Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 5 / 3

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.

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.

The printed symbol $E$ is not defined. Take $E=L^1(\mathbb{R})$, or more generally an $L^1$ subspace with its induced $L^1$ norm; density must be in that norm. Given $g\in E$, choose a sequence $f_n\in F$ withBy the assumed convergence on $F$, each $T_r{f}_n\to T_0{f}_n$ holds outside a null set $N_n$. Their countable union $N={\bigcup }_n{N}_n$ is still null. Thus outside $N$ the convergence statements for every member of this single approximation sequence hold simultaneously. No assertion of simultaneous convergence for the entire uncountable dense class is needed.

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.

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.

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.