24.7 The Lebesgue Differentiation Theorem
Lemma 24.7.1 (Vitali Covering Lemma).label 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:
- (1)
$\bracsn{B_{\real^d}(x_j, \eps_j)}_{j \in J}$ is pairwise disjoint.
- (2)
$m\paren{\bigsqcup_{j \in J}B_{\real^d}(x_j, \eps_j)}> 3^{-d}c$.
Proof, [Lemma 3.15, Fol99]. 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 $K \subset \bigcup_{j \in J_1}B_{\real^d}(x_{j}, \eps_{j})$.
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
As $J$ is finite, there exists $n \in \natp$ such that $J_{n+1}= \emptyset$. The above construction then yields a pairwise disjoint family $\bracsn{B_{\real^d}(x_{i_k}, \eps_{i_k})}_{1}^{n}$.
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
$\square$
Definition 24.7.2 (Average).label 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
is the average of $\mu$ on $B_{\real^d}(x, r)$.
Lemma 24.7.3.label 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.
Proof. 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 $n \to \infty$.
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 Dominated Convergence Theorem implies that $A_{r_n}\mu(x_{n}) \to A_{r}\mu(x)$ as $n \to \infty$.$\square$
Definition 24.7.4 (Hardy-Littlewood Maximal Function).label Let $\mu: \cb_{\real^d}\to [0, \infty]$ be a regular Borel measure on $\real^{d}$, then
is the Hardy-Littlewood maximal function of $\mu$.
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.
Proof. 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 Lemma 24.7.3. Since $H\mu$ is a supremum of continuous functions, it is lower semicontinuous by Proposition 5.22.3.$\square$
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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