> [!theoremb]- Theorem (1.15) > > Let $F: \real \to \real$ be an increasing and right [[Continuity|continuous]] [[Function|function]]. Extend $F$ to $\ol{F}: \ol{\real} \to \ol{\real}$ by > $ > \ol{F}(x) = \begin{cases} > F(x) & x \in \real \\ > \limv{x}F(x) &x = \infty \\ > \lim_{x \to -\infty} F(x) & x = -\infty > \end{cases} > $ > An h-interval is an interval of the form $(a, b]$ or $(a, \infty)$ where $-\infty \le a < b < \infty$. Let > $ > \begin{align*} > \ch_1 &= \bracs{ > (a, b]: -\infty \le a < b < \infty > } \\ > \ch_2 &= \bracs{(a, \infty): -\infty \le a} > \end{align*} > $ > where $\ch_1$ is the collection of h-intervals bounded from above, and $\ch_2$ is the collection of h-intervals unbounded from above. Take $\ce = \ch_1 \cup \ch_2 \cup \bracs{\emptyset}$, then $\ce$ is an [[Elementary Family|elementary family]], and the collection > $ > \alg = \bracs{\bigcup_{i = 1}^{n}I_i: I_i \in \ce, n \in \nat, \bracs{I_i}_1^{n} \text{ disjoint}} > $ > of finite-disjoint-unions (FDUs) of $\ce$ is an [[Algebra|algebra]]. > > Define $\mu_0: \ce \to [0, \infty]$ as > $ > \mu_0(I) = \begin{cases} > \ol F(b) - \ol F(a) &I = (a, b] \in \ch_1 \\ > \ol F(\infty) - \ol F(a) &I = (a, \infty) \in \ch_2 \\ > 0 &I = \emptyset > \end{cases} > $ > Since $F$ is increasing, $\ol{F}(-\infty)$ is finite or $-\infty$. Subtraction by $\ol{F}(-\infty)$ is equivalent to adding $\infty$. Therefore this definition does not involve undefined operations. > > Extend $\mu_0$ to all elements of $\alg$ by > $ > \mu_0\paren{\bigcup_{i = 1}^{n}I_i} = \sum_{i = 1}^{n}\mu_0(I_i) > $ > Then $\mu_0$ is a [[Premeasure|premeasure]] on the collection $\alg$. > > *Proof*. > > # Properties of h-intervals > > ### h-intervals form an elementary family > > $\ce$ forms an elementary family. > > *Proof.* Let $(a, b] \in \ch_1$, then > $ > (a, b]^c = (-\infty, a] \cup [b, \infty) > $ > where $(-\infty, a] \in \ch_1$ ($a > -\infty$) or $(-\infty, a] = \emptyset$ ($a = -\infty$). > > Let $(a, b], (c, d] \in \ch_1$, then $(a, b] \cap (c, d] = (\max(a, c), \min(b, d)]$, in $\ch_1$ if it's not empty. Let $(a, \infty), (c, \infty) \in \ch_2$, then $(a, \infty) \cap (c, \infty) = (\max(a, c), \infty)$, also in $\ch_2$. And $(a, \infty) \cap (c, d] = (\max(a, c), d]$, in $\ch_1$ if not empty. > > If $(a, \infty) \in \ch_2$, then $(a, \infty)^c = (-\infty, a]$ where $(-\infty, a] \in \ch_1$ or is the empty set, depending on the value of $a$. > > Lastly, $\emptyset^c = (-\infty, \infty) \in \ch_2$. Therefore $\ce$ is an elementary family, and $\alg$ forms an algebra. > > ### Unbounded FDUs of h-intervals contain unbounded h-intervals > > Let $E = \bigcup_{j = 1}^{n}I_j \in \alg$. If $E$ is unbounded from above, then there exists a unique $j$ such that $I_j$ is unbounded from above. If $E$ is unbounded from below, then there exists a unique $k$ such that $I_k$ is unbounded from below. > > *Proof*. Since the union is finite, the [[Supremum and Infimum|supremum and infimum]] can be reduced to the maximum and minimum > $ > \begin{align*} > \sup E &= \sup_{j}\sup I_j = \max_j \sup I_j \\ > \inf E &= \inf_{j}\inf I_j = \min_j \inf I_j \\ > \end{align*} > $ > If $\sup E = \infty$, then $\exists j: \sup I_j = \infty$ and $I_j = (a, \infty]$ is unbounded from above. Since the union is disjoint, $\bigcup_{i \ne j}I_j \subseteq (-\infty, a]$ is bounded from above. $I_j$ is the unique h-interval in the union unbounded from above. > > If $\inf E = -\infty$, then $\exists k: \inf I_k = -\infty$, $I_j$ is unbounded from below. Since the union is