The centered Hardy-Littlewood maximal operator is weak type $(1,1)$

Put $E_t=\{x:Mf(x)>t\}$. For a fixed $x\in E_t$, [L1] supplies a radius $r>0$ with ${\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }>t\lambda (B(x,r))$. Continuity [L2] with this fixed radius makes the inequality persist in a neighbourhood of $x$, so $E_t$ is open. Let $K\subseteq E_t$ be compact. If $K=\mathrm{\varnothing }$, the required bound for $\lambda (K)$ is immediate. Otherwise take the family of ALL balls $B(x,r)$ with $x\in K$, $r>0$ and the strict witness inequality. This family covers $K$, so it has a finite subcover $B(x_1,r_1),\dots ,B(x_m,r_m)$. No simultaneous choice of a radius for every point of $K$ is made.

Apply [L4] to that finite subcover. There are pairwise disjoint balls [step 1.1, L4, L5, algebra] $B_{i_1},\dots ,B_{i_{\mathrm{\ell }}}$ among $B(x_1,r_1),\dots ,B(x_m,r_m)$ such that $K\subseteq {\bigcup }_{j=1}^{m}B(x_j,r_j)\subseteq {\bigcup }_{k=1}^{\mathrm{\ell }}5B_{i_k}$ and therefore $\lambda (K)\le 5^n{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})\mathrm{.}$ Each chosen ball still satisfies the witness inequality from step 1.1, so $t{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})<{\sum }_{k=1}^{\mathrm{\ell }}{\int }_{B_{i_k}}\mathrm{\mid }f\mathrm{\mid }d\lambda \le {\int }_{{\mathbb{R}}^n}\mathrm{\mid }f\mathrm{\mid }d\lambda =\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ because the chosen balls are pairwise disjoint. Hence $\lambda (K)\le \frac{5^n}{t}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

By [L3], $\lambda (E_t)$ is the supremum of $\lambda (K)$ over compact $K\subseteq E_t$. Step 2.1 gives the required upper bound for nonempty $K$, and step 1.1 handles the empty set. Hence $\lambda (E_t)\le (5^n\mathrm{/}t)\mathrm{\parallel }f{\mathrm{\parallel }}_1$. The stated countable choice covers [L3] and the measure clause of [L4]; the finite-subcover argument introduces no stronger selection assumption.

This is exactly the weak type $(1,1)$ estimate for the centered maximal [step 3.1] operator.