The Centered Maximal Function and Interval Indicators
Maximal-function estimates turn information about averages into information about the size of a set. For an interval indicator, the averages can be computed exactly, so this is a useful setting for seeing how the level threshold, the interval length, and the covering constants interact. For finite sums of indicators, subadditivity gives immediate estimates; a covering selection explains a more general bound without counting overlapping integrals more than once.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. Functions in the examples are nonnegative real-valued and locally integrable. For such a function \(f\), define its centered Hardy–Littlewood maximal function by
The supremum is over all radii \(r>0\); the averaging intervals must be centered at \(x\). For a measurable set \(E\), its indicator is \(\chi_E\), equal to \(1\) on \(E\) and \(0\) off \(E\). For \(\lambda>0\), the superlevel set is \(\{x:Mf(x)>\lambda\}\), with a strict inequality. Values at interval endpoints will be considered explicitly where they affect membership.
Worked Example: The Exact Superlevel Set of One Interval
Let \(I=[a,b]\) have length \(L=b-a>0\), and take \(f=\chi_I\). If \(x\in(a,b)\), some sufficiently small centered intervals around \(x\) lie inside \(I\). Their averages equal \(1\), and no average can exceed \(1\), so \(M\chi_I(x)=1\).
Now take \(x<a\), and write \(d=a-x>0\). A centered interval of radius \(r\le d\) meets \(I\) in a set of measure zero or less, so its average is zero. For \(d<r<d+L\), the overlap has length \(r-d\), and the average is \((r-d)/(2r)\). This expression increases with \(r\), since its derivative is \(d/(2r^2)>0\). For \(r\ge d+L\), the interval contains all of \(I\), and the average is \(L/(2r)\), which decreases with \(r\). The maximum is therefore attained at \(r=d+L\), giving \(L/(2(d+L))\). The same calculation applies for \(x>b\), with \(d=x-b\).
At \(x=a\) or \(x=b\), small centered intervals have average \(1/2\), and larger radii cannot give a larger average. Thus the endpoint value is \(1/2\). It follows that for \(0<\lambda<1/2\), the strict superlevel set is the interval expanded by distance \(L(1/(2\lambda)-1)\) at both ends. For \(1/2\le\lambda<1\), it is \((a,b)\), and for \(\lambda\ge1\), it is empty. In particular, take \(I=[2,5]\), so \(L=3\), and \(\lambda=1/4\). The expansion distance is \(3(1/(2\cdot1/4)-1)=3(2-1)=3\), so
Check: at either outer endpoint, the best average is \(3/(2(3+3))=1/4\), so neither endpoint belongs to the strict superlevel set. The set has measure \(9\), while the one-dimensional weak-type bound proved below gives \(3\cdot3/(1/4)=36\), a valid but non-sharp estimate.
Finite Sums and Subadditivity
For nonnegative locally integrable \(f\) and \(g\), each radius satisfies \(\operatorname{avg}(f+g)=\operatorname{avg}(f)+\operatorname{avg}(g)\le Mf(x)+Mg(x)\). Taking the supremum over radii gives \(M(f+g)(x)\le Mf(x)+Mg(x)\). Repeating this argument proves subadditivity for every finite sum. This pointwise inequality converts a superlevel estimate for a sum into a union of superlevel sets for its terms.
Worked Example: Two Separated Interval Indicators
Let \(f=\chi_{[0,1]}+\chi_{[3,4]}\) on \(\mathbb{R}\), and estimate \(\{Mf>1/2\}\). The intervals are disjoint, so \(\int_{\mathbb{R}} f\,dm=1+1=2\). By subadditivity, \(Mf\le M\chi_{[0,1]}+M\chi_{[3,4]}\). If both terms on the right were at most \(1/4\), their sum would be at most \(1/2\); hence
For each interval, \(L=1\) and the threshold is \(1/4\), so the expansion distance is \(1(1/(2\cdot1/4)-1)=1\). The first strict superlevel set is \((-1,2)\), and the second is \((2,5)\). The point \(2\) is excluded from both: its distance from each interval is \(1\), giving maximal average \(1/(2(1+1))=1/4\) for either indicator. The union has measure \(3+3=6\). Therefore \(m\{Mf>1/2\}\le6\).
Check: the general weak-type estimate gives \(m\{Mf>1/2\}\le3\cdot2/(1/2)=12\). The subadditivity estimate \(6\) is better here. Neither calculation asserts that the superlevel set has measure exactly \(6\); they give upper bounds.
How a Finite Covering Selection Produces the Constant
Proof. Write \(U=\{x:Mf(x)>\lambda\}\). For each fixed \(r>0\), the function \(x\mapsto\int_{x-r}^{x+r}f(t)\,dt\) is continuous: it is the difference of values of a continuous indefinite integral of the locally integrable function \(f\). Thus \(\{x:(2r)^{-1}\int_{x-r}^{x+r}f>\lambda\}\) is open. Since \(U\) is the union of these open sets over \(r>0\), \(U\) is open.
Let \(K\subset U\) be compact. For every \(x\in K\), the definition of the supremum gives a radius \(r_x>0\) such that the centered interval \(I_x=[x-r_x,x+r_x]\) satisfies \(\int_{I_x}f\,dm>\lambda |I_x|\), where \(|I_x|=2r_x\). These intervals cover \(K\), so compactness gives a finite subcover. From this finite family, repeatedly choose an interval of greatest length and discard it and every remaining interval that intersects it. The chosen intervals are pairwise disjoint.