disjoint, $I_k$ is the unique h-interval in the union unbounded from below. > > ### FDUs of h-intervals that form one h-interval are "connected" > > Let $E = \bigcup_{j = 1}^{n}I_j \in \alg$, $n \ge 2$, and write > $ > E = \bigcup_{j = 1}^{m}(a_j, b_j] \cup \po{(d, \infty)} > $ > where $\po{(d, \infty)}$ is the unique interval unbounded from above (if $E$ is unbounded), and > $ > a_1 < b_1 \le a_2 < b_2 \le \cdots \le a_m < b_m \le \po{d} > $ > by relabelling the indices. Since this is a finite disjoint union, the intervals may be ordered by $a_j$s and $\po d$. > > If $E$ is an interval of any form, then > $ > a_1 < b_1 = a_2 < b_2 = \cdots = a_m < b_m = \po{d} > $ > > *Proof*. Consider $b_j, a_{j + 1}$, then since $b_j, a_{j + 1} \in E$, and $E$ is an interval, $[b_j, a_{j + 1}] \subset E$. Moreover, since $a_{j+1} \not\in (a_k, b_k]$ for any $k \ge j + 1$, $a_j \not\in \po{(b, \infty)}$, and $b_j \not\in (a_k, b_k]$ for any $k < j$, > $ > [b_j, a_{j + 1}] \subset (a_j, b_j] > $ > we have $b_j = a_{j + 1}$. > > For the unbounded part, we have > $ > [b_m, d] \subset (a_m, b_m] \subset E > $ > since $\po d \not\in \po{(d, \infty)}$ and $a_m \not\in (a_k, b_k]$ for any $k < m$. Therefore $\po d = b_m$. > > > ### Well-defined on h-intervals > > Let $E = \bigcup_{j = 1}^{n}I_j \in \alg$, $n \ge 2$. If $E = \emptyset$, $I_j = \emptyset \forall j$, and $\mu_0\paren{E} = \sum_{j = 1}^{m}\mu_0(\emptyset) = 0$. $\mu_0$ is well-defined on $\emptyset$. Since $\mu_0(\emptyset) = 0$ (does not affect the sum), we can assume that $I_j \ne \emptyset \forall j$. > > If $E$ is bounded from above, > $ > E = \bigcup_{j = 1}^{m}(a_j, b_j] = (a, b] \in \ch_1 > $ > Then since $a_j = b_{j + 1}$ for all $j < m$, > $ > \begin{align*} > \mu_0\paren{\bigcup_{j = 1}^{m}(a_j, b_j]} &= \sum_{j = 1}^{m}[\ol F(b_j) - \ol F(a_j)] \\ > &= \ol F(b_m) - \ol F(a_1) = \ol F(b) - \ol F(a) \\ > &= \mu_0((a, b]) > \end{align*} > $ > the sum telescopes and $\mu_0$ is well-defined on $\ch_1$. > > If $E$ is unbounded from above, > $ > E = \bigcup_{j = 1}^{m}(a_j, b_j] \cup \po{(d, \infty)} = (a, \infty) \in \ch_2 > $ > then using the previous result, > $ > \begin{align*} > &\mu_0\paren{\bigcup_{j = 1}^{m}(a_j, b_j] \cup \po{(d, \infty)}} \\ > &= \mu_0\paren{\bigcup_{j = 1}^{m}(a_j, b_j]} + \po{\ol F(\infty) - \ol F(d)} \\ > &= \ol F(b_m) - \ol F(a_1) + \ol F(\infty) - \ol F(d) \\ > &= \ol F(\infty) - \ol F(a_1) = \ol F(\infty) - \ol F(a) \\ > &= \mu_0((a, \infty)) > \end{align*} > $ > $\mu_0$ is well-defined on $\ch_2$. > > Therefore $\mu_0$ is well-defined on $\ce$. > > > ### Well-defined on FDUs of h-intervals > > Let $E \in \alg$, > $ > E = \bigcup_{i = 1}^{m} I_i = \bigcup_{i = 1}^{n}J_i > $ > then we can decompose > $ > E = \bigcup_{i = 1}^{m}\bigcup_{j = 1}^{n}I_i \cap J_j > $ > where each $I_i \cap J_j \in \ce$ is an h-interval. Since $\mu_0$ is well-defined on $\ce$, we can write any h-interval as a FDU of h-intervals: > $ > \begin{align*} > \mu_0\paren{\bigcup_{i = 1}^{m}I_i} &= \sum_{i = 1}^{m}\mu_0(I_i) \\ > &= \sum_{i = 1}^{m}\mu_0\paren{\bigcup_{j = 1}^{n}I_i \cap J_j} \\ > &= \sum_{i = 1}^{m}\sum_{j = 1}^{n}\mu_0(I_i \cap J_j) \\ > &= \sum_{j = 1}^{n}\sum_{i = 1}^{m}\mu_0(I_i \cap J_j) \\ > &= \sum_{j = 1}^{n}\mu_0\paren{\bigcup_{i = 1}^{m}I_i \cap J_j} \\ > &= \sum_{j = 1}^{n}\mu_0(J_j) \\ > &= \mu_0\paren{\bigcup_{j = 1}^{n}J_j} > \end{align*} > $ > and $\mu_0$ is well-defined on $\alg$. > > > # Properties of a finitely additive measure > > $\mu_0$ is finitely additive, monotone, and finitely