The circle maximal function is weak type one one for finite measures

First let $0<h<\frac{1}{2}$. If ${\zeta }_n\to \zeta$ in $\mathbb{T}$, then for every $\eta \in I_h(\zeta )$ the triangle inequality for the circular distance gives $d({\zeta }_n,\eta )\le d({\zeta }_n,\zeta )+d(\zeta ,\eta )<h$ for all large $n$, so the indicators satisfy ${\mathbf{1}}_{I_h(\zeta )}\le {\mathrm{lim\hspace{0.17em}inf}}_n{\mathbf{1}}_{I_h({\zeta }_n)}$ pointwise; applying [L3] to the finite measure $\mathrm{\mid }\mu \mathrm{\mid }$ of [L2] gives $\mathrm{\mid }\mu \mathrm{\mid }(I_h(\zeta ))\le {\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\mid }\mu \mathrm{\mid }(I_h({\zeta }_n))$, that is, the map $\zeta \mapsto \mathrm{\mid }\mu \mathrm{\mid }(I_h(\zeta ))$ is lower semicontinuous. For $h=\frac{1}{2}$, [L1] gives $I_h(\zeta )=\mathbb{T}$ for every centre, so the mass function is constant and hence lower semicontinuous. Since [L1] makes $m(I_h(\zeta ))=2h$ independent of $\zeta$ for every $0<h\le \frac{1}{2}$, the set $A_h:=\{\zeta \in \mathbb{T}:\mathrm{\mid }\mu \mathrm{\mid }(I_h(\zeta ))>\lambda m(I_h(\zeta ))\}$ is the superlevel set of a lower semicontinuous function and is therefore open.

Because $M_{\mathbb{T}}\mu$ is the supremum of the quotients over $0<h\le \frac{1}{2}$, the identity $\{M_{\mathbb{T}}\mu >\lambda \}={\bigcup }_{0<h\le 1\mathrm{/}2}{A}_h$ holds; it is a union of open sets, so $\{M_{\mathbb{T}}\mu >\lambda \}$ is open and in particular Borel measurable.

Let $K\subseteq \{M_{\mathbb{T}}\mu >\lambda \}$ be compact. If $K=\mathrm{\varnothing }$, then $m(K)=0\le 3\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\mathrm{/}\lambda$; assume henceforth that $K\mathrm{\ne }\mathrm{\varnothing }$. The family of all open arcs $I_h(\zeta )$ with $h\in (0,\frac{1}{2}]$ and $\mathrm{\mid }\mu \mathrm{\mid }(I_h(\zeta ))>\lambda m(I_h(\zeta ))$ is a family of open subsets of $\mathbb{T}$, described by a formula and hence requiring no selection, that covers $K$ by step 2.1; [L6] provides a finite subcover $I_1,\dots ,I_N$ of $K$ by such arcs, each satisfying $\mathrm{\mid }\mu \mathrm{\mid }(I_j)>\lambda m(I_j)$.

Relabel the finite list so that the radii satisfy $h_1\ge h_2\ge \cdots \ge h_N$, and pass through it once, keeping an arc exactly when it is disjoint from every previously kept arc. The kept arcs are pairwise disjoint and each still satisfies $\mathrm{\mid }\mu \mathrm{\mid }(I_i)>\lambda m(I_i)$. If $h_1=\frac{1}{2}$, the first kept arc is $\mathbb{T}$ and contains every arc of the subcover; set ${\widehat{I}}_1=\mathbb{T}$, whose measure is at most $3m(I_1)$. Otherwise all radii are strictly less than $\frac{1}{2}$. If $I_j$ with center $c_j$ is rejected, it meets a kept arc $I_i$ with center $c_i$ and $i<j$, so $h_i\ge h_j$ and $d(c_i,c_j)\le h_i+h_j\le 2h_i$; every $\eta \in I_j$ therefore satisfies $d(\eta ,c_i)\le d(\eta ,c_j)+d(c_j,c_i)<h_j+2h_i\le 3h_i$. Writing ${\widehat{I}}_i:=\{\eta :d(\eta ,c_i)<3h_i\}$, every arc of the subcover lies in ${\widehat{I}}_i$ for some kept arc $I_i$, and $m({\widehat{I}}_i)\le 3m(I_i)$: if $3h_i<\frac{1}{2}$ this reads $6h_i=3\cdot 2h_i$, while if $3h_i\ge \frac{1}{2}$ then $m({\widehat{I}}_i)=1\le 6h_i=3m(I_i)$; at equality the antipode is excluded but has measure zero.

The kept arcs are pairwise disjoint, so their $m$-measures add and their $\mathrm{\mid }\mu \mathrm{\mid }$-values add; by step 4.1 and [L2], $\lambda m\left({\bigcup }_i{I}_i\right)=\lambda {\sum }_{i}m(I_i)<{\sum }_i\mathrm{\mid }\mu \mathrm{\mid }(I_i)=\mathrm{\mid }\mu \mathrm{\mid }\left({\bigcup }_i{I}_i\right)\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\mathrm{.}$

The arcs $I_1,\dots ,I_N$ cover $K$ and each lies in some ${\widehat{I}}_i$ of a kept arc, so step 4.1 and step 5.1 give $m(K)\le m\left({\bigcup }_i{\widehat{I}}_i\right)\le {\sum }_{i}m({\widehat{I}}_i)\le 3{\sum }_{i}m(I_i)<\frac{3\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})}{\lambda }\mathrm{.}$

By [L4] the measure of the Borel set $\{M_{\mathbb{T}}\mu >\lambda \}$ is the supremum of $m(K)$ over compact $K\subseteq \{M_{\mathbb{T}}\mu >\lambda \}$; step 2.1 supplies the measurability and step 6.1 bounds every such $m(K)$ by $3\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\mathrm{/}\lambda$, so $m(\{M_{\mathbb{T}}\mu >\lambda \})\le 3\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\mathrm{/}\lambda$.

Let $f\in L^1(\mathbb{T},m)$ and apply step 7.1 to the finite complex measure $fm$: [L5] gives $\mathrm{\mid }fm\mathrm{\mid }(\mathbb{T})={\int }_{\mathbb{T}}\mathrm{\mid }f\mathrm{\mid }dm=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $M_{\mathbb{T}}(fm)=M_{\mathbb{T}}f$, so $m(\{M_{\mathbb{T}}f>\lambda \})\le 3\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{/}\lambda$.