Definition of a premeasure . A premeasure on an algebra \(\mathcal A \subseteq \mathcal{P}(X)\) of sets on \(X\) is a set function \(\mu_0:\mathcal A \to [0,\infty]\) satisfying
\(\mu_0(\varnothing)=0\),
if \((A_n)_{n\in \mathbb N}\) is a sequence of disjoint sets in \(\mathcal A\) such that \(\bigcup_{n=1}^{\infty}A_n \in \mathcal A\), then \[\mu_0\bigg(\bigcup_{n=1}^{\infty}A_n\bigg)=\sum_{n=1}^{\infty}\mu_0(A_n).\]
Remark.
In other words, a premeasure \(\mu_0:\mathcal A \to [0,\infty]\) on an algebra \(\mathcal A\) is a countably additive set function such that \(\mu_0(\varnothing)=0\).
A premeasure is always finitely additive.
The notions of finite and \(\sigma\)-finite premeasures are defined just as for measures.
Theorem. Let \(\rho:\mathcal{E} \to [0,\infty]\) be a set function on a semi-algebra \(\mathcal{E}\) of subsets of a set \(X\) and \(\rho(\varnothing)=0\). Suppose that \(\rho\) is finitely additive on \(\mathcal{E}\) and let \(\mathcal{A}\) be the algebra that consists of all finite disjoint unions of members of \(\mathcal{E}\). Define a set function \(\mu_0\) on \(\mathcal{A}\) by setting \[\mu_0(A)=\sum_{i=1}^{n}\rho(E_i)\] for \(A=\bigcup_{j=1}^{n}E_j \in \mathcal{A}\), where \(E_1,\ldots,E_n \in \mathcal{E}\) and \(E_i \cap E_j = \varnothing\) if \(i \neq j\).
Then \(\mu_0\) is a well-defined additive set function on \(\mathcal{A}\) such that \(\mu_0(\varnothing)=0\) and \(\mu_0=\rho\) on \(\mathcal{E}\).
If additionally \(\rho\) is countably subadditive on \(\mathcal{E}\) then \(\mu_0\) is a premeasure on \(\mathcal{A}\).
Proof of (a): To prove that \(\mu_0\) is well defined it suffices to show that for any finite collection of disjoint sets \((E_i)_{i=1}^{m}\subseteq \mathcal E\) and \((F_j)_{j=1}^{n}\subseteq \mathcal E\) such that \[\bigcup_{i=1}^{m}E_i=\bigcup_{j=1}^{n}F_j,\] one has \[\sum_{i=1}^{m}\rho(E_i)=\sum_{j=1}^{n}\rho(F_j).\] Let \(G_{i,j}=E_i \cap F_j\), then \(\{G_{i,j}: 1\le i\le m, \ 1\le j\le n\}\) is a collection of disjoint sets in \(\mathcal{E}\) and \[E_i=\bigcup_{j=1}^{n}G_{i,j}, \quad F_j=\bigcup_{i=1}^{m}G_{i,j}.\]
Since \(\rho\) is finitely additive on \(\mathcal{E}\) we obtain \[\rho(E_i)=\mu_0(E_i)=\sum_{j=1}^{n}\rho(G_{i,j}),\] and \[\rho(F_j)=\mu_0(F_j)=\sum_{i=1}^{m}\rho(G_{i,j}).\] Consequently \[\sum_{i=1}^{m}\rho(E_i)=\sum_{i=1}^{m}\sum_{j=1}^{n}\rho(G_{i,j}) =\sum_{j=1}^{n}\sum_{i=1}^{m}\rho(G_{i,j})=\sum_{j=1}^{n}\rho(F_j).\] Thus \(\mu_0\) is well defined finitely additive set function on \(\mathcal{A}\) such that \(\mu_0(E)=\rho(E)\) for every \(E \in \mathcal{E}\).
