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Suppose a family of linear operators, including its limiting operator, is pointwise dominated by a sublinear operator of weak type (1,1). Convergence almost everywhere on an $L^1$-dense subspace then extends to every $L^1$ input. Approximate the input by errors of norm at most $2^{-2n}$. The weak estimate and the Borel-Cantelli first lemma make the corresponding domination errors tend to zero almost everywhere. Linearity bounds the difference between the full and approximating convergence errors by twice that domination error. This argument does not need an uncountable supremum to be measurable.

Suppose a convolution operator has $L^2$ operator norm at most $A$, and its kernel satisfies the Hörmander integral kernel condition with constant $C$. Under hypotheses making its integrals well-defined, the Calderón–Zygmund decomposition gives weak type (1,1). Indeed, the bounded part has squared $L^2$ norm at most $2^d\lambda \Vert f{\Vert }_1$. The bad cubes occupy at most $\Vert f{\Vert }_1/\lambda$, while outside their $4\sqrt{d}$ dilations the cancellation estimate outside a dilated cube bounds the bad image in $L^1$. Applying distributional estimates to these three regions gives $|\{|Tf|>\lambda \}|\le ((4\sqrt{d}{)}^d+2^{d+2}{A}^2+4C)\Vert f{\Vert }_1/\lambda$.

This integral smoothness condition on a convolution kernel controls how translations change the kernel away from the translation scale. It need not imply differentiability. Coupled with an $L^2$ operator bound, it yields a weak type (1,1) bound through the Calderón–Zygmund decomposition. It is distinct from the bracket-generating condition for differential operators also named after Hörmander.

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 $L^1$.

The domination hypothesis includes $r=0$. For any approximation error $h$, linearity and the triangle inequality therefore give the uniform estimateFor a general input $g$ and an approximating $f$, this impliesThe first term converges on the dense class, while the weak type (1,1) estimate controls the exceptional set where the second is large.

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.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

A nonnegative sublinear operator satisfies $S(f+g)\le S(f)+S(g)$ and $S(cf)=|c|S(f)$ almost everywhere. It has weak type (1,1) if there is a finite constant $C$, independent of the function and threshold, such thatFor a signed or complex output use $|S(f)|$ in the level set. The bound controls the distribution of large values; it does not assert that $S(f)$ is integrable.

For $r>0$, define the forward and backward one-sided interval averaging operatorsand set $A_0^{\pm }f=f$. Each positive-radius average is bounded in absolute value by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.

To dominate the limiting identity too without assuming the differentiation result we are proving, use the sublinear operatorThe maximal estimate and Chebyshev inequality implyThus $S$ has weak type (1,1) and dominates every $A_r^{\pm }$, including $r=0$.

Smooth compactly supported functions are dense in $L^1$, obtained by truncating and then smoothing with a compactly supported approximate identity. For each such continuous function both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^{\pm }f(x)\to f(x)$ almost everywhere.

The forward difference quotient of the primitive is $A_r^{+}f(x)$, and the backward quotient is $A_r^{-}f(x)$:They have the same limit, proving differentiability almost everywhere, with the stronger conclusionThis proves differentiation of an indefinite Lebesgue integral directly from the maximal estimate and density, without assuming $m_u(f)\ge |f|$ as an unproved prerequisite.

If $k(x)=\varphi (|x|)\ge 0$ is decreasing in radius and has integral one, then $|k_t\ast f(x)|\le Mf(x)$, where $k_t(x)=t^{-d}k(x/t)$ and $M$ is the centered Hardy-Littlewood maximal function of $|f|dx$. The layer cake representation expresses $k_t$ as a positive mixture of ball indicators; each corresponding integral is bounded by its ball volume times $Mf(x)$. Integrating these volumes gives exactly one. This applies to the Poisson kernel for the upper half-space and supplies a weak type (1,1) domination for its boundary convergence.