subadditive. > > *Proof*. Let $\bracs{(a_j, b_j]}_1^n$ and $\bracs{(c_j, d_j]}_1^m$ be disjoint, then > $ > \begin{align*} > \mu_0\paren{\bigcup_{j = 1}^{n}(a_j, b_j] \cup \bigcup_{j = 1}^{m}(c_j, d_j]} &= \sum_{j = 1}^{n}[F(b_j) - F(a_j)] \\ > &+ \sum_{j = 1}^{m}[F(d_j) - F(c_j)] \\ > &= \mu_0\paren{\bigcup_{j = 1}^{n}(a_j, b_j]} + \mu_0\paren{\bigcup_{j = 1}^{m}(c_j, d_j]} > \end{align*} > $ > and $\mu_0$ is finitely additive. > > For any $E, F \in \alg, E \subseteq F$, > $ > \mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E) > $ > and $\mu_0$ is monotone. > > Lastly, let $\bracs{E}_1^n \subset \alg$, then > $ > \begin{align*} > \mu_0\paren{\bigcup_{i = 1}^{n}E_i} &= \mu_0\paren{\bigcup_{i = 1}^{n}\braks{E_i \setminus \bigcup_{j = 1}^{i - 1}E_j}} \\ > &= \sum_{i = 1}^{n}\mu_0\paren{E_i \setminus \bigcup_{j = 1}^{i - 1}E_j} \\ > &\le \sum_{i = 1}^{n}\mu_0(E_i) > \end{align*} > $ > and $\mu_0$ is finitely subadditive. > > > # Countable Additivity on Intervals > > ### Countable additivity on bounded sets > > Let $(a, b]$ be a bounded h-interval, and $\seq{(a_j, b_j]}$ be a [[Sequence|sequence]] of bounded disjoint h-intervals such that > $ > (a, b] = \bigcup_{j \in \nat}(a_j, b_j] > $ > Then > $ > \mu_0((a, b]) = \sum_{j \in \nat}\mu_0((a_j, b_j]) > $ > *Proof*. Let $J \subset \nat: |J| < |\nat|$, then > $ > \sum_{j \in J}\mu_0(a_j, b_j] = \mu_0\paren{\bigcup_{j \in J}(a_j, b_j]} \le \mu_0((a, b]) > $ > therefore $\sum_{j \in \nat}\mu_0((a_j, b_j]) \le \mu_0((a, b])$. > > ![[h_interval_cover.png]] > > Since $F$ is right-continuous, for any $\varepsilon < 0$, > $ > \exists \seq{\delta_j} \subset (0, \infty): F(b_j + \delta_j) - F(b_j) < \frac{\varepsilon}{2^{j + 1}} > $ > and $\delta_0 > 0: F(a_j + \delta_0) - F(a_j) < \varepsilon/2$. > > ![[h_interval_extend.png]] > Extend each $(a_j, b_j]$ into an open interval $(a_j, b_j + \delta_j)$. Then $\seq{(a_j, b_j + \delta_j)}$ is an [[Open Cover|open cover]] for $[a + \delta_0, b]$. Since the bounded closed interval is [[Compactness|compact]], > $ > \exists J_\varepsilon \subset \nat, |J_\varepsilon| < |\nat|: \bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j) \supset [a + \delta_0, b] > $ > it has a finite subcover $\bracs{(a_j, b_j + \delta_j): j \in J_\varepsilon}$. Now, we can close the right side and turn them into h-intervals: > $ > \bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j] \supset \bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j) \supset [a + \delta_0, b] > $ > > ![[h_interval_subcover.png]] > > Since the extended h-intervals $\bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j] \in \alg$ is a FDU of h-intervals that contains $[a + \delta_0, b] \supset (a + \delta_0, b]$, we have > $ > \mu_0\paren{\bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j]} \ge \mu_0((a + \delta_0, b]) > \mu_0((a, b]) - \varepsilon/2 > $ > where > $ > \begin{align*} > \mu_0((a, b]) &= F(b) - F(a) < F(b) - F(a + \delta_0) + \varepsilon/2 \\ > &= \mu_0((a + \delta_0, b]) + \varepsilon/2 > \end{align*} > $ > From finite subadditivity of $\mu_0$ and right-continuity of $F$, > $ > \begin{align*} > \mu_0\paren{\bigcup_{j \in J_\varepsilon}(a_j, b_j + \delta_j]} &\le \sum_{j \in J_\varepsilon}\braks{F(b_j + \delta_j) - F(a_j)} \\ > &< \sum_{j \in J_\varepsilon}\braks{F(b_j) - F(a_j) + \frac{\varepsilon}{2^{-j}}} \\ > &< \sum_{j \in J_\varepsilon}\braks{F(b_j) - F(a_j)} + \frac{\varepsilon}{2} > \end{align*} > $ > and we have > $ > \sum_{j \in J_\varepsilon}\braks{F(b_j) - F(a_j)} > \mu_0((a, b]) - \varepsilon > $ > This allows us to grab $\seq{J_n}$ such that > $ > \sum_{j \in J_n}\braks{F(b_j) - F(a_j)} \upto \mu_0((a, b]) > $ > which means that > $ > \sum_{j \in \nat}\mu_0((a_j, b_j]) = \sum_{j \in \nat}\braks{F(b_j) - F(a_j)} \ge \mu_0((a, b]) > $ > > ### Continuity from below on single h-intervals > > Let $E \in \ce$, then > $ > \mu_0(E) = \sup_{m \in \nat}(E \cap (-m, m]) > $ > *Proof.