[proofplan]
We apply the [Calderón-Zygmund Decomposition](/theorems/3154) at level $\lambda$ to obtain disjoint dyadic cubes $\{Q_j\}_{j \in J}$ with $|f| \le \lambda$ a.e. off $\Omega := \bigsqcup_j Q_j$, $\fint_{Q_j}|f|\, d\mathcal{L}^n \le 2^n\lambda$ on each cube, and $\sum |Q_j| \le \|f\|_{L^1}/\lambda$. Defining $\Omega^* := \bigcup_j 2Q_j$ (the union of doubled cubes), we show that for $\mathcal{L}^n$-a.e. $x \notin \Omega^*$, $Mf(x) \le C'_n\,\lambda$ for an explicit dimensional constant $C'_n = 1 + 2^n(1+2\sqrt n)^n$. The argument: for any ball $B(x, r)$, decompose $B = (B \setminus \Omega) \sqcup \bigsqcup_j (B \cap Q_j)$. Off $\Omega$, $|f| \le \lambda$. The geometric observation $x \notin 2Q_j$ together with the requirement $B \cap Q_j \ne \varnothing$ forces $\ell(Q_j) < 2r$, so the cubes meeting $B$ are contained in a ball of radius $(1 + 2\sqrt n) r$ centred at $x$, controlled by the upper average bound $\fint_{Q_j}|f| \le 2^n\lambda$. Combining gives $Mf(x) \le C'_n\,\lambda$ on the complement of $\Omega^*$, so $\{Mf > C'_n\,\lambda\} \subseteq \Omega^*$ modulo a null set. The volume bound $|\Omega^*| \le 2^n |\Omega| \le 2^n \|f\|_{L^1}/\lambda$ together with rescaling $\lambda \mapsto \lambda/C'_n$ gives the weak-(1,1) bound with constant $C_n = 2^n\,C'_n$.
[/proofplan]
[step:Apply the Calderón-Zygmund decomposition at level $\lambda$]
Let $f \in L^1(\mathbb{R}^n)$ and $\lambda > 0$ be as in the statement. Apply the [Calderón-Zygmund Decomposition](/theorems/3154) to $f$ at level $\lambda$: there exists a countable collection of disjoint dyadic cubes $\{Q_j\}_{j \in J} \subset \mathcal{D}$ such that, writing $\Omega := \bigsqcup_{j \in J} Q_j$,
\begin{align*}
|f(x)| &\le \lambda \quad \text{for $\mathcal{L}^n$-a.e. } x \notin \Omega, \\
\lambda < \frac{1}{|Q_j|}\int_{Q_j}|f|\, d\mathcal{L}^n &\le 2^n \lambda \quad \text{for every } j \in J, \\
\sum_{j \in J} |Q_j| &\le \frac{\|f\|_{L^1}}{\lambda}.
\end{align*}
The hypotheses of the [Calderón-Zygmund Decomposition](/theorems/3154) — $f \in L^1(\mathbb{R}^n)$ and $\lambda > 0$ — are exactly those given.
Define the **doubled-cube set**
\begin{align*}
\Omega^* := \bigcup_{j \in J} 2 Q_j,\qquad \text{where } 2Q_j := \{x : \operatorname{dist}_\infty(x, \text{centre}(Q_j)) \le \ell(Q_j)\}
\end{align*}
is the cube concentric with $Q_j$ and side length $2\ell(Q_j)$, hence volume $|2Q_j| = 2^n |Q_j|$.
[/step]
[step:Show that $Mf(x) \le \lambda$ for $\mathcal{L}^n$-a.e. $x \notin \Omega^*$]
Fix $x \notin \Omega^*$ at which the off-set bound $|f(x)| \le \lambda$ holds for a.e.\ point of $\mathbb{R}^n \setminus \Omega$ (the exceptional null set is negligible). We show $Mf(x) \le \lambda$.
Take any $r > 0$ and consider the Euclidean ball $B := B(x, r)$ of volume $|B| = \omega_n r^n$, where $\omega_n$ is the Lebesgue volume of the unit ball in $\mathbb{R}^n$.
[claim:Geometric bound on cubes touching $B$]
For any $j \in J$ with $Q_j \cap B \ne \varnothing$, the side length satisfies $\ell(Q_j) < 2r$.
[/claim]
[proof]
Pick $w \in Q_j \cap B$. Then $|x - w| < r$. Since $x \notin 2Q_j$, the $\ell^\infty$-distance from $x$ to the centre $c_j$ of $Q_j$ satisfies $\|x - c_j\|_\infty > \ell(Q_j)$. Since $w \in Q_j$, $\|w - c_j\|_\infty \le \ell(Q_j)/2$. By the reverse triangle inequality,
\begin{align*}
\|x - w\|_\infty \ge \|x - c_j\|_\infty - \|w - c_j\|_\infty > \ell(Q_j) - \frac{\ell(Q_j)}{2} = \frac{\ell(Q_j)}{2}.
\end{align*}
The Euclidean norm dominates the $\ell^\infty$-norm: $|x - w| \ge \|x - w\|_\infty > \ell(Q_j)/2$. Combined with $|x - w| < r$, $\ell(Q_j)/2 < r$, i.e., $\ell(Q_j) < 2r$.
[/proof]
[claim:Volume bound on the union of cubes meeting $B$]
The disjoint union $\bigsqcup_{j: Q_j \cap B \ne \varnothing} Q_j$ has Lebesgue volume at most $|B|$.
