Added the definition of the Lebesgue-Stieltjes measure.

src/measure/measure/index.tex

@@ -6,3 +6,5 @@
 \input{./src/measure/measure/semifinite.tex}
 \input{./src/measure/measure/sigma-finite.tex}
 \input{./src/measure/measure/outer.tex}
+\input{./src/measure/measure/regular.tex}
+\input{./src/measure/measure/lebesgue-stieltjes.tex}

src/measure/measure/lebesgue-stieltjes.tex

@@ -0,0 +1,104 @@
+\section{Lebesgue-Stieltjes Measures}
+\label{section:lebesgue-stieltjes}
+
+\begin{definition}[Stieltjes Function]
+\label{definition:stieltjes-function}
+    Let $F: \real \to \real$, then $F$ is a \textbf{Stieltjes function} if $F$ is right-continuous and non-decreasing.
+\end{definition}
+
+\begin{definition}[h-interval]
+\label{definition:h-interval}
+    A \textbf{h-interval} is an interval of the form $(a, b] \subset \real$ where $-\infty < a \le b < \infty$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:h-interval-ring}
+    Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the collection of all $h$-intervals, then:
+    \begin{enumerate}
+        \item $\ci$ is an elementary family.
+        \item The collection
+        \[
+        \alg = \bracs{\bigsqcup_{j = 1}^n I_j \bigg | \seqf{I_j} \subset \ci, n \in \natp}
+        \]
+        is a ring.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $(a, b], (c, d] \in \ci$ and assume without loss of generality that $a < b$ and $c < d$, then
+    \begin{enumerate}
+        \item[(P2)] $(a, b] \cap (c, d] = (\max(a, c), \min(b, d)] \in \ci$.
+        \item[(E)] $(a, b] \setminus (c, d] = (a, \min(b, c)] \sqcup (\max(a, d), b]$.
+    \end{enumerate}
+
+    (2): By \ref{proposition:elementary-family-algebra}, $\alg$ is a ring.
+\end{proof}
+
+\begin{definition}[Lebesgue-Stieltjes Measure, {{\cite[Theorem 1.16]{Folland}}}]
+\label{definition:lebesgue-stieltjes-measure}
+    Let $\mu: \cb_\real \to [0, \infty]$ be a Borel measure on $\real$ such that for any $K \subset \real$ compact, $\mu(K) < \infty$, then
+    \begin{enumerate}
+        \item There exists a Stieltjes function $F: \real \to \real$ such that for all $-\infty < a < b < \infty$,
+        \[
+        \mu((a, b]) = F(b) - F(a)
+        \]
+        \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
+    \end{enumerate}
+    Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
+\end{definition}
+\begin{proof}
+    (1): Let
+    \[
+    F: \real \to \real \quad x \mapsto \begin{cases}
+        \mu((0, x]) &x \ge 0 \\
+        -\mu((x, 0]) &x \le 0
+    \end{cases}
+    \]
+    then $F$ is a Stieltjes function by monotonicity and continuity from above (\ref{proposition:measure-properties}). For any $-\infty < a < b < \infty$,
+    \[
+    \mu((a, b]) = \begin{cases}
+        \mu((0, b]) - \mu((0, a]) &a, b \ge 0 \\
+        \mu((a, 0]) + \mu((0, b]) &a \le 0, b \ge 0 \\
+        \mu((a, 0]) - \mu((b, 0]) &a, b \le 0
+    \end{cases}
+    \]
+    In all three cases, $\mu((a, b]) = F(b) - F(a)$.
+
+    (U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
+
+
+    Converse: Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
+    \[
+    \alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
+    \]
+    Define
+    \[
+    \mu_0: \alg \to [0, \infty) \quad \bigsqcup_{j = 1}^n (a_j, b_j] \mapsto \sum_{j = 1}^n [F(b_j) - F(a_j)]
+    \]
+    Let $(a, b] \in \ci$ with $(a, b] = \bigsqcup_{j = 1}^n (a_j, b_j]$. Assume without loss of generality that $\seqf{a_j}$ is non-decreasing, then $b_j = a_{j+1}$ for each $1 \le j \le n - 1$, $b = b_n$, and $a = a_1$. Thus
+    \begin{align*}
+    F(b) - F(a) &= F(b_n) + \sum_{j = 1}^{n-1} [F(b_{j}) - F(b_j)] - F(a_1) \\
+    &= F(b_n) + \sum_{j = 1}^{n-1} [F(b_j) - F(a_{j+1})] - F(a_1) = \sum_{j = 1}^n[F(b_j) - F(a_j)]
+    \end{align*}
+    and $\mu_0$ is well-defined and finitely-additive on $\alg$.
+
+    Let $\seq{(a_n, b_n]} \subset \ci$ such that $\bigcup_{n \in \natp}(a_n, b_n] = (a, b]$. By monotonicity,
+    \[
+    \sum_{k = 1}^n \mu_0((a_k, b_k]) = \mu_0 \paren{\bigsqcup_{k = 1}^n (a_k, b_k]} \le \mu_0((a, b])
+    \]
+    Let $\eps > 0$. By right-continuity of $F$, there exists $\delta > 0$ such that $F(a + \delta) - F(a) < \eps/2$, and $\seq{\delta_n} \subset \real_{> 0}$ such that $F(b_n + \delta_n) - F(b_n) < \eps/2^{n+1}$ for all $n \in \natp$. The intervals $\seq{(a_n, b_n + \delta_n)}$ forms an open cover for $[a + \delta, b]$. By compactness, there exists $N \in \natp$ such that
+    \[
+    [a + \delta, b] \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n) \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n]
+    \]
+    Thus
+    \begin{align*}
+        \mu_0((a, b]) &\le F(a + \delta) - F(a) + \mu_0([a + \delta, b]) \\
+        &\le \eps/2 + \sum_{n = 1}^N \mu_0((a_n, b_n + \delta_n]) \\
+        &\le \eps/2 + \sum_{n = 1}^N [\mu_0((a_n, b_n]) + F(b_n + \delta_n) - F(b_n)] \\
+        &\le \eps/2 + \sum_{n = 1}^N [\mu_0((a_n, b_n]) + \eps/2^{n+1}] \le \eps + \sum_{n = 1}^N\mu_0((a_n, b_n])
+    \end{align*}
+    Since such an $N$ exists for each $\eps > 0$,
+    \[
+    \mu_0((a, b]) = \sum_{n \in \natp}\mu_0((a_n, b_n])
+    \]
+    Therefore $\mu_0$ is a premeasure on $\alg$. By Carathéodory's Extension Theorem (\ref{theorem:caratheodory-extension}), $\mu_0$ extends uniquely to a Borel measure on $\real$, which satisfies (1).
+\end{proof}