* Since $E \in \ce$, $E \cap (-m, m] \in \ch_1$ is a bounded elementary set. If $E \in \ch_1$, then > $ > \begin{align*} > \mu_0(E) &= \ol F(b) - \ol F(a) \\ > &= \sup_{m \in \nat} \ol F(\min(b, m)) - \inf \ol F(\max(a, -m)) \\ > &= \sup_{m \in \nat} [\ol F(\min(b, m)) - \ol F (\max (a, -m))] \\ > &= \sup_{m \in \nat}\mu_0(E \cap (-m, m]) > \end{align*} > $ > If $E \in \ch_2$, then the same thing applies, except we remove the min part on $\ol F(\min (b, m))$ and simply replace it with $m$. > > > ### Countable additivity on the elementary family > > Let $E \in \ce$ and $\seq{E_j} \subset \ce$ be a sequence of disjoint, bounded h-intervals such that > $ > E = \bigcup_{j \in \nat}E_j > $ > then > $ > \mu_0(E) = \sum_{j \in \nat}\mu_0(E_j) > $ > *Proof*. > $ > \begin{align*} > \mu_0(E \cap (-m, m]) &= \sum_{j \in \nat}\mu_0(E_j \cap (-m, m]) \\ > \sup_{m \in \nat}\mu_0(E \cap (-m, m]) &= \sup_{m \in \nat}\sum_{j \in \nat}\mu_0(E_j \cap (-m, m]) \\ > \mu_0(E) &= \mu_0(E_j) > \end{align*} > $ > > # Countable additivity on FDUs of h-intervals > > Let $E = \bigcup_{j = 1}^{n}I_j \in \alg$ be a finite disjoint union of h-intervals, and $E_j = \bigcup_{k = 1}^{n_j}I_{j, k} \in \alg$ be a disjoint sequence of FDUs of h-intervals such that $\bigcup_{j \in \nat}E_j = E$. Then since $\ce$ is closed under finite intersections, > $ > E_j \cap I_i = \bigcup_{k = 1}^{n_j}I_{j, k} \cap I_i > $ > each $I_{j, k} \cap I_i \in \ce$, which allows pulling the unions out as sums (since countable union of finite sets are still countable), > $ > \begin{align*} > \mu_0\paren{\bigcup_{j \in \nat}E_j \cap I_i} &= > \mu_0\paren{\bigcup_{j \in \nat}\bigcup_{k = 1}^{n_j}I_{j, k} \cap I_i} \\ > &= \sum_{j \in \nat}\sum_{k = 1}^{n_j}\mu_0(I_{j, k} \cap I_i) \\ > &= \sum_{j \in \nat}\mu_0(E_j \cap I_i) > \end{align*} > $ > Now we can interchange the sums to obtain > $ > \begin{align*} > \mu_0(E) &= \mu_0\paren{\bigcup_{i = 1}^{n}I_i} \\ > &= \sum_{i = 1}^{n}\mu_0(I_i) \\ > &= \sum_{i = 1}^{n}\mu_0\paren{\bigcup_{j \in \nat}E_j \cap I_i} \\ > &= \sum_{i = 1}^{n}\sum_{j \in \nat}\mu_0(E_j \cap I_i) \\ > &= \sum_{j \in \nat}\sum_{i = 1}^{n}\mu_0(E_j \cap I_i) \\ > &= \sum_{j \in \nat}\mu_0(E_j) > \end{align*} > $ > and $\mu_0$ is countably additive on elements of $\alg$. > [!definitionb] Definition > > Let $F: \real \to \real$ be an increasing, right [[Continuity|continuous]] [[Function|function]], then there is a unique [[Borel Measure|Borel measure]] $\mu_F$ on $\real$ such that $\mu_F((a, b]) = F(b) - F(a)$ for all $a, b$. If $G$ is another such function, then $\mu_F = \mu_G$ if and only if $F - G$ is a constant. > > Let $\mu$ be a Borel [[Measure Space|measure]] on $\real$ that is finite on all bounded [[Borel Sigma Algebra|Borel sets]], and define > $ > F(x) = \begin{cases} > \mu((0, x]) & x > 0 \\ > 0 & x = 0 \\ > -\mu((x, 0]) &x < 0 > \end{cases} > $ > then $F$ is increasing and right-continuous, and $\mu = \mu_F$. > > *Proof*. Since $F: \real \to \real$ is increasing and right continuous, it induces a [[Premeasure|premeasure]] on $\alg$, the collection of FDUs of h-intervals. By [[Carathéodory's Extension Theorem]], $\mu_F$ is an extension