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For a positive Borel measure on ${\mathbb{R}}^d$, let $E_{\alpha }=\{M_u\nu >\alpha \}$, using open balls in the uncentered maximal function of a finite measure. Every ball witnessing the inequality lies entirely in $E_{\alpha }$. Cover a compact subset by finitely many witnessing balls and select disjoint balls greedily by decreasing radius. Their threefold dilations cover the compact set. Their total measure is at most $\nu (E_{\alpha })$, giving the bound by inner regularity of Lebesgue measure. For a signed or complex measure, apply this to its total variation measure.

The exact criterion isNecessity follows already from one approximation with error less than one: a compact set $K\subseteq \mathbb{R}$ is bounded, hence has finite Lebesgue measure, so $m(E)=m(K)+m(E\setminus K)<\infty$.

Conversely, suppose $m(E)<\infty$. By continuity from above of a measure, the decreasing sets $E\setminus [-N,N]$ have measures tending to zero. Choose $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.

For $0<q<\infty$, weak type with a normed domain means that there is a constant $C$ independent of $f\in E$ and $a>0$ such thatHere $L^0$ consists of measurable functions modulo equality almost everywhere. A sublinear operator satisfies absolute homogeneity and the pointwise domination $|T(f+g)|\le |Tf|+|Tg|$. In particular the definition controls the output's superlevel measures, not necessarily its strong $L^q$ norm. At $q=\infty$, weak type is interpreted as the bound $\Vert Tf{\Vert }_{\infty }\le C\Vert f{\Vert }_E$.

A useful theorem is dense approximation under a weak-type maximal bound. Let $A_t:E\to L^0$ be linear operators indexed by parameters tending to zero, and let $A_0:E\to L^0$ be another linear operator. Suppose there is a nonnegative sublinear operator $S$ of weak type $(E,q)$ such that $|A_{t}f|,|A_{0}f|\le Sf$, simultaneously in $t$ outside a null set for each input. Assume the suprema and limiting suprema used below are measurable. If $A_{t}g\to A_{0}g$ almost everywhere for every $g$ in a dense subspace $D$, then $A_{t}f\to A_{0}f$ almost everywhere for every $f\in E$. For a sequence of operators all suprema are automatically countable; continuously parametrized averages below also have measurable suprema.

To prove the theorem, fix $f$ and $g\in D$. Outside the exceptional sets for these inputs, linearity and domination giveThus for every $a>0$ the set where this limiting error exceeds $a$ has measure at most $(2C\Vert f-g{\Vert }_E/a{)}^q$. Choose a sequence $g_k\in D$ converging to $f$ in norm and take the union of their null exceptional sets. Letting $k\to \infty$ makes the bound zero. Taking countably many positive rational $a$ proves convergence almost everywhere. This proves the requested sufficient condition rather than merely invoking a convergence theorem.

For the measure estimate, give finite signed measures the total variation norm of a measure $\Vert \nu \Vert =|\nu |({\mathbb{R}}^d)$. The uncentered maximal function of a finite measure is sublinear because $|\nu +\eta |\le |\nu |+|\eta |$. Its set $E_a=\{m_u\nu >a\}$ is open: it is precisely the union of open balls whose variation mass exceeds $a$ times their volume.

Fix a compact subset $K\subset E_a$ and select a finite cover of $K$ by these high-density balls. Order that finite collection by decreasing radius. Keep a ball if it is disjoint from all previously retained balls. Every discarded ball intersects a retained ball of at least its radius and is contained in that retained ball's triple dilation. The retained balls $B_j$ are disjoint and their triples cover $K$. ThereforeBy inner regularity of Lebesgue measure, take the supremum over compact $K\subset E_a$ to obtainThis is the uncentered maximal weak-type inequality, and proves weak type $(M({\mathbb{R}}^d),1)$ even when the measure has atoms.

Finally work in dimension one. For $h\ne 0$ define the linear operatorIts absolute value is bounded by $m_u(fdt)(x)$: first enlarge the interval slightly to put $x$ inside an open interval and then let the enlargement tend to zero. The endpoint masses vanish because $fdt$ is absolutely continuous. Taking $A_{0}f=f$, both operators are dominated by $Sf=m_u(fdt)+|f|$. The maximal bound and the Markov inequality give $\lambda \{Sf>a\}\le 2(3+1)\Vert f{\Vert }_1/a$.

