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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.

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

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.

Let $S$ be a sublinear operator with positive constants and $\mu \{Sf>a\}\le C_1\Vert f{\Vert }_1/a$ and $\Vert Sf{\Vert }_{\infty }\le C_{\infty }\Vert f{\Vert }_{\infty }$. Split $f$ into its part above $a/(2C_{\infty })$ and its bounded remainder. The latter contributes at most $a/2$, so the weak bound controls the superlevel measure by $2C_1{a}^{-1}{\int }_{|f|>a/(2C_{\infty })}|f|$. Integrating this estimate with layer cake representation proves the displayed bound for $1<p<\infty$ and $q=p/(p-1)$. In particular it gives the strong $L^p$ estimate for the Hardy-Littlewood maximal function.

For a nonnegative sublinear operator, weak type $(1,1)$ is the displayed level-set bound for all integrable inputs and positive thresholds. It supplies an exceptional-set estimate for approximation errors; the uncentered maximal weak-type inequality is a central example.

For a normed space $E$ and $0<q<\infty$, an operator $T:E\to L^0$ is of weak type $(E,q)$ when the displayed distribution bound holds uniformly over $f\in E$ and $a>0$. No completeness of $E$ is needed. A sublinear operator has pointwise triangle domination and absolute homogeneity; its outputs may be understood modulo almost everywhere equality.