From 3a81f8833aa580638c215a566b15e18e246b096c Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 16 Sep 2026 11:23:13 -0400 Subject: [PATCH] Housekeeping with the maximal theorem. --- src/fa/norm/normed.tex | 13 ++++++ src/measure/notation.tex | 1 + src/measure/vector/index.tex | 1 + src/measure/vector/ld.tex | 84 ++++++++++++++++++++++++++++++++++++ 4 files changed, 99 insertions(+) create mode 100644 src/measure/vector/ld.tex diff --git a/src/fa/norm/normed.tex b/src/fa/norm/normed.tex index a79b18d..62a06f5 100644 --- a/src/fa/norm/normed.tex +++ b/src/fa/norm/normed.tex @@ -148,3 +148,16 @@ Thus $d(x_N, \text{span}\bracsn{x_n|1 \le n < N}) = \norm{x_N + F_N}_{E/F} > \alpha_N$. \end{proof} +\begin{proposition} +\label{proposition:finite-dim-norm} + Let $K \in \RC$ and $d \in \natp$, then every norm on $K^d$ is equivalent to each other. +\end{proposition} +\begin{proof} + Let $[\cdot]: K^d \to [0, \infty)$ be an arbitrary and + \[ + \norm{\cdot}_{1}: K^d \to [0, \infty) \quad x \mapsto \sum_{j = 1}^n |x_j| + \] + + then the identity $(K^d, \norm{\cdot}_1) \to (K^d, [\cdot])$ is continuous. Let $S = \bracsn{x \in K^d|\ \norm{x}_1 = 1}$, then $S$ is compact with respect to $\norm{\cdot}_1$ and hence compact with respect to $[\cdot]$. By continuity of $[\cdot]$, $\min_{x \in S}[x]$ exists and is non-zero. +\end{proof} + diff --git a/src/measure/notation.tex b/src/measure/notation.tex index ad1976a..3ec9d15 100644 --- a/src/measure/notation.tex +++ b/src/measure/notation.tex @@ -32,6 +32,7 @@ $\mathcal{L}^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, without quotient. & \autoref{definition:lp-unequivalence} \\ $\|f\|_{L^p}$, & $L^p$ norm of $f$. & \autoref{definition:esssup} \\ $L^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, modulo equality almost everywhere. & \autoref{definition:lp} \\ + $A_r\mu(x)$ & Average of $\mu$ at $B_{\real^d}(x, r)$. & \autoref{definition:hl-average} \end{tabular} diff --git a/src/measure/vector/index.tex b/src/measure/vector/index.tex index 88ec265..c97d6cb 100644 --- a/src/measure/vector/index.tex +++ b/src/measure/vector/index.tex @@ -7,5 +7,6 @@ \input{./ac.tex} \input{./ms.tex} \input{./rn.tex} +\input{./ld.tex} \input{./fin.tex} diff --git a/src/measure/vector/ld.tex b/src/measure/vector/ld.tex new file mode 100644 index 0000000..f60ede1 --- /dev/null +++ b/src/measure/vector/ld.tex @@ -0,0 +1,84 @@ +\section{The Lebesgue Differentiation Theorem} +\label{section:lebesgue-differentiation} + +\begin{lemma}[Vitali Covering Lemma] +\label{lemma:vitali-covering} + 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: + \begin{enumerate} + \item $\bracsn{B_{\real^d}(x_j, \eps_j)}_{j \in J}$ is pairwise disjoint. + \item $m\paren{\bigsqcup_{j \in J}B_{\real^d}(x_j, \eps_j)} > 3^{-d}c$. + \end{enumerate} +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma 3.15]{Folland}}}. ] + 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 + \[ + J_{n+1} = \bracs{j \in J \bigg | B_{\real^d}(x_j, \eps_j) \cap \bigcup_{k = 1}^n B_{\real^d}(x_{i_k}, \eps_{i_k}) = \emptyset} + \] + + 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 + \begin{align*} + 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})} \\ + &\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})) + \end{align*} +\end{proof} + +\begin{definition}[Average] +\label{definition:hl-average} + 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 + \[ + (0, \infty) \times \real^d \to [0, \infty) \quad (r, x) \mapsto A_r\mu(x) = \frac{\mu(B_{\real^d}(x, r))}{m(B_{\real^d}(x, r))} + \] + + is the \textbf{average} of $\mu$ on $B_{\real^d}(x, r)$. +\end{definition} + +\begin{lemma} +\label{lemma:hl-average-continuous} + 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. +\end{lemma} +\begin{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 \hyperref[Dominated Convergence Theorem]{theorem:dct} implies that $A_{r_n}\mu(x_n) \to A_{r}\mu(x)$ as $n \to \infty$. +\end{proof} + + +\begin{definition}[Hardy-Littlewood Maximal Function] +\label{definition:hl-maximal-function} + Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$, then + \[ + H\mu: \real^d \to [0, \infty) \quad x \mapsto \sup_{r > 0}A_r\mu(x) + \] + + is the \textbf{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. +\end{definition} +\begin{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 \autoref{lemma:hl-average-continuous}. Since $H\mu$ is a supremum of continuous functions, it is lower semicontinuous by \autoref{proposition:semicontinuous-properties}. +\end{proof} + +\begin{theorem}[The Maximal Theorem] +\label{theorem:hl-maximal-theorem} + 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 $\alpha > 0$, + \[ + m^*\bracsn{H\mu > \alpha} \le \frac{3^d \mu(\real^d)}{\alpha} + \] +\end{theorem} +\begin{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 \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, + \begin{align*} + 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} + \end{align*} + + As the above holds for all $c \in [0, m(U))$, $m^*\bracsn{H\mu > \alpha} \le m(U) \le 3^d\mu(\real^d)/\alpha$. +\end{proof} + + +