[/claim]
[proof]
Each touching cube has $\ell(Q_j) < 2r$, so $\operatorname{diam}(Q_j) = \sqrt{n}\,\ell(Q_j) < 2\sqrt{n}\,r$. For any point $u$ in such a cube, the Euclidean distance from $u$ to $x$ satisfies
\begin{align*}
|u - x| \le |u - w| + |w - x| \le \operatorname{diam}(Q_j) + r < 2\sqrt{n}\,r + r = (1 + 2\sqrt{n})\,r,
\end{align*}
where $w \in Q_j \cap B$. Hence $\bigcup_{j: Q_j \cap B \ne \varnothing} Q_j \subseteq B(x, (1 + 2\sqrt{n})\,r)$. By disjointness of the $Q_j$,
\begin{align*}
\sum_{j: Q_j \cap B \ne \varnothing} |Q_j| \le \bigl|B(x, (1 + 2\sqrt{n})\,r)\bigr| = \omega_n\,(1 + 2\sqrt{n})^n\,r^n = (1 + 2\sqrt{n})^n |B|.
\end{align*}
[/proof]
Now compute the average of $|f|$ over $B$. Decompose
\begin{align*}
\int_B |f|\, d\mathcal{L}^n = \int_{B \setminus \Omega}|f|\, d\mathcal{L}^n + \sum_{j: Q_j \cap B \ne \varnothing}\int_{B \cap Q_j}|f|\, d\mathcal{L}^n.
\end{align*}
The first term: since $|f| \le \lambda$ a.e.\ on $\mathbb{R}^n \setminus \Omega$,
\begin{align*}
\int_{B \setminus \Omega}|f|\, d\mathcal{L}^n \le \lambda\,|B \setminus \Omega| \le \lambda\,|B|.
\end{align*}
The second term: enlarging the integration domain from $B \cap Q_j$ to $Q_j$ and using the CZ upper bound $\int_{Q_j}|f|\, d\mathcal{L}^n \le 2^n\lambda |Q_j|$,
\begin{align*}
\sum_{j: Q_j \cap B \ne \varnothing}\int_{B \cap Q_j}|f|\, d\mathcal{L}^n \le \sum_{j: Q_j \cap B \ne \varnothing}\int_{Q_j}|f|\, d\mathcal{L}^n \le 2^n\lambda \sum_{j: Q_j \cap B \ne \varnothing}|Q_j| \le 2^n\lambda\,(1 + 2\sqrt{n})^n |B|.
\end{align*}
Combining,
\begin{align*}
\frac{1}{|B|}\int_B |f|\, d\mathcal{L}^n \le \lambda + 2^n(1 + 2\sqrt{n})^n\,\lambda = C'_n\,\lambda,
\end{align*}
where $C'_n := 1 + 2^n(1 + 2\sqrt{n})^n$ is a dimensional constant.
Taking the supremum over $r > 0$,
\begin{align*}
Mf(x) \le C'_n\,\lambda \quad \text{for $\mathcal{L}^n$-a.e. } x \notin \Omega^*.
\end{align*>
[/step]
[step:Bound the volume of $\Omega^*$ via $|2Q_j| = 2^n |Q_j|$]
By the volume scaling of the doubled cubes and subadditivity (the $2Q_j$ may overlap),
\begin{align*}
|\Omega^*| = \left|\bigcup_{j \in J} 2Q_j\right| \le \sum_{j \in J}|2Q_j| = \sum_{j \in J} 2^n|Q_j| = 2^n \sum_{j \in J}|Q_j| \le 2^n\,\frac{\|f\|_{L^1}}{\lambda},
\end{align*}
using the CZ volume bound $\sum_j |Q_j| \le \|f\|_{L^1}/\lambda$ in the final inequality. Hence $|\Omega^*| \le 2^n \|f\|_{L^1}/\lambda$.
[/step]
[step:Rescale and combine to obtain the weak-(1,1) bound with constant $C_n = 2^n\,C'_n$]
Step 2 shows that $\{Mf > C'_n\,\lambda\} \subseteq \Omega^*$ modulo a $\mathcal{L}^n$-null set, where $\Omega^*$ is associated to the CZ decomposition at level $\lambda$ and $C'_n = 1 + 2^n(1 + 2\sqrt{n})^n$. Combined with the volume bound from step 3, $|\Omega^*| \le 2^n \|f\|_{L^1}/\lambda$,
\begin{align*}
\bigl|\{x \in \mathbb{R}^n : Mf(x) > C'_n\,\lambda\}\bigr| \le |\Omega^*| \le \frac{2^n}{\lambda}\,\|f\|_{L^1(\mathbb{R}^n)}.
\end{align*}
Now we rescale to obtain a bound at any level $\mu > 0$. Substitute $\lambda := \mu/C'_n$ (i.e., apply the CZ decomposition at level $\mu/C'_n$ instead of $\mu$). The above bound becomes
\begin{align*}
\bigl|\{x \in \mathbb{R}^n : Mf(x) > \mu\}\bigr| \le \frac{2^n}{\mu/C'_n}\,\|f\|_{L^1} = \frac{2^n\,C'_n}{\mu}\,\|f\|_{L^1}.
\end{align*}
Defining $C_n := 2^n\,C'_n = 2^n\,(1 + 2^n(1 + 2\sqrt{n})^n)$, we conclude
\begin{align*}
\bigl|\{x \in \mathbb{R}^n : Mf(x) > \mu\}\bigr| \le \frac{C_n}{\mu}\,\|f\|_{L^1(\mathbb{R}^n)} \qquad \text{for all } \mu > 0,
\end{align*}
which is the desired weak-$(1, 1)$ bound. The level set $\{Mf > \mu\}$ is Borel (since $Mf$ is Borel as the supremum of a continuous family of averages of $|f|$, which is itself Borel), so $|\cdot|$ refers to the Lebesgue measure of a Borel set, and the chain of inequalities is well-defined.
[/step]