Proof of (b): To prove the second part assume that \(\rho\) is countably subadditive on \(\mathcal{E}\). We first prove that \(\rho\) is countably additive on \(\mathcal{E}\). Let \((E_n)_{n \in \mathbb{N}}\subseteq \mathcal{E}\) be a collection of disjoint sets such that \(\bigcup_{n \in \mathbb{N}}E_n \in \mathcal{E}\). Consider \(\mu_0\) corresponding to \(\rho\) as in part (a). For every \(N \in \mathbb{N}\) we have \[\bigcup_{n=1}^{N}E_n \subseteq \bigcup_{n \in \mathbb{N}}E_n\in \mathcal{E} \subseteq \mathcal{A}.\] Since \(\bigcup_{n=1}^{N}E_n \in \mathcal{A}\), we conclude \[\rho\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg)=\mu_0\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg) \geq \mu_0\bigg(\bigcup_{n=1}^{N}E_n\bigg)=\sum_{n=1}^{N}\mu_0(E_n)=\sum_{n=1}^{N}\rho(E_n).\]
Passing with \(N \to \infty\) we have \[\rho\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg) \geq \sum_{n=1}^{\infty}\rho(E_n).\] Since \(\rho\) is also countably subadditivity we have \[\rho\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg) \leq \sum_{n=1}^{\infty}\rho(E_n).\] Thus \(\rho\) is countably additive on \(\mathcal{E}\).
We now have to show that countable additivity of \(\rho\) on \(\mathcal{E}\) ensures countable additivity of \(\mu_0\) on \(\mathcal{A}\). In other words, we have to prove that \(\mu_0\) is a premeasure on \(\mathcal{A}\).
Let \((A_n)_{n \in \mathbb{N}}\subseteq \mathcal{A}\) be a sequence of disjoint sets so that \(A=\bigcup_{n \in \mathbb{N}} A_n \in \mathcal{A}\). By the definition of \(\mathcal{A}\) for each \(n \in \mathbb{N}\) there is a finite disjoint collection \((F_{n,j_n})_{j_n=1}^{k_n}\subseteq \mathcal E\) for some \(k_n\in\mathbb N\) such that \[A_n=\bigcup_{j_n=1}^{k_n}F_{n,j_n}.\] Let \(\mathcal{F}=\{F_{n,j_n}: n\in\mathbb N, \ 1\le j_n\le k_n\}\subseteq \mathcal E\). Since \(A \in \mathcal{A}\) thus there exists a finite disjoint collection \(\mathcal{G}=(E_i)_{i=1}^m \subseteq \mathcal{E}\) so that \[A=\bigcup_{i=1}^{m}E_i.\] We see that \(F_{n,j_n} \cap E_{i} \in \mathcal{E}\).
If we let \[\mathcal{H}=\{F_{n,j_n} \cap E_i: n \in \mathbb{N},\ 1\le j_n\le k_n, \ 1\le i\le m\}\] then by the disjointness of the collection \(\mathcal{F}\) and the disjointness of the collection \(\mathcal{G}\) we see that \(\mathcal{H}\) is a disjoint collection in \(\mathcal{E}\) as well. \[\begin{align*} A_n \cap E_i&=\bigcup_{j=1}^{k_n}F_{n,j_n}\cap E_i,\\ E_{i}=A \cap E_{i}&=\bigcup_{n \in \mathbb{N}}\bigcup_{j_n=1}^{k_n}F_{n,j_n} \cap E_i. \end{align*}\] Since \(A_n \in \mathcal{A}\), and \(E_i \in \mathcal{E} \subseteq \mathcal{A}\) we deduce that \(A_n \cap E_i \in \mathcal{A}\).
By the definition of \(\mu_0\) and disjointness of collection \(\mathcal{H}\) we obtain \[\mu_0(A_n \cap E_i)=\sum_{j_n=1}^{k_n}\rho(F_{n,j_n} \cap E_i),\] and by countable additivity of \(\rho\) we have \[\rho(E_i)=\sum_{n \in \mathbb{N}}\sum_{j_n=1}^{k_n}\rho(F_{n,j_n} \cap E_v)=\sum_{n \in \mathbb{N}}\mu_0(A_n \cap E_i)\] hence \[\mu_0(A)=\sum_{i=1}^{m}\rho(E_i)=\sum_{n \in \mathbb{N}}\sum_{i=1}^{m}\mu_0(A_n \cap E_i)=\sum_{n \in \mathbb{N}}\mu_0(A_n).\] Hence \(\mu_0\) is countably additive on \(\mathcal{A}\) as desired.