of $\mu_0$ on $\cm(\alg)$. As the h-intervals generate the [[Borel Sigma Algebra|Borel sigma-algebra]], $\mu_F$ is a [[Borel Measure|Borel measure]]. Moreover, since > $ > \mu_F(\real) = \mu_0\paren{\bigcup_{n \in \integer}(n, n+1]} > $ > and > $ > \mu_F((n, n + 1]) = F(n + 1) - F(n) < \infty > $ > The measure $\mu_F$ is $\sigma$-finite, and therefore unique. > > Let $G: \real \to \real$ such that $\mu_F = \mu_G$, then for any $x > 0$, > $ > \begin{align*} > \mu_F((0, x]) &= \mu_G((0, x]) \\ > F(x) - F(0) &= G(x) - G(0) \\ > F(x) - G(x) &= F(0) - G(0) > \end{align*} > $ > and for any $x < 0$, > $ > \begin{align*} > \mu_F((x, 0]) &= \mu_G((x, 0]) \\ > F(0) - F(x) &= G(0) - G(x) \\ > F(0) - G(0) &= F(x) - G(x) > \end{align*} > $ > Therefore $F(x) - G(x) = F(0) - G(0)$ for all $x \in \real$, $F$ and $G$ differ by a constant. > > Let $\mu$ be a Borel measure on $\real$ that is finite on all bounded Borel sets, and let > $ > F(x) = \begin{cases} > \mu((0, x]) & x > 0 \\ > 0 &x = 0 \\ > -\mu((x, 0]) &x < 0 > \end{cases} > $ > $F: \real \to \real$ since $\mu$ is finite on all bounded Borel sets. Then for any $0 \le a \le b$, > $ > F(a) = \mu((0, a]) \le \mu((0, b]) = F(b) > $ > and for any $a \le b \le 0$, > $ > F(a) = -\mu((a, 0]) \le -\mu((b, 0]) = F(b) > $ > Lastly, $F(x) \ge 0$ for any $x > 0$, and $F(x) \le 0$ for any $x \le 0$. So $F$ is increasing. > > Now, let $\seq{x_n} \subset [0, \infty)$, $x_n \downto x$, then if $x > 0$, we can use continuity from above since all of these sets are finite: > $ > \begin{align*} > \limv{n}F(x_n) &= \limv{n}\mu((0, x_n]) \\ > &= > \mu\paren{\bigcap_{n \in \nat}(0, x_n]} \\ > &= \mu((0, x]) = F(x) > \end{align*} > $ > If $x = 0$, then $\bigcap_{n \in \nat}(0, x_n] = \emptyset$ and the limit evaluates to $F(0)$. > > Let $\seq{x_n} \subset (-\infty, 0)$, $x_n \downto x$, then > $ > \begin{align*} > \limv{n}F(x_n) &= \limv{n}-\mu((x_n, 0]) \\ > &= -\mu\paren{\bigcup_{n \in \nat}(x_n, 0]} \\ > &= -\mu((x, 0]) = F(x) > \end{align*} > $ > Note that for any $x_n \downto x$, if $x < 0$, then $x_n < 0$ eventually, so the assumption can be made. Therefore $F$ is right-continuous. > > For any $(a, b]$, we have a disgusting amount of cases, > $ > \begin{align*} > \mu_F((a, b]) &= F(b) - F(a) \\ > &= \begin{cases} > \mu((0, b]) + \mu((a, 0]) &a < 0, b > 0 \\ > \mu((0, b]) - \mu((0, a]) &a > 0, b > 0 \\ > \mu((a, 0]) - \mu((b, 0]) &a < 0, b < 0 \\ > \mu((a, 0]) & a < 0, b = 0\\ > \mu((0, b]) & a = 0, b > 0 > \end{cases} \\ > &= \mu((a, b]) > \end{align*} > $ > and $\mu_F = \mu$ on the collection of bounded h-intervals. Any unbounded interval can be approximated as > $ > \begin{align*} > \mu_F((-\infty, b]) &= \mu_F\paren{\bigcup_{n \in \nat}(-n, b]} \\ > &= \limv{n}\mu_F((-n, b]) \\ > &= \limv{n}\mu((-n, b]) = \mu((-\infty, b]) \\ > \mu_F((a, \infty]) &= \mu_F\paren{\bigcup_{n \in \nat}(a, \infty]} \\ > &= \limv{n}\mu_F((a, n]) \\ > &= \limv{n}\mu((a, n]) = \mu((a, n]) \\ > \mu_F(\real) &= \mu_F((-\infty, 0]) + \mu_F((0, \infty)) \\ > &= \mu((-\infty, 0]) + \mu((0, \infty)) \\ > &= \mu(\real) > \end{align*} > $ > and $\mu_F|_\ce = \mu|_\ce$. Since $\mu_F$ is finitely additive, we have $\mu_F|_\alg = \mu|_\alg$ by > $ > \mu_F\paren{\bigcup_{i = 1}^{n}E_i} = \sum_{i = 1}^{n}\mu_F(E_i) = \sum_{i = 1}^{n}\mu(E_i) = \mu\paren{\bigcup_{i = 1}^{n}E_i} > $ > for any finite disjoint union of elementary sets. Since $\mu_F$ is the unique Borel measure induced by $\mu_F|_\alg = \mu|_\alg$, $\mu_F = \mu$ by the extension theorem. > [!theorem]- Alternate Approximations > > Let $F: \real \to \real$ be an increasing, right-continuous function and $\mu$ be the Lebesgue-Stieltjes measure associated with $F$. Then since $\mu$ is induced by [[Carathéodory's Theorem]] as an [[Outer Measure|outer measure]], $\mu$ uses > $ > \Sigma_E = \bracs{ > \sum_{n \in \nat}\mu_0(E_n): E_n \in \alg, E \subseteq \bigcup_{n \in \nat}E_n > } > $ > and takes the value $\mu(E) = \inf \Sigma_E$ (see [[Upper and Lower Approximations]]). Take > $ > \begin{align*} > \sigma_E &= \bracs{\sum_{n \in \nat}\mu_0(E_n): E_n \in \ce, E \subseteq \bigcup_{n \in \nat}E_n} \\ > S_E &= \bracs{\sum_{n \in \nat}\mu_0((a_n, b_n]):E \subseteq \bigcup_{n \in \nat}(a_n, b_n]} \\ > s_E &= \bracs{\sum_{n \in \nat}\mu((a_n, b_n)):E \subseteq \bigcup_{n \in \nat}(a_n, b_n)} > \end{align*} > $ > where $\ce$ is the collection of h-intervals, and $\alg$ is the [[Algebra|algebra]] of FDUs of h-intervals. Then > $ > \mu(E) = \inf \Sigma_E = \inf \sigma_E = \inf S_e = \inf s_E > $ > *Proof*. First note that $\alg \supset \ce \supset \ch_{1}$, and $\ch_1$ contains all bounded h-intervals used in $S_E$, so we have $\Sigma_E \supseteq \sigma_E \supseteq S_E$ since they are simply countable unions from those sets. Now, for any $\seq{E_n} \subset \alg$, where $E_n = \bigcup_{i = 1}^{m_n}E_{n, i}$ is a FDU of $\ce$, then using the fact that $\ce$ is closed under finite intersections, > $ > \begin{align*} > E_n &= \bigcup_{i = 1}^{m_n}E_{n, i} = \bigcup_{i = 1}^{m_n}\bigcup_{k \in \integer}E_{n, i} \cap (k, k + 1] \\ > \mu_0(E_n) &= \sum_{i = 1}^{m_n}\mu_0(E_{n, i}) = \sum_{i = 1}^{m_n}\sum_{k \in \integer}\mu_0(E_{n, i} \cap (k, k + 1])\\ > \sum_{n \in \nat}\mu_0(E_n) &= \sum_{n \in \nat}\sum_{i = 1}^{m_n}\mu_0(E_{n, i}) = \sum_{n \in \nat}\sum_{i = 1}^{m_n}\sum_{k \in \integer}\mu_0(E_{n, i} \cap (k, k + 1]) > \end{align*} > $ > we can rewrite any countable collection of $\alg$ sets as a countable collection of $\ce$ sets or of bounded $\ch_1$ sets. This shows that $\Sigma_E \subseteq \sigma_E \subseteq S_E$ the approximation values calculated from these three families are the same. Therefore > $ > \mu(E) = \inf \Sigma_E = \inf \sigma_E = \inf S_E > $ > For any $\bigcup_{n \in \nat}(a_n, b_n) \supset E$, > $ > \sum_{n \in \nat}\mu(a_n, b_n) \ge \mu\paren{\bigcup_{n \in \nat}(a_n, b_n)} \ge \mu(E) > $ > so $\mu(E) \le \inf s_E$. If $\mu(E) = \infty$, then we're done. Otherwise, since $F$ is right continuous, > $ > \forall \varepsilon > 0, \exists \delta_n > 0: F(b_n + \delta_n) - F(b_n) < \varepsilon/2^{n} > $ > As $\mu(E) = \inf S_E$, for any $\seq{(a_n, b_n]}: E \subseteq \bigcup_{n \in \nat}(a_n, b_n]$, we can create a new family $\seq{(a_n, b_n + \delta_n)}: E \subset \bigcup_{n \in \nat}(a_n, b_n + \delta_n)$ where > $ > \mu_0((a_n, b_n]) + \varepsilon/2^{n}> \mu_0((a_n, b_n + \delta_n]) \ge \mu((a_n, b_n + \delta_n)) > $ > and > $ > \sum_{n \in \nat}\mu((a_n, b_n + \delta_n)) < \sum_{n \in \nat}\mu_0((a_n, b_n]) + \varepsilon > $ > Since $\varepsilon$ is arbitrary, we can approach any $S_E$ approximation from above by a sequence of approximations in $s_E$. Therefore > $ > \mu(E) \le \inf s_E \le \inf S_E = \mu(E) > $ > [!theorem] > > Let $\mu$ be a Lebesgue-Stieltjes measure and $\cm_\mu$ be the collection of $\mu$-measurable sets. Then for any $E \in \cm_\mu$, > $ > \begin{align*} > \mu(E) &= \inf\bracs{\mu(U): U \supset E, U \text{ open}} \\ > &= \sup\bracs{\mu(K): K \subset E, K \text{ compact}} > \end{align*} > $ > *Proof*. For any $U \supset E$, $\mu(U) \ge \mu(E)$. If $\mu(E) = \infty$, then