Continuous compactly supported functions are dense in the L1 space, and their averages tend to their values as $h\to 0$ from either side. For a fixed $x$, the interval integral varies continuously for $h\ne 0$, so the required suprema can be taken over rational $h$ and are measurable. Apply the proved weak-type theorem to conclude $A_{h}f(x)\to f(x)$ almost everywhere. Hence

For a positive Borel measure $\nu$, the centered Hardy-Littlewood maximal function and the uncentered maximal function of a finite measure areWe use open balls and allow infinite values. For signed or complex measures the definition uses their total variation measure. Since a centered ball is also an admissible uncentered ball, and every ball $B(c,r)$ containing $x$ lies in $B(x,2r)$,For functions, $m(f)$ means the centered Hardy-Littlewood maximal function of $|f|dx$.

The set $E_{\alpha }$ is the union of the open balls $B$ with $\nu (B)>\alpha |B|$. In particular, each witnessing ball lies entirely inside $E_{\alpha }$. If $K\subset E_{\alpha }$ is compact, take a finite witnessing cover of $K$. Choose its largest-radius ball, discard all balls meeting it, and repeat. The selected balls $B_j$ are disjoint. Every discarded ball has no larger radius than the selected ball that discarded it and is contained in its threefold dilation. HenceTaking the supremum over compact $K$ by inner regularity of Lebesgue measure proves the stronger, localized estimateIf the right side is infinite there is nothing to prove. Positivity of the measure is important: a signed version uses $|\nu |(E_{\alpha })$, not $\nu (E_{\alpha })$.

For the radial-kernel inequality, integrability implies $\varphi (r)\to 0$. For $0<s<\varphi (0)$, let $r(s)$ be the unique radius with $\varphi (r(s))=s$. The layer cake representation and Tonelli theorem giveThe last integral equals the total mass of $\varphi (|\cdot |)$, namely one. Taking the absolute value of the original convolution therefore proves the claim with constant exactly one. This is radial decreasing kernel domination by the maximal function.

An application is boundary convergence of the Poisson integral. Its normalized radial profile is $c_d(1+r^2{)}^{-(d+1)/2}$, so ${\sup }_{t>0}|P_t\ast f|\le m(f)$. For a continuous compactly supported $g$, the approximate identity property gives $P_t\ast g\to g$ pointwise. For arbitrary $f\in L^1$, put $h=f-g$; thenThe preceding weak type (1,1) estimate and the elementary integral bound for $|h|$ show that the set where this exceeds $2a$ has measure at most $(3^d+1)\Vert h{\Vert }_1/a$. Continuous compactly supported functions are dense in $L^1$, so this measure is zero. Thus the Poisson integrals of an integrable function converge to it almost everywhere; the same approximate identity also gives convergence in $L^1$.

Let $U_{\alpha }=\{x:m_u(f)(x)>\alpha \}$. It is open: it is the union of all open intervals $I$ for which ${\int }_I|f|>\alpha |I|$. This also proves measurability of the uncentered maximal function. Triangle inequality inside each average proves sublinearity and absolute homogeneity.

Take a compact subset $K\subset U_{\alpha }$, and choose finitely many such intervals covering it. The one-dimensional Wiener covering lemma selects pairwise disjoint intervals $I_1,\dots ,I_N$ whose concentric triples cover $K$. For a finite family this follows by choosing a longest remaining interval and discarding all intervals intersecting it: each discarded interval has no greater length and lies in the triple of the chosen one. ThusTaking the supremum over compact subsets, by inner regularity of Lebesgue measure, givesThis establishes weak type (1,1). It is the one-dimensional uncentered maximal weak-type inequality for the finite measure $|f(t)|dt$; the constant need not be optimal for this question.