Proposition. Let \(X\neq\varnothing\) be a set and let \(\mu_0\) be a premeasure on an algebra \(\mathcal A\subseteq \mathcal P(X)\) then an outer measure \(\mu^{*}\) induced by \(\mu_0\) is defined by \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\] Then one has that
\(\mu_0(E)=\mu^{*}(E)\) for all \(E \in \mathcal A\),
\(\mathcal A\subseteq \mathcal M(\mu^*)\). In other words, every set in \(\mathcal A\) is \(\mu^{*}\)-measurable.
Proof of (a). Suppose \(E \in \mathcal A\), we will show that \(\mu_0(E)=\mu^{*}(E)\).
Taking \(E_1=E\) and \(E_2=E_3=\ldots=\varnothing\) we see that \(E \subseteq \bigcup_{j=1}^{\infty}E_j\) and consequently \(\mu^{*}(E) \leq \mu_0(E)\).
We now reverse this inequality and show that \(\mu_0(E) \leq \mu^{*}(E)\).
If \(E \subseteq \bigcup_{j=1}^{\infty}A_j\) with \(A_j \in \mathcal A\), then we let \[B_n=E \cap \bigg(A_n \setminus \bigcup_{j=1}^{n-1}A_j\bigg)\in\mathcal A.\] Then \(B_n \cap B_m =\ \varnothing\) if \(n \neq m\) and \(E\subseteq \bigcup_{j=1}^{\infty}B_j=\bigcup_{j=1}^{\infty}A_j\). Now \[\begin{align*} \mu_0(E)\le \sum_{j=1}^{\infty}\mu_0(B_j) \leq \sum_{j=1}^{\infty}\mu_0(A_j). \end{align*}\] It follows that \(\mu_0(E) \leq \mu^{*}(E)\) as claimed.
Proof of (b). We now prove that \(\mathcal A\subseteq \mathcal M(\mu^*)\).
Let \(C \in \mathcal A\) and \(E \subseteq X\) it suffices to show that \[\mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c).\] Let \(\varepsilon>0\), then there is a sequence \((B_n)_{n=1}^{\infty}\subseteq \mathcal A\) such that \[E \subseteq \bigcup_{n =1}^{\infty}B_n \quad \text{ and } \quad \sum_{n=1}^{\infty}\mu_0(B_n) \leq \mu^{*}(E)+\varepsilon.\] \(\mu_0\) is additive on \(\mathcal A\), thus \[\begin{align*} \mu^{*}(E)+\varepsilon \geq \sum_{j=1}^{\infty}\mu_0(B_j)= \sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c)). \end{align*}\]
Since \(\mu_0 \geq \mu^{*}\) and \(B=\bigcup_{j=1}^{\infty}B_j\) we obtain \[\begin{align*} \mu^{*}(E)+\varepsilon &\geq\sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c))\\& \geq \sum_{j=1}^{\infty}(\mu^{*}(B_j \cap C)+\mu^{*}(B_j \cap C^c))\\&\geq (\mu^{*}(B \cap C)+\mu^{*}(B \cap C^c)) \\&\geq (\mu^{*}(E \cap C)+\mu^{*}(E \cap C^c)) \end{align*}\] by subadditivity of \(\mu^*\). Letting \(\varepsilon \to 0\) we conclude \[\begin{align*} \mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c), \end{align*}\] thus \(C\) is \(\mu^{*}\)-measurable and \(\mathcal A\subseteq \mathcal M(\mu^*)\) as desired.
Carathéodory’s Theorem. Let \(\mu^{*}\) be an outer measure on \(X\), and let \[\mathcal M(\mu^*) =\big\{A\subseteq X: \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \quad \text{ for all } \quad E \subseteq X\big\}.\] be the collection of \(\mu^{*}\)-measurable sets. Then
\(\mathcal M(\mu^*)\) is a \(\sigma\)-algebra, (Carathéodory’s \(\sigma\)-algebra).
\((X, \mathcal M(\mu^*), \mu^{*})\) is a complete measure space.