we're done. Otherwise, since $\mu(E) = \inf s_E$, > $ > \begin{align*} > \forall \varepsilon > 0, \exists \seq{(a_n, b_n)}: E &\subset \bigcup_{n \in \nat}(a_n, b_n), \\ > \mu\paren{\bigcup_{n \in \nat}(a_n, b_n)} \le \sum_{n \in \nat}\mu((a_n, b_n)) &< \mu(E) + \varepsilon > \end{align*} > $ > the value of $\mu(E)$ can be approximated by [[Open Set|open sets]] from above. Therefore $\mu(E) \le \inf \bracs{\mu(U): U \supset E, U \text{ open}}$. > > For any $K \subset E$, $\mu(K) \le \mu(E)$. First suppose that $E$ is bounded. If $E$ is also [[Closed Set|closed]], then $E$ is [[Compactness|compact]] ([[Heine-Borel Theorem]]) and we get the result for free. If $E$ is not closed, then there exists an open set $U \supset \ol{E} \setminus E$ such that $\mu(U) < \mu(\ol{E} \setminus E) + \varepsilon$. This is possible since $\mu$ is finite on all bounded sets. Then > $ > \mu(\ol{E}) = \mu(U) + \mu(U^c \cap \ol {E}) > $ > where $U^c \cap \ol{E}$ is closed and bounded, and therefore compact. Moreover, since $U \supset \ol{E} \setminus E$, $U^c \cap \ol{E}$ is a compact subset of $E$, and > $ > \begin{align*} > \mu(U^c \cap \ol{E}) &= \mu(\ol{E}) - \mu(U)\\ > &= \mu(E) + \mu(\ol{E} \setminus E) - \mu(U) \\ > &> \mu(E) - \varepsilon > \end{align*} > $ > Note that this is where we use the fact that $E$ is outer-measurable. Since $\varepsilon$ is arbitrary, the value of $\mu(E)$ can be approximated by compact sets from below. Therefore > $ > \mu(E) \le \sup\bracs{\mu(K): K \subset E, K \text{ compact}} > $ > > Now let $E \in \cm_\mu$ be arbitrary, then > $ > \begin{align*} > \mu(E) &= \mu\paren{\bigcup_{n \in \nat}E \cap [-n, n]} \\ > &= \limv{n}\mu(E \cap [-n, n])\\ > &= \sup_{n \in \nat}\mu(E \cap [-n, n]) > \end{align*} > $ > by continuity from below. For any $n \in \nat$, > $ > \exists K_n \text{ compact}: K_n \subset (E \cap [-n, n]), \mu(E \cap [-n, n]) < \mu(K_n) + 1/n > $ > If $\mu(E) < \infty$, then > $ > \begin{align*} > \mu(E \cap [-n, n]) - 1/n &< \mu(K_n) \le \mu(E \cap [-n, n]) \\ > \limv{n}\mu(E \cap [-n, n]) &\le \limv{n}\mu(K_n) \le \limv{n}\mu(E \cap [-n, n]) > \end{align*} > $ > and we have found a sequence $\seq{K_n}$ of compact sets where $\mu(K_n) \upto \mu(E)$. > > If $\mu(E) = \infty$, then $\mu(E \cap [-n, n]) \upto \infty$. Since $\mu(K_n) > \mu(E \cap [-n, n]) - 1$, $\mu(K_n) \upto \infty$ as well. Therefore > $ > \mu(E) = \sup\bracs{\mu(K): K \subset E, K \text{ compact}} > $ > [!theoremb] Theorem > > Let $\mu$ be a Lebesgue-Stieltjes measure and $\cm_\mu$ be the collection of $\mu$-measurable sets, then the following are equivalent > - $E \in \cm_\mu$ > - $E = V \setminus N_1$, where $V$ is a $G_\delta$ set and $N_1$ is a $\mu$-[[Null Set|null set]]. > - $E = H \cup N_2$, where $N$ is a $F_\sigma$ set and $N_2$ is a $\mu$-null set. > > *Proof*. Since $\mu$ is induced using [[Carathéodory's Theorem]], $\mu$ is a [[Complete Measure|complete measure]] whose domain includes the [[Borel Sigma Algebra|Borel sigma-algebra]]. If $E = V \setminus N_1$ or $H \cup N_2$, then $V, H, N_1, N_2 \in \cm_\mu$ implies that $E \in \cm_\mu$. > > Now suppose that $E \in \cm_\mu$. First assume that $\mu(E) < \infty$. Since > $ > \mu(E) = \inf\bracs{\mu(U): U \supset E, U \text{ open}} > $ > we can find a sequence $\seq{U_n}$ such that $\mu(U_n) \downto \mu(E)$. Then using continuity from above, > $ > \limv{n}\mu(U_n) = \mu\paren{\bigcap_{n \in \nat}U_n} = \mu(E) > $ > Since $E$ is outer-measurable, > $ > \mu\paren{\bigcap_{n \in \nat}U_n \setminus E} = \mu\paren{\bigcap_{n \in \nat}U_n} - \mu(E) = 0 > $ > and $E = V \setminus N_1$ where $V = \bigcap_{n \in \nat}U_n$ is a $G_\delta$ set, and $N_2 = \bigcap_{n \in \nat}U_n \setminus E$ is a null set. > > As > $ > \mu(E) = \sup\bracs{\mu(K): K \subset E, K \text{ compact}} > $ > we can find a sequence $\seq{K_n}$ such that $\mu(K_n) \upto \mu(E)$. Using continuity from below, > $ > \limv{n}\mu(K_n) = \mu\paren{\bigcup_{n \in \nat}K_n} = \mu(E) > $ > and > $ > \mu\paren{E \setminus \bigcup_{n \in \nat}K_n} = \mu\paren{E} - \mu\paren{\bigcup_{n \in \nat}K_n} = 0 > $ > so $E = H \cup N_2$ where $H = \bigcup_{n \in \nat}K_n$ (since each $K_n$ is closed and bounded) is a $F_\sigma$ set, and $N_2 = E \setminus \bigcup_{n \in \nat}K_n$ is a null set. > > Now suppose that $E \in \cm_\mu$ is any set, then > $ > \mu(E) = \mu\paren{\bigcup_{n \in \integer}E \cap (n, n + 1]} = \sum_{n \in \integer}E \cap (n, n + 1] > $ > Since $\mu$ is finite on any bounded sets, $\mu(E \cap (n, n + 1]) < \infty$. Therefore > $ > E \cap (n, n + 1] = \bigcap_{k \in \nat}U_{n, k} \setminus M_n > $ > where $U_{n, k}$ are [[Open Set|open sets]] and $M_n$ are null sets. Also, > $ > E \cap (n \in n + 1] = \bigcup_{k \in \nat}K_{n, k} \cup N_n > $ > where $K_{n, k}$ are closed sets and $N_n$ are null sets. Then since the first union is disjoint, > $ > \begin{align*} > E &= \bigcup_{n \in \integer}\bigcap_{k \in \nat}U_{n, k} \setminus M_n \\ > &= \paren{\bigcup_{n \in \integer}\bigcap_{k \in \nat}U_{n, k}} \setminus \paren{\bigcup_{n \in \integer}M_n} \\ > &= \paren{\bigcap_{k \in \nat}\bigcup_{n \in \integer}U_{n, k}} \setminus \paren{\bigcup_{n \in \integer}M_n} > \end{align*} > $ > As $\bigcup_{n \in \integer}U_{n, k}$ is a union of open sets, it's also an open set, so $\bigcap_{k \in \nat}\bigcup_{n \in \integer}U_{n, k}$ is a $G_\delta$ set. As $\bigcup_{n \in \integer}M_n$ is a countable union of null sets, it's also a null set. > > On the other side, > $ > E = \bigcup_{n \in \integer}\bigcup_{k \in \nat}K_{n, k} \cup N_n = \paren{\bigcup_{n \in \integer}\bigcup_{k \in \nat}K_{n, k}} \cup \paren{\bigcup_{n \in \integer}N_n} > $ > where $\bigcup_{n \in \integer}\bigcup_{k \in \nat}K_{n, k}$ is still a countable union of closed sets, and therefore a $F_\sigma$ set. $\bigcup_{n \in \integer}N_n$ is a countable union of null sets, so it's also a null set. > [!theorem] > > Let $\mu$ be a Lebesgue-Stieltjes measure, and $E \in \cm_\mu: \mu(E) < \infty$ be a finite $\mu$-measurable set. Then for any $\varepsilon > 0$, there exists a finite union $A$ of bounded open intervals such that $\mu(E \Delta A) < \varepsilon$. > > *Proof*. Fix $\varepsilon > 0$, then > $ > \begin{align*} > \exists U \text{ open}&: U \supset E, \mu(U) < \mu(E) + \varepsilon/2 \\ > \exists K \text{ compact}&: K \subset E, \mu(E) < \mu(K) + \varepsilon/2 > \end{align*} > $ > Since $U$ is open, $\forall x \in U, \exists \delta_x > 0: B(x, \delta_x) \subseteq U$, we can write $U = \bigcup_{x \in U}B(x, \delta_x)$ as a union of open intervals. The collection of these open intervals $\Sigma = \bracs{B(x, \delta_x): x \in U}$ forms an [[Open Cover|open cover]] of the compact set $K$, and therefore has a finite subcover. Let $A$ be the union of the finite subcover, then > $ > \varepsilon/2 > \mu(E \setminus K) \ge \mu(E \setminus A) > $ > and > $ > \mu(A \setminus E) \le \mu(U \setminus E) < \varepsilon /2 > $ > so > $ > \mu(E \Delta A) = \mu(E \setminus A) + \mu(A \setminus E) < \varepsilon > $ > Since each $\delta_x$ is finite, each $B(x, \delta_x)$ is a bounded open interval.