Write $dm=dt/(2\pi )$ in angular coordinates on the unit circle. The centered Hardy-Littlewood maximal function iswhere $I_r(\omega )$ is the centered circular arc of angular radius $r$, with $r=\pi$ giving the whole circle. The Hardy-Littlewood maximal inequality in this one-dimensional setting is the weak-type estimateThe precise constant is not important for differentiation, but we prove this one explicitly.

Let $E=\{Mf>\alpha \}$ and choose a compact subset $F\subset E$. For each point of $F$, choose a centered arc whose average of $|f|$ exceeds $\alpha$. A finite subfamily covers $F$. Order these arcs by decreasing length and retain an arc precisely when it is disjoint from the previously retained arcs. Every rejected arc meets a retained arc of at least its length and lies in the latter's triple, interpreted on the circle and truncated to the whole circle if necessary. If the retained disjoint arcs are $I_1,\dots ,I_N$, thenThis is the interval version of the Wiener covering lemma. The superlevel set is measurable: the average at each fixed radius is continuous in the center, by translation continuity in $L^1$, so $E$ is open. Inner regularity of Lebesgue measure now passes the compact-set bound to $E$, proving the Hardy-Littlewood maximal inequality. The same finite-interval argument gives the line version $|\{Mf>\alpha \}|\le 3\Vert f{\Vert }_1/\alpha$ for integrable functions on the real line.

To prove differentiation, first recall that continuous functions are dense in $L^1(m)$ on the circle. Approximate an integrable function by simple functions; for each of their measurable sets, inner and outer regularity give a compact subset and an open superset with arbitrarily small measure difference. A continuous cutoff between these sets approximates the indicator in $L^1$. Combining finitely many cutoffs proves the claimed density.

For a representative of $f$ finite almost everywhere, define its local absolute oscillation byIf $h$ is continuous and $u=f-h$, the triangle inequality and continuity of $h$ giveThus the Hardy-Littlewood maximal inequality and the elementary integral bound for a superlevel set yield, for every $\epsilon >0$,Choose continuous $h$ with $\Vert f-h{\Vert }_1$ arbitrarily small. It follows that $\Delta f=0$ outside a null set, by using positive rational $\epsilon$. In particular,for almost every $\omega$. This proves the requested Lebesgue differentiation theorem, including the stronger absolute-oscillation formulation, rather than assuming pointwise convergence of the approximating continuous functions.

For failure at a specified point, use coordinates $-\pi <t\le \pi$ and put $a_n=2^{-2^n}$. Let $f$ be one on the annuli $a_{2j+1}<|t|\le a_{2j}$ for $j\ge 0$, and zero elsewhere, including at zero. This bounded measurable function is integrable. Since $a_{n+1}/a_n\to 0$, the average $A_n$ over the centered interval $[-a_n,a_n]$ satisfiesIndeed, the outermost annulus occupies all but the fraction $a_{n+1}/a_n$ of that interval, and has value one or zero according to parity. The even-radius subsequence tends to one and the odd-radius subsequence to zero, so no centered-average limit exists at $\omega =1$. The normalization of $m$ cancels from these averages. This is how oscillating annuli obstruct differentiation at a point without contradicting almost-everywhere differentiation.

One potential form of the Hardy–Littlewood–Sobolev inequality is: if $0<\alpha <n$, $1<p<n/\alpha$, and $1/r=1/p-\alpha /n$, thenA normalization constant for the Riesz potential only changes $C$. Equivalently, for $0<\lambda <n$, $p,s>1$ and $1/p+1/s+\lambda /n=2$,The exponent relation is necessary under dilation. We prove the potential estimate and then derive the bilinear version.

First establish the needed maximal estimate, so it is not an unproved step in the argument. Define the centered Hardy-Littlewood maximal function by $Mf(x)={\sup }_{t>0}|B_t{|}^{-1}{\int }_{B_t(x)}|f|$. For integrable $f$, cover a compact subset of $\{Mf>a\}$ by finitely many balls whose averages exceed $a$. Order these balls by decreasing radius and greedily retain a ball if it is disjoint from those already retained. Every discarded ball intersects a retained ball of at least its radius, so lies in its triple. HenceThe passage from compact subsets uses inner regularity of Lebesgue measure; the superlevel set is open because fixed-radius local averages are continuous. For $f\in L^p$, split $f=f{1}_{\{|f|>a/2\}}+f{1}_{\{|f|\le a/2\}}$. The second summand's maximal function is at most $a/2$, and the first is integrable. ThereforeThe layer cake representation and Tonelli theorem giveThus the Strong Lp bound for the Hardy-Littlewood maximal function is proved for every $p>1$.