src/measure/measure/outer.tex

@@ -16,13 +16,15 @@
 
 \begin{proposition}[{{\cite[Proposition 1.10]{Folland}}}]
 \label{proposition:outer-measure-inf}
-    Let $X$ be a set, $\ce \subset 2^X$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset, X \in \ce$ and $\rho(\emptyset) = 0$. Define
+    Let $X$ be a set, $\ce \subset 2^X$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset \in \ce$ and $\rho(\emptyset) = 0$. Define
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}
     \]
     then $\mu^*$ is an outer measure.
 \end{proposition}
 \begin{proof}
+    (OM1): $\emptyset \in \ce$ and $\rho(\emptyset) = 0$.
+
     (OM2): Let $\seq{E_n} \subset 2^X$ and assume without loss of generality that $\sum_{n \in \natp}\mu^*(E_n) < \infty$. Let $\eps > 0$. For each $n \in \nat$, let $\bracs{F_{n, k}}_{k = 1}^\infty \subset \ce$ such that $\bigcup_{k \in \natp}F_{n, k} \supset E_n$ and $\sum_{n \in \natp}\rho(F_{n, k}) < \mu^*(E_n) + 2^{-n}\eps$, then
     \begin{align*}
     \mu^*\paren{\bigcup_{n \in \natp}E_n} &\le \sum_{(n, k) \in \natp \times \natp}\rho(F_{n, k}) = \sum_{n \in \natp}\sum_{k \in \natp}\rho(F_{n, k}) \\
@@ -84,7 +86,7 @@
 
 \begin{definition}[Premeasure]
 \label{definition:premeasure}
-    Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$, then $\mu_0$ is a \textbf{premeasure} if:
+    Let $X$ be a set, $\alg \subset 2^X$ be a ring, and $\mu_0: \alg \to [0, \infty]$, then $\mu_0$ is a \textbf{premeasure} if:
     \begin{enumerate}
         \item[(M1)] $\mu_0(\emptyset) = 0$.
         \item[(PM)] For any $\seq{A_n} \subset \alg$ pairwise disjoint with $\bigsqcup_{n \in \natp}A_n \in \alg$,
@@ -96,7 +98,7 @@
 
 \begin{theorem}[Carathéodory's Extension Theorem, {{\cite[Theorem 1.14]{Folland}}}]
 \label{theorem:caratheodory-extension}
-    Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
+    Let $X$ be a set, $\alg \subset 2^X$ be a ring, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
     \begin{enumerate}
         \item $\cm \supset \alg$.
         \item $\mu|_\alg = \mu_0$.
@@ -116,11 +118,11 @@
     \]
     then $\mu^*$ is an outer measure by \ref{proposition:outer-measure-inf}. By Carathéodory's theorem (\ref{theorem:caratheodory}), there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
 
-    (1): Let $E \in \alg$ and $F \subset X$. For any $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$,
+    (1): Let $E \in \alg$ and $F \subset X$. If there exists $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$, then
     \[
     \sum_{n \in \natp}\mu_0(F_n) = \sum_{n \in \natp}\mu_0(F_n \cap E) + \sum_{n \in \natp}\mu_0(F_n \setminus E) \ge \mu^*(F \cap E) + \mu^*(F \setminus E)
     \]
-    As this holds for all such $\seq{F_n}$, $\mu_0^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$.
+    for all such $\seq{F_n}$, so $\mu^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$. Otherwise, $\mu^*(F) = \infty$ and the result holds directly.
 