Theorem. Let \(X\neq\varnothing\) be a set, let \(\mathcal A \subseteq \mathcal{P}(X)\) be an algebra, \(\mu_0\) a premeasure on \(\mathcal A\), and \(\mathcal M= \sigma(\mathcal A)\). There exists a measure \(\mu\) on \(\mathcal M\) whose restriction to \(\mathcal A\) is \(\mu_0\) and \(\mu(E)=\mu^{*}(E)\) for all \(E \in \mathcal M\), where \(\mu^{*}\) is given by \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\]
If \(\nu\) is another measure on \(\mathcal M\) that extends \(\mu_0\), then \(\nu(E) \leq \mu(E)\) for all \(E \in \mathcal M\), with equality when \(\mu(E)<\infty\).
If \(\mu_0\) is \(\sigma\)-finite, then \(\mu\) is the unique extension of \(\mu_0\) to a measure on \(\mathcal M\).
Proof. For the first part of the theorem note that:
\((X, \mathcal M(\mu^*), \mu^*)\) is a complete measure space, which follows from the Carathéodory theorem.
\(\mathcal A\subseteq \mathcal M(\mu^*)\) by the previous proposition and consequently we obtain that \(\mathcal M=\sigma(\mathcal A)\subseteq \mathcal M(\mu^*)\).
Thus it suffices to define \(\mu\) to be the restriction of \(\mu^*\) to \(\mathcal M=\sigma(\mathcal A)\). Then we see that \(\mu(E)=\mu_0(E)\) for all \(E\in \mathcal A\).
Proof of (i). Assume that \(\nu\) is another measure on \(\mathcal M=\sigma(\mathcal A)\) such that \(\nu(E)=\mu_0(E)\) for all \(E \in \mathcal A\). We show that
(a) \(\nu(E) \leq \mu(E)\) for all \(E \in \sigma(\mathcal A)\);
(b) \(\nu(E)=\mu(E)\) if \(E \in \sigma(\mathcal A)\) and \(\mu(E)<\infty\).
For (a): If \(E \in \mathcal M\) and \(E \subseteq \bigcup_{j \in \mathbb{N}}A_j\), where \((A_j)_{j\in\mathbb N} \in \mathcal A\) then \[\nu(E) \leq \sum_{j=1}^{\infty}\nu(A_j)=\sum_{j=1}^{\infty}\mu_0(A_j)\] and we conclude that \({\color{red}\nu(E) \leq \mu(E)}\).
For (b): If we set \(A=\bigcup_{j=1}^{\infty}A_j\), then we have \[\color{blue} \nu(A)=\lim_{n \to \infty}\nu\big(\bigcup_{j=1}^{n}A_j\big) =\lim_{n \to \infty}\mu_0\big(\bigcup_{j=1}^{n}A_j\big)=\lim_{n \to \infty}\mu\big(\bigcup_{j=1}^{n}A_j\big)=\mu(A).\]
If \(\mu(E)<\infty\), let \(\varepsilon>0\) and choose \(A_j\)’s from \(\mathcal A\) such that \[\mu(A) \leq \mu(E)+\varepsilon,\] hence \(\mu(A \setminus E)<\varepsilon\) and \[\mu(E) \leq {\color{blue}\mu(A) =\nu(A)}=\nu(E)+{\color{red}\nu(A \setminus E)} \leq \nu(E)+{\color{red}\mu(A \setminus E)} \leq \nu(E)+\varepsilon.\]
Letting \(\varepsilon \to 0\) we see \(\mu(E) \leq \nu(E)\) if \(E \in \mathcal M=\sigma(\mathcal A)\) and \(\mu(E)<\infty\) and we conclude in this case that \[\nu(E)=\mu(E).\]
Proof of (ii). Finally suppose that \(X=\bigcup_{j=1}^{\infty}A_j\), with \(\mu_0(A_j)<\infty\) and \(A_i \cap A_j=\varnothing\) if \(i \neq j\). Then for any \(E \in \mathcal M\) we have \[\mu(E)=\sum_{j=1}^{\infty}\mu(E \cap A_j)=\sum_{j=1}^{\infty}\nu(E \cap A_j)=\nu(E).\] This competes the proof of Carathéodory’s extension theorem.
Definition. Let \(F:\mathbb R\to \mathbb R\) be an increasing and right-continuous function. The Lebesgue–Stieltjes outer measure is defined for any \(E \subseteq \mathbb{R}\) by \[\begin{align*} \mu_{F}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}(F(b_n)-F(a_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n]\Big\}. \end{align*}\]
The Lebesgue–Stieltjes measure \(\mu_F\) is the restriction of \(\mu_F^*\) to the Carathéodory \(\sigma\)-algebra \(\mathcal{L}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) of \(\mu^{*}_F\)-measurable sets.
