a Compact Set

Maximal-function estimates begin with a set of points where some local average is large. At each such point there is at least one ball that witnesses the inequality. The collection of all witnessing balls may be infinite, but a compact subset of the superlevel set can be covered by finitely many of them. This compact reduction is the step that makes finite covering arguments available.

Work on \(\mathbb{R}^n\) with Lebesgue measure \(m\). Let \(f:\mathbb{R}^n\to\mathbb{R}\) be Lebesgue measurable and integrable, so \(\int_{\mathbb{R}^n}|f|\,dm<\infty\). For an open ball \(B(x,r)=\{y:|y-x|<r\}\), with \(r>0\), define its average by \[ \frac{1}{m(B(x,r))}\int_{B(x,r)}|f|\,dm. \] Here \(m(B(x,r))=\omega_n r^n\), where \(\omega_n=m(B(0,1))>0\). The centered maximal function and its strict superlevel set are \[ Mf(x)=\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f|\,dm, \qquad E_\lambda=\{x\in\mathbb{R}^n:Mf(x)>\lambda\}, \] for \(\lambda>0\). These functions and sets are understood up to the stated pointwise definitions; in particular, \(Mf(x)>\lambda\) means that at least one radius has average strictly greater than \(\lambda\).

Proof. If \(x\in E_\lambda\), the definition of supremum gives some \(r>0\) for which the average over \(B(x,r)\) is strictly greater than \(\lambda\). For this fixed \(r\), the map \(z\mapsto\int_{B(z,r)}|f|\,dm\) is continuous. Indeed, if \(z_k\to z\), then \(\chi_{B(z_k,r)}(y)\to\chi_{B(z,r)}(y)\) for every \(y\) outside the boundary sphere of \(B(z,r)\); that sphere has Lebesgue measure zero. For all sufficiently large \(k\), the balls \(B(z_k,r)\) lie in one bounded set \(Q\). The functions \(|f|\chi_{B(z_k,r)}\) are bounded by the integrable function \(|f|\chi_Q\), and converge almost everywhere to \(|f|\chi_{B(z,r)}\). The Dominated Convergence Theorem therefore gives convergence of their integrals. Since the denominator \(m(B(z,r))=\omega_n r^n\) is fixed, the average at radius \(r\) is continuous in the center. Its value at \(x\) is greater than \(\lambda\), so it remains greater than \(\lambda\) for all centers in a neighborhood of \(x\). Each such center belongs to \(E_\lambda\), proving that \(E_\lambda\) is open. \(\square\)

Inner Regularity and Finite Subcovers

For open sets, inner regularity says that their measure can be recovered by compact subsets. In the present setting this fact can be seen directly, including when the open set has infinite measure.

Proof. If \(O=\varnothing\), the only compact subset needed is \(\varnothing\), and both sides are zero. If \(O=\mathbb{R}^n\), the compact cubes \([-j,j]^n\) have measures \((2j)^n\), which tend to infinity, so the supremum is \(m(O)=\infty\). In the remaining case define, for integers \(j\ge1\), \[ K_j=\{x\in\mathbb{R}^n:|x|\le j,\ \operatorname{dist}(x,O^c)\ge 1/j\}. \] The distance to the closed set \(O^c\) is continuous, so \(K_j\) is closed and bounded, hence compact. The sets are increasing: if \(x\in K_j\), then \(|x|\le j+1\) and \(\operatorname{dist}(x,O^c)\ge 1/j>1/(j+1)\), so \(x\in K_{j+1}\). Each \(K_j\) lies in \(O\). Conversely, for every \(x\in O\), openness gives \(\operatorname{dist}(x,O^c)>0\); choosing \(j\) sufficiently large gives both \(|x|\le j\) and \(\operatorname{dist}(x,O^c)\ge1/j\). Thus \(O=\bigcup_jK_j\). Continuity from below of the measure \(m\) now gives \(m(O)=\lim_jm(K_j)\), which proves the assertion. \(\square\)

Apply this to the open set \(E_\lambda\). Its measure is the supremum of the measures of its compact subsets. For each compact \(K\subseteq E_\lambda\), choose for every \(x\in K\) a ball \(B(x,r_x)\) whose average is strictly greater than \(\lambda\). These open balls cover \(K\). Compactness gives a finite subcover. The radii and centers may vary with \(x\); no single radius is being asserted to work at every point.

