Large Averages and Small Exceptional Sets
A maximal function records the largest average of a function over balls centered at each point. Its value can be large even where the function itself is small, because a ball may collect mass from elsewhere. The weak-type estimate says that for an integrable function, the set of points where this largest average exceeds a fixed threshold has measure controlled by the total integral divided by that threshold. The geometry behind the estimate is a selection of disjoint balls: each selected ball must contain enough integral to pay for its volume.
Throughout, \(n\ge1\), and \((\mathbb{R}^n,\mathcal{L},m)\) is Euclidean space with Lebesgue measure. This measure space is complete and sigma-finite, but not finite. Functions \(f:\mathbb{R}^n\to\mathbb{C}\) are Lebesgue measurable and integrable, meaning \(\int_{\mathbb{R}^n}|f|\,dm<\infty\); real-valued functions are included. For \(x\in\mathbb{R}^n\) and \(r>0\), let \(B(x,r)\) be the open ball and let \(|B(x,r)|=m(B(x,r))\). Define the centered Hardy–Littlewood maximal function by
It takes values in \([0,\infty]\). We use the measurability of \(Mf\), established in the earlier material on measurability of maximal functions. Thus \(E_\lambda=\{x\in\mathbb{R}^n:Mf(x)>\lambda\}\) is Lebesgue measurable for every \(\lambda>0\). The strict inequality matters: if \(x\in E_\lambda\), the definition of the supremum guarantees at least one radius \(r_x>0\) for which the average is strictly greater than \(\lambda\).
The Weak-Type Estimate
The proof uses two results from the earlier material on covering and compact reductions. The finite fivefold covering lemma says that, from any finite family of balls, one can select pairwise disjoint balls from that family such that the union of the original family is contained in the union of the selected balls dilated by a factor of \(5\) about their centers. The compact reduction for a measurable set in \(\mathbb{R}^n\) says its measure is the supremum of the measures of its compact subsets; for an unbounded set this follows by first intersecting with expanding bounded cubes. We also use the Lebesgue scaling formula \(m(5B)=5^n m(B)\).
Proof. Fix a compact set \(K\subset E_\lambda\). For every \(x\in K\), choose a radius \(r_x>0\) such that $$ \frac{1}{|B(x,r_x)|}\int_{B(x,r_x)}|f|\,dm>\lambda. $$ The open balls \(B(x,r_x)\), for \(x\in K\), cover \(K\). Compactness gives a finite subcover, say \(B_1,\ldots,B_N\). Apply the finite fivefold covering lemma to this finite family. It gives pairwise disjoint selected balls \(B_{i_1},\ldots,B_{i_q}\) such that \(K\subset\bigcup_{j=1}^q5B_{i_j}\). Therefore, by countable subadditivity and the scaling formula, $$ m(K)\le\sum_{j=1}^q m(5B_{i_j}) =5^n\sum_{j=1}^q m(B_{i_j}). $$ Each selected ball belongs to the original family, so its average exceeds \(\lambda\). Consequently, $$ \lambda\sum_{j=1}^q m(B_{i_j}) <\sum_{j=1}^q\int_{B_{i_j}}|f|\,dm \le \int_{\mathbb{R}^n}|f|\,dm. $$ For the last inequality, the selected open balls are pairwise disjoint, so their indicators sum to at most \(1\) everywhere. Combining the inequalities yields \(m(K)\le 5^n\lambda^{-1}\int|f|\,dm\). Taking the supremum over compact \(K\subset E_\lambda\), using compact reduction for the measurable set \(E_\lambda\), proves the theorem. \(\square\)
Worked Example: A One-Dimensional Indicator
Let \(f=\chi_{[-1,1]}\) on \((\mathbb{R},\mathcal{L},m)\), and take \(\lambda=\tfrac14\). We compute the superlevel set exactly. If \(|x|<1\), a sufficiently small ball centered at \(x\) lies in \([-1,1]\), so \(Mf(x)=1\). If \(x>1\), write a centered ball as \((x-r,x+r)\). For \(x-1\le r\le x+1\), its intersection with \([-1,1]\) has length \(r-(x-1)\), and the average is $$ \frac{r-x+1}{2r}=\frac12-\frac{x-1}{2r}. $$ This increases with \(r\), so its largest value on this range is \(1/(x+1)\), at \(r=x+1\). For \(r\ge x+1\), the ball contains the whole interval and its average is \(1/r\le1/(x+1)\); for \(r<x-1\), the intersection is empty. Hence \(Mf(x)=1/(x+1)\) when \(x>1\). Reflection gives \(Mf(x)=1/(|x|+1)\) when \(x<-1\). At \(x=\pm1\), the supremum is \(1/2\), consistent with the same formula. Thus \(Mf(x)>\tfrac14\) exactly when \(|x|<3\), and \(m(E_{1/4})=6\).