\(\mathcal{L}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) will be called the Lebesgue–Stieltjes \(\sigma\)-algebra.
The members of \(\mathcal{L}_{\mu_F}\) are the Lebesgue–Stieltjes measurable sets.
By Carathéodory’s theorem \((\mathbb R, \mathcal{L}_{\mu_F}, \mu_F)\) is a complete measure space.
Moreover, \((\mathbb R, \mathcal{L}_{\mu_F}, \mu_F)\) is a \(\sigma\)-finite measure space and \({\rm Bor}(\mathbb{R})\subseteq\mathcal{L}_{\mu_F}\) and \(\mu_F((a, b])=F(b)-F(a)\) for any \(-\infty\le a\le b\le \infty\).
Set functions on \(J_{oc}\) induced by increasing and right-continuous functions. Let \(F:\mathbb R\to \mathbb R\) be an increasing and right-continuous function. We define a set function \(\rho_F:J_{oc} \to [0,\infty]\) by setting \(\rho_F(\varnothing)=0\) and \[\rho_F(I)= F(b)-F(a) \quad\text{ whenever } \quad I=(a, b]\in J_{oc},\] with the understanding that \((a, \infty]=(a, \infty)\) and \(F(\infty)=\lim_{x\to \infty}F(x)\).
Remark.
(a) Let \(\mathcal{A}\) be the algebra of finite disjoint unions of elements from \(J_{oc}\). Then it is clear that \(\sigma(\mathcal{A})=\sigma(J_{oc})={\rm { Bor}}(\mathbb{R}).\)
(b) It is easy to see that \(\rho_F\) is finitely additive on the semi-algebra \(J_{oc}\), i.e. for any disjoint family \((I_j)_{j=1}^n \subseteq J_{oc}\) such that \(\bigcup_{j=1}^nI_j\in J_{oc}\) we have \[\rho_F\bigg(\bigcup_{j=1}^{n}I_j\bigg)=\sum_{j=1}^{n}\rho_F(I_j).\]
A finitely additive set function on \(\mathcal A\). Define a set function \(\mu_{F, 0}\) on \(\mathcal{A}\) by setting \[\mu_{F, 0}(E)=\sum_{i=1}^{n}\rho_F(I_i)\] for \(E=\bigcup_{j=1}^{n}I_j \in \mathcal{A}\), where \((I_j)_{j=1}^n\subseteq J_{oc}\) and \(I_i \cap I_j = \varnothing\) if \(i \neq j\).
By the first theorem and last remark \(\mu_{F, 0}\) is well defined and finitely additive on \(\mathcal{A}\).
Our goal is to show that \(\rho_F\) is countably subadditive on \(J_{oc}\). Then by the first theorem we conclude that \(\rho_F\) is countably additive on \(J_{oc}\), which in turn implies that its extension \(\mu_{F, 0}\) to \(\mathcal{A}\) is also countably additive on \(\mathcal{A}\). In other words, \(\mu_{F, 0}\) is a premeasure on \(\mathcal{A}\).
Theorem. The set function \(\rho_F:J_{oc} \to [0,\infty]\) defined above is countably subadditive on \(J_{oc}\), which ensures that its extension \(\mu_{F, 0}\) to \(\mathcal{A}\) is a premeasure on \(\mathcal A\).
Proof. We have to prove that for any \((I_n)_{n \in \mathbb{N}} \subseteq J_{oc}\) such that \(I=\bigcup_{n \in \mathbb{N}}I_n\in J_{oc}\) we have \[\rho_F(I) \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\] We first assume that \(\rho_F(I)<\infty\).
If \(I =\varnothing\) then of course \(\rho_F(I)=0 \leq \sum_{n \in \mathbb{N}}\rho_F(I_n)\).
If \(\rho_F(I_n)=\infty\) for some \(n \in \mathbb{N}\), then there is nothing to prove, since \[\rho_F(I)<\infty=\sum_{n \in \mathbb{N}}\rho_F(I_n).\]
We may assume that \(\rho_F(I_n)<\infty\) for all \(n \in \mathbb{N}\).
