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\section{The Lebesgue Differentiation Theorem}
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\label{section:lebesgue-differentiation}
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\begin{lemma}[Vitali Covering Lemma]
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\label{lemma:vitali-covering}
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Let $\bracsn{B_{\real^d}(x_i, \eps_i)}_{i \in I}$ be open balls in $\real^d$, $U = \bigcup_{i \in I}B_{\real^d}(x_i, \eps)$, and $m$ be the Lebesgue measure on $\real^d$, then for any $c \in [0, m(U))$, there exists $J \subset I$ finite such that:
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\begin{enumerate}
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\item $\bracsn{B_{\real^d}(x_j, \eps_j)}_{j \in J}$ is pairwise disjoint.
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\item $m\paren{\bigsqcup_{j \in J}B_{\real^d}(x_j, \eps_j)} > 3^{-d}c$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma 3.15]{Folland}}}. ]
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By inner regularity of the Lebesgue measure, there exists $K \subset U$ compact such that $m(K) > c$. In which case, there exists $J_1 \subset I$ finite such that $K \subset \bigcup_{j \in J_1}B_{\real^d}(x_j, \eps_j)$.
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For each $n \in \natp$, if $J_n \ne \emptyset$, let $j_n \in J_n$ such that $\eps_{j_n} = \max_{j \in J_n}\eps_j$, and
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\[
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J_{n+1} = \bracs{j \in J \bigg | B_{\real^d}(x_j, \eps_j) \cap \bigcup_{k = 1}^n B_{\real^d}(x_{i_k}, \eps_{i_k}) = \emptyset}
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\]
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As $J$ is finite, there exists $n \in \natp$ such that $J_{n+1} = \emptyset$. The above construction then yields a pairwise disjoint family $\bracsn{B_{\real^d}(x_{i_k}, \eps_{i_k})}_1^n$.
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For each $j \in J$, let $1 \le k \le n$ be the smallest $k$ such that $B_{\real^d}(x_j, \eps_j) \cap B_{\real^d}(x_{i_k}, \eps_{i_k}) \ne \emptyset$, then $j \in J_k$, and $\eps_j \le \eps_{i_k}$. Thus $B_{\real^d}(x_j, \eps_j) \subset B_{\real^d}(x_{i_k}, 3\eps_{i_k})$. Therefore
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\begin{align*}
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c < m(K) &\le m\paren{\bigcup_{j \in J}B_{\real^d}(x_j, \eps_j)} \le m\paren{\bigcup_{k = 1}^n B_{\real^d}(x_{i_k}, 3\eps_{i_k})} \\
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&\le \sum_{k = 1}^n m(B_{\real^d}(x_{i_k}, 3\eps_{i_k})) = 3^d\sum_{k = 1}^n m(B_{\real^d}(x_{i_k}, \eps_{i_k}))
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\end{align*}
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\end{proof}
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\begin{definition}[Average]
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\label{definition:hl-average}
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Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$ and $m$ be the Lebesgue measure on $\real^d$, then
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\[
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(0, \infty) \times \real^d \to [0, \infty) \quad (r, x) \mapsto A_r\mu(x) = \frac{\mu(B_{\real^d}(x, r))}{m(B_{\real^d}(x, r))}
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\]
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is the \textbf{average} of $\mu$ on $B_{\real^d}(x, r)$.
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\end{definition}
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\begin{lemma}
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\label{lemma:hl-average-continuous}
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Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$ such that $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then the average $(r, x) \mapsto A_r\mu(x)$ is jointly continuous.
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\end{lemma}
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\begin{proof}
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Let $\seq{(r_n, x_n)} \subset (0, \infty) \times \real^d$ and $(r, x) \in (0, \infty) \times \real^d$ such that $(r_n, x_n) \to (r, x)$. Given that $\mu(\partial B_{\real^d}(y, s)) = 0$ for all $x \in \real^d$ and $s > 0$, $\one_{B_{\real^d}(x_n, r_n)} \to \one_{B_{\real^d}(x, r)}$ $\mu$-a.e. and Lebesgue-a.e. as $n \to \infty$.
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Since $\seq{(r_n, x_n)}$ is convergent, there exists $K \subset \real^d$ compact such that $B_{\real^d}(x, r) \cup \bigcup_{n \in \natp}B_{\real^d}(x_n, r_n) \subset K$. By regularity of $\mu$, $\mu(K) < \infty$. Thus the \hyperref[Dominated Convergence Theorem]{theorem:dct} implies that $A_{r_n}\mu(x_n) \to A_{r}\mu(x)$ as $n \to \infty$.
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\end{proof}
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\begin{definition}[Hardy-Littlewood Maximal Function]
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\label{definition:hl-maximal-function}
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Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$, then
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\[
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H\mu: \real^d \to [0, \infty) \quad x \mapsto \sup_{r > 0}A_r\mu(x)
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\]
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is the \textbf{Hardy-Littlewood maximal function} of $\mu$.
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If $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then $H\mu$ is lower semicontinuous, and in particular Borel measurable.
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\end{definition}
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\begin{proof}
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If $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then $(r, x) \mapsto A_{r}\mu(x)$ is jointly continuous by \autoref{lemma:hl-average-continuous}. Since $H\mu$ is a supremum of continuous functions, it is lower semicontinuous by \autoref{proposition:semicontinuous-properties}.
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\end{proof}
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\begin{theorem}[The Maximal Theorem]
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\label{theorem:hl-maximal-theorem}
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Let $\mu: \cb_{\real^d} \to [0, \infty)$ be a finite Borel measure, and $m^*$ be the Lebesgue outer\footnote{Measurability of $H\mu$ is not guaranteed.} measure on $\real^d$, then for each $\alpha > 0$,
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\[
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m^*\bracsn{H\mu > \alpha} \le \frac{3^d \mu(\real^d)}{\alpha}
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\]
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\end{theorem}
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\begin{proof}
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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 $U \supset \bracsn{H\mu > \alpha}$.
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Let $c \in [0, m(U))$, then by the \hyperref[Vitali Covering Lemma]{lemma:vitali-covering}, 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,
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\begin{align*}
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c &\le 3^d \sum_{x \in A}m(B_{\real^d}(x, r_x)) \le \frac{3^d}{\alpha}\sum_{x \in A}\mu(B_{\real^d}(x, r_x)) \le \frac{3^d\mu(\real^d)}{\alpha}
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\end{align*}
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As the above holds for all $c \in [0, m(U))$, $m^*\bracsn{H\mu > \alpha} \le m(U) \le 3^d\mu(\real^d)/\alpha$.
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\end{proof}
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Bokuan Li