[proofplan]
We first prove the weak-type $(1,1)$ inequality using the [Vitali Covering Theorem](/theorems/3020). For each point $x$ where $\mathcal{M}f(x) > \lambda$, we select a witnessing ball $B(x, r_x)$ where the average of $|f|$ exceeds $\lambda$. The Vitali theorem extracts a disjoint subcollection whose five-fold dilates cover the super-level set, and the doubling property of $\mathcal{L}^n$ converts the dilation factor into the constant $C(n) = 5^n$. The strong-type $(p, p)$ bound for $1 < p \leq \infty$ then follows from the Marcinkiewicz interpolation theorem, interpolating between the weak $(1,1)$ bound and the immediate $L^\infty \to L^\infty$ bound.
[/proofplan]
[step:Define the super-level set and select witnessing balls]
Let $\lambda > 0$ and define the super-level set
\begin{align*}
E_\lambda := \{x \in \mathbb{R}^n : \mathcal{M}f(x) > \lambda\},
\end{align*}
where the Hardy-Littlewood maximal function is
\begin{align*}
\mathcal{M}f(x) := \sup_{r > 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} |f(y)| \, d\mathcal{L}^n(y).
\end{align*}
For each $x \in E_\lambda$, since $\mathcal{M}f(x) > \lambda$, there exists $r_x > 0$ such that
\begin{align*}
\frac{1}{\mathcal{L}^n(B(x, r_x))} \int_{B(x, r_x)} |f(y)| \, d\mathcal{L}^n(y) > \lambda.
\end{align*}
Rearranging:
\begin{align*}
\mathcal{L}^n(B(x, r_x)) < \frac{1}{\lambda} \int_{B(x, r_x)} |f(y)| \, d\mathcal{L}^n(y).
\end{align*}
[/step]
[step:Apply the Vitali Covering Theorem to extract a disjoint subcollection]
We first show the super-level set can be treated locally. Fix any compact $K \subset E_\lambda$. The collection $\mathcal{F}_K := \{\overline{B}(x, r_x) : x \in K\}$ has bounded diameters: since $f \in L^1(\mathbb{R}^n)$, the condition $\int_{B(x,r_x)} |f| \, d\mathcal{L}^n > \lambda \, \mathcal{L}^n(B(x, r_x)) = \lambda \alpha(n) r_x^n$ forces $r_x^n < \|f\|_{L^1} / (\lambda \alpha(n))$, hence $r_x < (\|f\|_{L^1} / (\lambda \alpha(n)))^{1/n}$. So the diameters are uniformly bounded.
Apply the [Vitali Covering Theorem](/theorems/3020) to $\mathcal{F}_K$: there exists a countable pairwise disjoint subcollection $\{B_j\}_{j=1}^\infty = \{\overline{B}(x_j, r_{x_j})\}_{j=1}^\infty$ such that
\begin{align*}
\bigcup_{B \in \mathcal{F}_K} B \subseteq \bigcup_{j=1}^\infty 5 B_j.
\end{align*}
Since $K \subset \bigcup_{B \in \mathcal{F}_K} B$ (each $x \in K$ is the center of $\overline{B}(x, r_x) \in \mathcal{F}_K$):
\begin{align*}
K \subseteq \bigcup_{j=1}^\infty 5 B_j.
\end{align*}
[/step]
[step:Estimate $\mathcal{L}^n(K)$ using the doubling property and the witnessing inequality]
From the covering $K \subset \bigcup_j 5B_j$ and the subadditivity of $\mathcal{L}^n$:
\begin{align*}
\mathcal{L}^n(K) \leq \sum_{j=1}^\infty \mathcal{L}^n(5 B_j).
\end{align*}
Each $5B_j = \overline{B}(x_j, 5 r_{x_j})$ has Lebesgue measure $\mathcal{L}^n(5B_j) = 5^n \mathcal{L}^n(B_j)$ (since Lebesgue measure scales as $\mathcal{L}^n(B(x, cr)) = c^n \mathcal{L}^n(B(x, r))$ for any $c > 0$). Therefore
\begin{align*}
\mathcal{L}^n(K) \leq 5^n \sum_{j=1}^\infty \mathcal{L}^n(B_j).
\end{align*}
Substituting the witnessing inequality $\mathcal{L}^n(B_j) < \frac{1}{\lambda} \int_{B_j} |f| \, d\mathcal{L}^n$:
\begin{align*}
\mathcal{L}^n(K) < \frac{5^n}{\lambda} \sum_{j=1}^\infty \int_{B_j} |f(y)| \, d\mathcal{L}^n(y).
\end{align*}
Since the balls $\{B_j\}$ are pairwise disjoint:
\begin{align*}
\sum_{j=1}^\infty \int_{B_j} |f(y)| \, d\mathcal{L}^n(y) = \int_{\bigcup_j B_j} |f(y)| \, d\mathcal{L}^n(y) \leq \int_{\mathbb{R}^n} |f(y)| \, d\mathcal{L}^n(y) = \|f\|_{L^1(\mathbb{R}^n)}.
\end{align*}
Therefore
\begin{align*}
\mathcal{L}^n(K) \leq \frac{5^n}{\lambda} \|f\|_{L^1(\mathbb{R}^n)}.