A Finite Selection of Balls

A finite subcover is not generally disjoint, so to estimate its union one uses a simple selection procedure. The following finite lemma explains why enlarging a disjoint selection by a fixed factor still covers all the original balls.

Proof. If the family is empty, choose no balls; the conclusion holds. Otherwise, arrange the balls in nonincreasing order of radius. Choose the first ball, then proceed through the list: retain a ball if it is disjoint from every previously retained ball, and discard it otherwise. The retained balls are pairwise disjoint. Each discarded ball \(B(c,r)\) intersects a previously retained ball \(B(c_j,r_j)\). Since the list is ordered, \(r\le r_j\). Intersection implies \(|c-c_j|<r+r_j\). For any \(y\in B(c,r)\), the triangle inequality gives \[ |y-c_j|\le |y-c|+|c-c_j|<r+(r+r_j)=2r+r_j\le3r_j. \] Thus \(B(c,r)\subset B(c_j,3r_j)\). Each retained ball is also contained in its own threefold enlargement. This proves the claim, including the case of a one-ball family. \(\square\)

The Compact Covering Estimate

Proof. For nonempty \(K\), cover it by witnessing balls and take a finite subcover. Apply the finite selection lemma to this finite family. The selected balls \(B_1,\ldots,B_N\) have pairwise disjoint interiors and their threefold enlargements cover every ball in the finite subcover, hence cover \(K\). Lebesgue measure scales by \(m(3B_j)=3^n m(B_j)\). Subadditivity therefore gives \(m(K)\le\sum_jm(3B_j)=3^n\sum_jm(B_j)\). Each selected ball witnesses a strict inequality \(\lambda m(B_j)<\int_{B_j}|f|\,dm\); summing the finitely many strict inequalities gives the second bound. Since the selected balls are pairwise disjoint, countable additivity of the integral of the nonnegative function \(|f|\) gives \(\sum_j\int_{B_j}|f|\,dm=\int_{\cup_jB_j}|f|\,dm\le\int_{\mathbb{R}^n}|f|\,dm\). Combining these inequalities proves the assertion. \(\square\)

The role of inner regularity is now precise: \(E_\lambda\) is open, so its measure is the supremum of \(m(K)\) over compact \(K\subseteq E_\lambda\). A bound established uniformly for every such compact \(K\) can therefore be passed to the whole superlevel set by taking that supremum. The finite-cover argument supplies the compact-set bound; inner regularity explains why compact sets suffice, even when the superlevel set itself is unbounded or has infinite measure.

A Common Endpoint Pitfall

A finite collection of open balls can fail to cover a compact set even when the intervals appear to meet at their endpoints. For example, intervals centered at \(\tfrac14,\tfrac12,\tfrac34\) with radius \(\tfrac18\) are \[ (\tfrac18,\tfrac38),\qquad(\tfrac38,\tfrac58),\qquad(\tfrac58,\tfrac78). \] Their union omits \(\tfrac38\) and \(\tfrac58\). In particular, it does not cover \(K=[\tfrac14,\tfrac34]\), despite covering points on both sides of each omitted endpoint. The correction is to increase the radius, for instance to \(\tfrac16\): the intervals then overlap, cover \(K\), and remain inside \([0,1]\). Since \(f=\chi_{[0,1]}\), each corrected ball still has average \(1>\tfrac12\). Always check both endpoint coverage and the strict witness inequality before using compactness to extract a finite subcover.