The integral is \(\int|f|\,dm=2\), so the theorem gives \(m(E_{1/4})\le 5\cdot2/(\tfrac14)=40\). Check: the exact measure \(6\) is below \(40\). At \(x=3\) and \(x=-3\), the maximal value is exactly \(\tfrac14\), so neither endpoint belongs to the strict superlevel set; this does not change its measure.
Worked Example: Two Disjoint Disks in the Plane
In \((\mathbb{R}^2,\mathcal{L},m)\), let $$ f=\chi_{B((0,0),1)}+\chi_{B((4,0),1)} \quad\text{and}\quad \lambda=\tfrac12. $$ The two unit disks are disjoint because their centers are distance \(4\) apart, greater than the sum of their radii. Each has area \(\pi\), so \(\int|f|\,dm=2\pi\). At every point strictly inside either disk, some sufficiently small centered ball is contained in that disk. On that ball \(f=1\), so its average is \(1>\tfrac12\). The superlevel set therefore contains both unit disks and has measure at least \(2\pi\).
Here \(n=2\), so the theorem gives $$ m(E_{1/2})\le\frac{5^2}{1/2}(2\pi)=50\cdot2\pi=100\pi. $$ Check: the lower bound from the two disks is \(2\pi\), and \(2\pi\le100\pi\). The theorem does not assert that the superlevel set consists only of those disks; it supplies an upper bound for the entire set.
Worked Example: Amplitude and Strict Endpoints
Let \(f=3\chi_{[0,1]}\) on \((\mathbb{R},\mathcal{L},m)\), with \(\lambda=1\). For \(0<x<1\), small centered intervals lie inside \([0,1]\), giving \(Mf(x)=3\). At \(x=0\) and \(x=1\), intervals shrinking to the endpoint have half their length in \([0,1]\), so the supremum is \(3/2\). For \(x>1\), an interval centered at \(x\) first collects mass when its radius exceeds \(x-1\); its average increases up to radius \(x\), where it contains all of \([0,1]\), and equals \(3/(2x)\). Larger radii give a smaller average. Thus \(Mf(x)=3/(2x)\) for \(x>1\). By reflection about \(1/2\), \(Mf(x)=3/(2(1-x))\) for \(x<0\).
For \(x<0\), the strict condition \(3/(2(1-x))>1\) is equivalent to \(x>-\tfrac12\). For \(x>1\), \(3/(2x)>1\) is equivalent to \(x<\tfrac32\). Including the interval \([0,1]\), the full strict superlevel set is \((-\tfrac12,\tfrac32)\), whose measure is \(2\). Meanwhile, \(\int|f|\,dm=3\), and the theorem gives \(m(E_1)\le5\cdot3=15\). Check: at both exterior endpoints, \(Mf=1\), so they are excluded, and the computed measure \(2\) satisfies the bound.
Why Disjointness Is the Key
The local-average condition gives a useful inequality for each ball individually: $$ \lambda\,m(B)<\int_B|f|\,dm. $$ But it does not permit summing over an arbitrary cover. A cover can contain heavily overlapping balls, and then the same integral of \(|f|\) may be counted repeatedly. The covering lemma solves this by selecting disjoint balls from a finite cover. Their total integral is at most the integral over all of \(\mathbb{R}^n\). The price for replacing the original cover by the selected balls is that the selected balls must be enlarged to cover the compact set; the volume of each enlargement is exactly \(5^n\) times the original volume.
Common mistake. Suppose a point is covered by \(N\) identical balls \(B\), each satisfying \(\lambda m(B)<\int_B|f|\,dm\). A tempting but invalid step is to add these \(N\) inequalities and then bound the sum of the integrals by \(\int|f|\,dm\). The left side becomes \(N\lambda m(B)\), but the right side is \(N\int_B|f|\,dm\), which counts the same region \(N\) times and need not be at most the total integral. The correction is not to sum an arbitrary cover: first apply finite fivefold selection, then sum only over the selected disjoint balls, and account for the enlargement using \(m(5B)=5^n m(B)\).
The estimate is called weak type because it controls the measure of a strict superlevel set by the \(L^1\) size of \(f\), with a factor \(1/\lambda\). In particular, increasing the threshold forces the bound on the exceptional set to decrease. This is the basic quantitative role of the maximal function in differentiation arguments: it turns an integral bound into control of the size of the set where local averages are unusually large.
Check Your Understanding
Use the hypotheses and strict inequalities in the theorem when answering.
- Why does \(x\in E_\lambda\) guarantee a ball centered at \(x\) whose average is strictly greater than \(\lambda\)?
- In dimension \(n=3\), what factor replaces \(5^n\) in the weak-type estimate?
- For \(f=3\chi_{[0,1]}\) on \(\mathbb{R}\), why are the points \(-\tfrac12\) and \(\tfrac32\) excluded from \(E_1\), and what is \(m(E_1)\)?
- Where does disjointness enter the proof when bounding the sum of the selected-ball integrals?
- Explain why the fivefold enlargement contributes a factor \(5^n\), rather than \(5\), to the measure estimate.