\end{align*}
[/step]
[step:Pass to the full super-level set via inner regularity]
The bound $\mathcal{L}^n(K) \leq \frac{5^n}{\lambda} \|f\|_{L^1}$ holds for every compact $K \subset E_\lambda$. The set $E_\lambda = \{\mathcal{M}f > \lambda\}$ is open, since $\mathcal{M}f$ is lower semicontinuous (the supremum of continuous functions $x \mapsto \frac{1}{\mathcal{L}^n(B(x,r))} \int_{B(x,r)} |f| \, d\mathcal{L}^n$ is lower semicontinuous). By inner regularity of Lebesgue measure on open sets:
\begin{align*}
\mathcal{L}^n(E_\lambda) = \sup\{\mathcal{L}^n(K) : K \subset E_\lambda, \, K \text{ compact}\} \leq \frac{5^n}{\lambda} \|f\|_{L^1(\mathbb{R}^n)}.
\end{align*}
This establishes the weak-type $(1, 1)$ inequality with $C(n) = 5^n$:
\begin{align*}
\mathcal{L}^n(\{x \in \mathbb{R}^n : \mathcal{M}f(x) > \lambda\}) \leq \frac{5^n}{\lambda} \|f\|_{L^1(\mathbb{R}^n)}.
\end{align*}
[/step]
[step:Establish the $L^\infty \to L^\infty$ bound]
If $f \in L^\infty(\mathbb{R}^n)$, then for every $x \in \mathbb{R}^n$ and every $r > 0$:
\begin{align*}
\frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} |f(y)| \, d\mathcal{L}^n(y) \leq \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \|f\|_{L^\infty} \, d\mathcal{L}^n(y) = \|f\|_{L^\infty}.
\end{align*}
Taking the supremum over $r > 0$:
\begin{align*}
\mathcal{M}f(x) \leq \|f\|_{L^\infty} \quad \text{for all } x \in \mathbb{R}^n.
\end{align*}
Hence $\|\mathcal{M}f\|_{L^\infty} \leq \|f\|_{L^\infty}$. The maximal operator $\mathcal{M}$ is bounded from $L^\infty(\mathbb{R}^n)$ to $L^\infty(\mathbb{R}^n)$ with operator norm at most $1$.
[/step]
[step:Deduce the strong-type $(p, p)$ bound for $1 < p < \infty$ via Marcinkiewicz interpolation]
We have established:
- **Weak-type $(1,1)$**: $\mathcal{L}^n(\{\mathcal{M}f > \lambda\}) \leq \frac{5^n}{\lambda} \|f\|_{L^1}$ for all $\lambda > 0$.
- **Strong-type $(\infty, \infty)$**: $\|\mathcal{M}f\|_{L^\infty} \leq \|f\|_{L^\infty}$.
The maximal operator $\mathcal{M}$ is sublinear: $\mathcal{M}(f + g)(x) \leq \mathcal{M}f(x) + \mathcal{M}g(x)$ (since averages are linear and the supremum of a sum is at most the sum of suprema) and $\mathcal{M}(\alpha f) = |\alpha| \mathcal{M}f$ for all $\alpha \in \mathbb{R}$.
By the [Marcinkiewicz Interpolation Theorem](/theorems/???), any sublinear operator that is weak-type $(1, 1)$ and strong-type $(\infty, \infty)$ is strong-type $(p, p)$ for all $1 < p < \infty$. Applying this to $\mathcal{M}$: for each $1 < p < \infty$, there exists $C(n, p) > 0$ depending only on $n$ and $p$ such that
\begin{align*}
\|\mathcal{M}f\|_{L^p(\mathbb{R}^n)} \leq C(n, p) \|f\|_{L^p(\mathbb{R}^n)}
\end{align*}
for all $f \in L^p(\mathbb{R}^n)$.
The explicit dependence on $p$ is $C(n, p) \lesssim (p/(p-1))^{1/p} \cdot 5^{n/p}$, which blows up as $p \to 1^+$, reflecting the failure of the strong-type $(1,1)$ bound.
[/step]
[step:Conclude with the case $p = \infty$]
For $f \in L^\infty(\mathbb{R}^n)$, the strong-type bound was already established: $\|\mathcal{M}f\|_{L^\infty} \leq \|f\|_{L^\infty}$. Combined with the weak-type $(1,1)$ inequality and the strong-type $(p, p)$ bounds for $1 < p < \infty$, this completes the proof of both parts of the theorem.
The constant $C(n) = 5^n$ in the weak-type bound and $C(n, p)$ in the strong-type bound depend only on the stated parameters. In summary:
\begin{align*}
&\mathcal{L}^n(\{\mathcal{M}f > \lambda\}) \leq \frac{5^n}{\lambda} \|f\|_{L^1} \quad \text{for all } f \in L^1(\mathbb{R}^n), \, \lambda > 0, \\
&\|\mathcal{M}f\|_{L^p} \leq C(n, p) \|f\|_{L^p} \quad \text{for all } f \in L^p(\mathbb{R}^n), \, 1 < p \leq \infty.
\end{align*}
[/step]