For $I_{\alpha }|f|(x)$, split at radius $R$. Dyadic shells with radii $2^{-j}R$ giveFor the outer part, Hölder's inequality givesThe last integral is finite exactly because $p<n/\alpha$. Balance the two bounds by $R^{n/p}=\Vert f{\Vert }_p/Mf(x)$ to obtain the Hedberg inequalityThe zero cases follow directly or by a limit; $Mf$ is finite almost everywhere by the maximal estimate. Since $r(1-\theta )=p$, integration and the proved maximal bound yield $\Vert I_{\alpha }f{\Vert }_r\le C\Vert f{\Vert }_p$. Taking $\alpha =n-\lambda$ and pairing $I_{\alpha }|f|$ with $|g|\in L^{r^{\prime }}$ proves the bilinear form by Hölder's inequality, because $s=r^{\prime }$. All integrals are justified first for nonnegative truncations and then by monotone convergence and the resulting finite bound. The strong estimate does not include $p=1$ or $p=n/\alpha$.

Start with a finite subcover of the compact set $K$. Among its remaining open balls, retain one of greatest radius and discard every ball intersecting it. Repeat until no balls remain. The retained family $\mathcal{F}$ is finite and pairwise disjoint. If a discarded ball $B(c,r)$ met the retained ball $B(c_0,R)$, then $r\le R$ and $|c-c_0|<r+R$. Therefore, for any $y\in B(c,r)$,Every original ball lies in the concentric triple of one retained ball. This proves the Wiener covering lemma for the finite subcover. By Lebesgue measure scaling and subadditivity,Disjointness givesThe empty compact set is covered by the empty selection.

The spherical derivative of a measure at $x$ is the limitwhen it exists, where $v_d$ is the unit-ball volume. We prove that this limit is zero outside a Lebesgue-null set for the singular measure.

First derive the needed maximal estimate from the selection lemma. For a finite positive Borel measure $\eta$, putThis is the uncentered maximal function of a finite measure. For $\alpha >0$, its strict superlevel set is the union of all balls with $\eta (B)>\alpha {\lambda }_d(B)$, so it is open. Cover any compact subset $H$ of this set by such balls and apply the finite Wiener covering lemma. The selected disjoint balls giveBy inner regularity of Lebesgue measure, this proves the uncentered maximal weak-type inequality

Since $\mu$ and ${\lambda }_d$ are mutually singular measures, there is a Borel set $N$ with ${\lambda }_d(N)=0$ and $\mu ({\mathbb{R}}^d\setminus N)=0$. By regularity of finite Borel measures on Euclidean space, for each $\epsilon >0$ choose compact $L\subset N$ with $\mu ({\mathbb{R}}^d\setminus L)<\epsilon$. Define the remainder measure ${\eta }_{\epsilon }(A)=\mu (A\setminus L)$.

For $x\notin L$, distance from $x$ to this compact set is positive. Thus sufficiently small centered balls avoid $L$, andWrite ${\overline{D}}_s\mu (x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u{\eta }_{\epsilon }(x)>\alpha$. HenceThe right side has Lebesgue measure at most $3^d\epsilon /\alpha$, since ${\lambda }_d(L)=0$. This even bounds the outer measure of the left side, without a separate measurability argument for the upper density. Let $\epsilon \downarrow 0$, and then take the countable union over positive rational $\alpha$. We obtain ${\overline{D}}_s\mu =0$ Lebesgue almost everywhere. The ratios are nonnegative, so their lower limit is also zero andThis proves that the spherical derivative of a singular measure vanishes. It is important to approximate the null carrier by compact subsets: a null carrier can be dense, so points outside it need not have any neighbourhood avoiding it.

