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

Put $E_t:=\{x\in {\mathbb{R}}^n:Mf(x)>t\}$. If $x\in E_t$, then [L1] gives a radius $r_x>0$ such that ${\int }_{B(x,r_x)}\mathrm{\mid }f(y)\mathrm{\mid }d\lambda (y)>t\lambda (B(x,r_x))\mathrm{.}$ By [L2] applied to $\mathrm{\mid }f\mathrm{\mid }$ at $(x,r_x)$, there is ${\delta }_x>0$ such that the same strict inequality holds with $x$ replaced by every $y\in B(x,{\delta }_x)$ and the radius kept equal to $r_x$. Hence $B(x,{\delta }_x)\subseteq E_t$, so $E_t$ is open. Now let [L1, L2, given, choose] $K\subseteq E_t$ be compact. The balls $B(x,r_x)$ with $x\in K$ cover $K$, so compactness yields a finite subcover $B(x_1,r_1),\dots ,B(x_m,r_m)$.

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], the open set $E_t$ is the supremum of the measures of its compact [step 2.1, L3, algebra] subsets. Step 2.1 gives the same upper bound for every compact $K\subseteq E_t$, so $\lambda (E_t)\le \frac{5^n}{t}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

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