Every discarded interval \(J\) intersects a chosen interval \(I\) whose length is at least \(|J|\). If their centers are \(c_I,c_J\), intersection gives \(|c_I-c_J|\le(|I|+|J|)/2\). For any \(y\in J\), it follows that \(|y-c_I|\le(|I|+|J|)/2+|J|/2\le3|I|/2\). Thus \(J\subseteq3I\), where \(3I\) is the interval with the same center as \(I\) and three times its length. Consequently the intervals \(3I\) from the selection cover \(K\), and
The last inequality holds because the chosen intervals are pairwise disjoint and \(f\ge0\). Lebesgue measure is inner regular on open subsets of \(\mathbb{R}\): the measure of \(U\) is the supremum of \(m(K)\) over compact \(K\subset U\), including when \(m(U)=\infty\). Taking this supremum proves the stated bound. \(\square\)
Worked Example: Tracking the Selected Intervals
Continue with \(f=\chi_{[0,1]}+\chi_{[3,4]}\), and take \(\lambda=1/4\). Consider the compact set \(K=[-2/5,-3/10]\cup[9/5,11/5]\cup[43/10,44/10]\). Its measure is \(1/10+2/5+1/10=3/5\). We exhibit three centered witness intervals, one covering each component.
For the first component, use the interval centered at \(-7/20\) with radius \(7/5\). It is \([-7/4,21/20]\), contains \([-2/5,-3/10]\), and intersects the support of \(f\) in \([0,1]\). Its length is \(14/5\), its integral of \(f\) is \(1\), and its average is \(1/(14/5)=5/14>1/4\).
For the middle component, use the interval centered at \(2\) with radius \(11/5\), namely \([-1/5,21/5]\). It contains \([9/5,11/5]\), and its integral is \(2\), since it contains both unit intervals. Its length is \(22/5\), so its average is \(2/(22/5)=5/11>1/4\).
For the last component, use the interval centered at \(87/20\) with radius \(7/5\). It is \([59/20,115/20]\), contains \([43/10,44/10]\), and its integral is \(1\). Its length is \(14/5\), so its average is \(5/14>1/4\). All three are valid witnesses because each average is strictly greater than the threshold.
The middle interval has the greatest length. It intersects both other intervals, so the finite selection keeps only this interval. Its triple has center \(2\), length \(66/5\), and endpoints \(-23/5\) and \(43/5\); it contains all three witness intervals and therefore covers \(K\). The covering estimate gives \(m(K)\le66/5\), while the disjoint-integral calculation for the selected interval is \((1/4)(22/5)=11/10<2=\int_{[-1/5,21/5]}f\,dm\). The bound is deliberately coarse for this \(K\), whose measure is \(3/5\); the example shows how the factor \(3\) and the strict average inequality enter the argument.
Check: the selected family contains just one interval, so disjointness is automatic. Its integral is \(2\), not \(3\): the interval contains each support interval once.
A Common Covering Mistake
A common error is to take a finite cover by witness intervals and claim that the sum of their integrals is at most \(\int f\,dm\). That inequality is justified only after passing to pairwise disjoint intervals. In the preceding example, the first and middle witness intervals overlap. Their integrals are \(1\) and \(2\), so their sum is \(3\), even though \(\int_{\mathbb{R}}f\,dm=2\). The overlap counts part of the support twice. The correction is to apply the finite selection, retain disjoint intervals, and use nonnegativity to bound the sum of their integrals by the full integral.
There is a separate centered-versus-one-sided pitfall. For \(f=\chi_{[0,1]}\) at \(x=-1\), the average over \([-1,1]\) is not \(1/2\): its intersection with \([0,1]\) has length \(1\), but the centered interval has length \(2\), so this particular average is \(1/2\). However, this interval is not the maximizing one. For \(1<r<2\), the centered interval \([-1-r,-1+r]\) overlaps \([0,1]\) in length \(r-1\), giving average \((r-1)/(2r)\); at \(r=2\) the average is \(1/4\), and for \(r>2\) it is \(1/(2r)\), which decreases. The maximal value is \(1/4\), in agreement with the exact formula. One must optimize over centered intervals rather than substitute a convenient interval that is not centered at the point being tested.
Check Your Understanding
In each problem, use strict superlevel membership and state the measure of the resulting set or bound.
- For \(I=[-2,2]\), compute \(M\chi_I(x)\) for \(x=5\), and determine whether \(x=5\) belongs to \(\{M\chi_I>1/6\}\).
- Find the exact measure of \(\{M\chi_{[0,2]}>1/3\}\), including the effect of the strict inequality at the outer endpoints.
- For \(f=\chi_{[0,1]}+\chi_{[4,5]}\), use subadditivity to bound \(m\{Mf>1/4\}\) by the union of two individual superlevel sets.
- In the finite selection proof, explain why the selected intervals are disjoint and why each discarded interval lies in the triple of an interval that caused it to be discarded.
- Give a concrete reason that the integrals over an overlapping finite cover cannot in general be summed and bounded by \(\int_{\mathbb{R}}f\,dm\).