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

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.