= Solution
Use the approximants just chosen. The <weak type (1,1)> bound gives
$$
\lambda\{S(g-f_n)>2^{-n}\}
\leq C\,2^n\|g-f_n\|_1\leq C2^{-n}.
$$
The <measures> of these exceptional sets are summable. Indeed, the <measure> of their union for $n\geq m$ is at most $C\sum_{n\geq m}2^{-n}\to0$. Hence their limsup is null; this is the <Borel-Cantelli first lemma>. Outside it, $S(g-f_n)\to0$.
Remove also the countable <null set> from part (b). At each remaining point and for every $n$, the estimate in part (a) yields
$$
\limsup_{r\downarrow0}|T_rg-T_0g|
\leq2S(g-f_n).
$$
Letting $n\to\infty$ makes the right side zero. Therefore
$$
\boxed{T_rg\longrightarrow T_0g\quad\text{almost everywhere as }r\downarrow0.}
$$
This 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.
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