Prove that the Hardy-Littlewood maximal function \(f^*\) for an integrable function \(f\) satisfies
\[m(\{x \in \mathbb{R}^d : f^*(x) > \alpha\}) \leq \frac{3^d}{\alpha} \|f\|_{L^1(\mathbb{R}^d)}\]
where \(\alpha > 0\). Recall that
\[f^*(x) = \sup_{x \in B} \frac{1}{m(B)} \int_B |f(y)| \, dy, \quad x \in \mathbb{R}^d\]
where the supremum is taken over all balls containing the point \(x\). You may assume the Vitali 3-times Covering Lemma. State it clearly if you use it.
Solution
1. State the covering lemma.
Vitali \(3\)-times covering lemma: every finite collection of balls in \(\mathbb R^d\) has a pairwise disjoint subcollection \(B_1,\dots,B_N\) such that the union of the whole collection is contained in \(\bigcup_{j=1}^N3B_j\), where \(3B_j\) is the concentric ball with three times the radius.
2. The superlevel set is open and is a union of balls with large averages.
Take the balls in the definition of \(f^*\) to be open. A closed ball containing \(x\) lies in a slightly larger concentric open ball, whose average of \(|f|\) is arbitrarily close, so this does not change \(f^*\).
Fix \(\alpha>0\) and set \[ E_\alpha:=\{x\in\mathbb R^d:f^*(x)>\alpha\}. \] For each \(x\in E_\alpha\), choose an open ball \(B_x\ni x\) such that \[ \frac1{m(B_x)}\int_{B_x}|f(y)|\,dy>\alpha. \] Every point of \(B_x\) then lies in \(E_\alpha\), so \(E_\alpha=\bigcup_{x\in E_\alpha}B_x\) is open, hence measurable.
3. Every compact subset of \(E_\alpha\) satisfies the bound.
Let \(K\subset E_\alpha\) be compact. Finitely many of the balls \(B_x\) cover \(K\). Step 1 gives pairwise disjoint balls \(B_1,\dots,B_N\) among them with \[ K\subseteq\bigcup_{j=1}^N3B_j. \] Each \(B_j\) satisfies \(\alpha m(B_j)<\int_{B_j}|f|\), and the \(B_j\) are pairwise disjoint. Therefore \[ m(K) \le\sum_{j=1}^N m(3B_j) =3^d\sum_{j=1}^N m(B_j) <\frac{3^d}{\alpha}\sum_{j=1}^N\int_{B_j}|f| \le\frac{3^d}{\alpha}\|f\|_1. \]
4. Inner regularity gives the bound for \(E_\alpha\).
Lebesgue measure is inner regular on measurable sets, so \(m(E_\alpha)=\sup\{m(K):K\subset E_\alpha\text{ compact}\}\) [@Fol13]. Step 3 then gives \[ \boxed{ m(\{x:f^*(x)>\alpha\}) \le\frac{3^d}{\alpha}\|f\|_{L^1(\mathbb R^d)}.} \]