In one dimension, consider the cumulative distribution function $F$. For either sign of $h\ne 0$, monotonicity gives a nonnegative difference quotient, and the relevant half-open interval lies inside $B(x,2|h|)$. In detail, the numerator is $\mu ((x,x+h])$ for $h>0$, while after reversing both signs it is $\mu ((x+h,x])$ for $h<0$. ThusAt each point where the spherical derivative of a measure is zero, this bound tends to zero from both sides. The enlarged open interval avoids any endpoint ambiguity from atoms. Therefore the ordinary two-sided derivative exists andThis is why singular distribution functions have zero derivative almost everywhere without having to be constant: almost-everywhere differentiation recovers increments only under extra hypotheses such as absolute continuity of a function.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

The relevant version of Wiener covering lemma is the finite ball selection lemma: from a finite collection of open Euclidean balls one can select pairwise disjoint balls $B_j$ such that the original union is contained in ${\bigcup }_{j}3B_j$, where $3B_j$ has the same centre and three times the radius. To prove it, repeatedly retain a largest-radius remaining ball and discard every ball meeting it. If a discarded ball has radius $r\le R$ and meets a retained ball of radius $R$, every point of the former is at distance less than $r+(r+R)\le 3R$ from the latter's centre. This proves the containment and, by scaling Lebesgue measure,This is a covering result; no Fourier-algebra version of Wiener's lemma is involved.

We use it for a disjoint ball decomposition modulo null sets of $V$. The empty set needs only the empty family. Otherwise put $V_0=V$. For an open residual set $V_k$ of positive finite measure, inner regularity of Lebesgue measure supplies a compact $K_k\subseteq V_k$ with ${\lambda }_d(K_k)\ge {\lambda }_d(V_k)/2$. Cover $K_k$ by finitely many open balls whose closures lie in $V_k$, and use the Wiener covering lemma to select disjoint balls. Their total measure is at least ${\lambda }_d(V_k)/(2\cdot 3^d)$. Remove their closures to obtain the open residual set $V_{k+1}$. Ball boundaries are null sets, soAll retained balls, including those from different stages, are pairwise disjoint and contained in $V$. They form a countable family. The uncovered set is contained in ${\bigcap }_k{V}_k$ together with their countably many boundaries, and both have measure zero. Enumerating the retained balls gives

Next consider the uncentered maximal function of a finite measure. For every real $\alpha$, its strict superlevel set isIndeed, membership in the ball is exactly the strict condition $\Vert x-y\Vert <r$ in the supremum. This union is open, proving that $m_u$ is lower semicontinuous as an extended nonnegative function; it may take the value infinity.

For $\alpha >0$, take a compact subset $K$ of this superlevel set and choose a finite cover by balls satisfying the displayed density inequality. The Wiener covering lemma selects disjoint balls $B_j$ whose triples cover $K$. Since $\mu$ is a probability measure,Take the supremum over compact $K$ using inner regularity of Lebesgue measure. This works even if the open superlevel set was initially unbounded, and proves the uncentered maximal weak-type inequalityFor a finite positive measure of mass $M$, scaling gives $3^{d}M/\alpha$ instead.

One important use is the Lebesgue differentiation theorem. For $f\in L^1({\mathbb{R}}^d)$, approximate it in $L^1$ by a compactly supported continuous $g$, and set $h=|f-g|$. The limiting mean oscillation of $f$ at $x$ is at most $m_{h{\lambda }_d}(x)+h(x)$, since that of $g$ vanishes by continuity. The preceding weak-type estimate and the elementary integral bound on $\{h>t\}$ giveLet the approximation error tend to zero and then take countably many $t\downarrow 0$. The local averages of $f$ converge to $f(x)$ almost everywhere. Localizing extends this to locally integrable functions. Thus the maximal estimate turns norm approximation by continuous functions into almost-everywhere information about local averages.

For a finite positive measure on Euclidean space, cover a compact subset of the maximal superlevel set by finitely many high-density balls. Wiener covering lemma selects disjoint balls whose triples cover it. Their total measure mass is at most the mass of the original measure. Inner regularity of Lebesgue measure yields the displayed bound. Applied to approximation errors, it proves the Lebesgue differentiation theorem.