     (2): Let $E \in \alg$ and $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$, then
     \[

src/measure/measure/regular.tex

@@ -0,0 +1,23 @@
+\section{Regular Measures}
+\label{section:regular-measure}
+
+\begin{definition}[Inner Regular]
+\label{definition:inner-regular}
+    Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a measure, then $\mu$ is \textbf{inner regular} if for any $E \in \cb_X$,
+    \[
+    \mu(E) = \sup\bracs{\mu(K)| K \subset E, K \text{ compact}}
+    \]
+\end{definition}
+
+\begin{definition}[Outer Regular]
+\label{definition:outer-regular}
+    Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a measure, then $\mu$ is \textbf{outer regular} if for any $E \in \cb_X$,
+    \[
+    \mu(E) = \sup\bracs{\mu(U)| U \supset A, U \text{ open}}
+    \]
+\end{definition}
+
+\begin{definition}[Regular]
+\label{definition:regular-measure}
+    Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a measure, then $\mu$ is \textbf{regular} if it is inner regular and outer regular.
+\end{definition}

src/measure/measure/sigma-finite.tex

@@ -8,4 +8,5 @@
         \item There exists $\seq{E_n} \subset \cm$ pairwise disjoint such that $\bigsqcup_{n \in \nat}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
         \item There exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
     \end{enumerate}
+    If the above holds, then $\mu$ is a \textbf{$\sigma$-finite measure}.
 \end{definition}

src/measure/sets/algebra.tex

@@ -11,6 +11,17 @@
     \end{enumerate}
 \end{definition}
 
+\begin{definition}[Ring]
+\label{definition:set-ring}
+    Let $X$ be a set and $\alg \subset 2^X$, then $\alg$ is a \textbf{ring} if:
+    \begin{enumerate}
+        \item[(A1)] $\emptyset \in \alg$.
+        \item[(A2')] For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.
+        \item[(A3)] For any $A, B \in \alg$, $A \cup B \in \alg$.
+    \end{enumerate}
+\end{definition}
+
+
 \begin{definition}[$\sigma$-Algebra]
 \label{definition:sigma-algebra}
     Let $X$ be a set and $\cm \subset 2^X$, then $\cm$ is a \textbf{$\sigma$-algebra} if:

src/measure/sets/elementary.tex

@@ -5,31 +5,45 @@
 \label{definition:elementary-family}
     Let $X$ be a set and $\ce \subset 2^X$, then $\ce$ is an \textbf{elementary family} if:
     \begin{enumerate}
-        \item[(P1)] $\emptyset \in \mathcal{P}$.
+        \item[(P1)] $\emptyset \in \ce$.
-        \item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
+        \item[(P2)] For any $A, B \in \ce$, $A \cap B \in \ce$.
-        \item[(E1)] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.
+        \item[(E)] For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j} \subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^n E_j$.
+    \end{enumerate}
+    If $X \in \ce$, then (E) may be replaced with
+    \begin{enumerate}
+        \item[(E')] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.
     \end{enumerate}
 \end{definition}
 
 \begin{proposition}[{{\cite[Proposition 1.7]{Folland}}}]
 \label{proposition:elementary-family-algebra}
-    Let $X$ be a set and $\ce \subset 2^X$ be an elementary family, then
+    Let $X$ be a set and $\ce \subset 2^X$ be an elementary family and
     \[
     \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
     \]
-    is an algebra.
+    then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
 \end{proposition}
 \begin{proof}
+    Firstly, let $A, B \in \alg$ with $\seqf{A_j}, \seqf[m]{B_j} \subset \ce$ and $A = \bigsqcup_{j = 1}^n A_j$ and $B = \bigsqcup_{j = 1}^mB_j$. If $A \cap B = \emptyset$, then
+    \[
+    A \sqcup B = \bigsqcup_{j = 1}^n A_j \sqcup  \bigsqcup_{j = 1}^mB_j \in \alg
+    \]
+    so $\alg$ is closed under disjoint unions.
+
     (A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
 
-    (A2): Let $A = \bigsqcup_{j = 1}^n A_j \in \alg$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $A_i^c = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
+    (A2'): Let $A = \bigsqcup_{j = 1}^n A_j \in \alg$ and $B \in \ce$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $B \setminus A_i = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
     \[
-    A^c = \bigcap_{i = 1}^n A_i^c = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
+    B \setminus A = \bigcap_{i = 1}^n B \setminus A_i = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
+    \]
+    Thus if $B \in \alg$ with $B = \bigsqcup_{j = 1}^n B_j$, then
+    \[
+    B \setminus A = \bigsqcup_{j = 1}^n B_j \setminus A \in \alg
     \]
 
     (A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
     \[
-    A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
+    A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
     \]
-    so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = (A^c \cap B^c)^c \in \alg$.
+    so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.
 \end{proof}