Let \(I=(a,b]\) for some \(a<b\) and let \(H_x=(x, \infty)\) and define \[A=\Big\{x \in [a, b]: F(b)-F(x) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_{x})\Big\}.\]
Note that if \(I_j=(c_j,d_j]\) then \(I_j \cap H_x=(\max\{c_j, x\},d_j]\), so the quantity \(\rho_F(I_j \cap H_x)=F(d_j)-F(\max\{c_j, x\})\) is always defined.
We have that \(b \in A\) since \[F(b)-F(b)=0 \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) \quad \text{ and } \quad A \subseteq [a,b].\]
Thus \(c=\inf A\) is defined and \(c=\inf A \in [a,b]\). We show that \(c \in A\). Indeed, right-continuity of \(F\) implies \[F(b)-F(c)=\inf_{x \in A}(F(b)-F(x)) \leq \inf_{x \in A}\sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c),\] since \(\rho_F(I_j \cap H_x)\le \rho_F(I_j \cap H_c)\) as \(c\le x\in A\).
We now prove that \(c=a\). If \(a<c\), then \(c \in (a,b]\) so there is \(k \in \mathbb{N}\) such that \(c \in I_k=(c_k,d_k]\), then \(x=\max\{c_k,a\}<c\le b\).
For each \(j \in \mathbb{N}\) we have \[\rho_F(I_j \cap H_c) \leq \rho_F(I_j \cap H_x)\] while \[\rho_F(I_k \cap H_x)=\rho_F(I_k \cap H_c)+(F(c)-F(x))\] and consequently \(x\in A\), since \[\begin{align*} \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) &\geq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c)+(F(c)-F(x)) \\ &\geq (F(b)-F(c))+(F(c)-F(x))=F(b)-F(x). \end{align*}\]
But \(x<c=\inf A\), which is impossible. Thus \(a=c \in A\), so \[\rho_F(I)=F(b)-F(a) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j).\]
Suppose now that \(\rho_F(I)=\infty\).
Assume that \(I=(a,\infty)=\bigcup_{n \in \mathbb{N}}I_n\in J_{oc}\) for some \(a \in \mathbb{R}\). For any \(n \in \mathbb{N}\) such that \(n>a\) we have \((a,n] \subseteq (a,\infty)\) and \[\rho_F((a,n])\le \rho_F((a,\infty))\] then \[F(n)-F(a) \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\]
Passing with \(n \to \infty\) we get \[\rho_F(I)=\infty \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\]
This completes the proof of the theorem.
The outer measure \(\mu_F^{*}\) is induced by the premeasure \(\mu_{F, 0}\) on the algebra \(\mathcal A\) of finite disjoint unions of elements from \(J_{oc}\). The premeasure \(\mu_{F, 0}\) is \(\sigma\)-finite, since \(\mathbb R=\bigcup_{n\in\mathbb N}(n, n+1]\).
\((\mathbb R, \mathcal{L}_{\mu_F}, \mu_F)\) is a complete \(\sigma\)-finite measure space.
By the Carathéodory extension theorem we conclude that
\(\mu_F(E)=\mu_{F}^{*}(E)=\mu_{F, 0}(E)\) for any \(E \in \mathcal{A}\);
\(\mathcal{A} \subseteq \sigma(J_{oc})=\sigma(\mathcal{A})={\rm Bor}({\mathbb{R}}) \subseteq \mathcal{L}_{\mu_F}\).
Therefore it follows that \(\mu_F\) is a \(\sigma\)-finite Borel measure and \[\mu((a, b])=F(b)-F(a)\] for any \(-\infty\le a\le b\le \infty\).
Moreover \(\mu_F\) is unique in the sense that if \(G\) is another such function, we have \(\mu_F=\mu_G\) if and only if \(F-G\) is constant.
Conversely, if \(\mu\) is a Borel measure on \(\mathbb R\) that is finite on all bounded Borel sets and we define \[F(x)=\ \begin{cases} \mu((-\infty, x]) & \text{ if } x>0,\\ 0 & \text{ if } x=0,\\ -\mu((x, 0])& \text{ if } x<0,\\ \end{cases}\] then \(F\) is increasing and right continuous and \(\mu=\mu_F\).