Theorem 24.7.5 (The Maximal Theorem).label Let $\mu: \cb_{\real^d}\to [0, \infty)$ be a finite Borel measure, and $m^{*}$ be the Lebesgue outer[1] measure on $\real^{d}$, then for each $\alpha > 0$,
Proof. For each $x \in \bracsn{H\mu > \alpha}$, there exists $r_{x} > 0$ such that $A_{r_x}\mu(x) > \alpha$. Let $U = \bigcup_{x \in B}B_{\real^d}(x, r_{x})$, then $U \supset \bracsn{H\mu > \alpha}$.
Let $c \in [0, m(U))$, then by the Vitali Covering Lemma, there exists $A \subset B$ finite such that $\bracsn{B_{\real^d}(x, r_x)}_{x \in A}$ is pairwise disjoint and $3^{d}\sum_{x \in A}m(B_{\real^d}(x, r_{x})) \ge c$. In which case,
As the above holds for all $c \in [0, m(U))$, $m^{*}\bracsn{H\mu > \alpha}\le m(U) \le 3^{d}\mu(\real^{d})/\alpha$.$\square$
- Measurability of $H\mu$ is not guaranteed.keyboard_return
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