5.6 Lebesgue-Stieltjes Measure

Proof. $\frak{S}$ has an action of $(\R,+)$ by translation $c*F=F+c$, for $F\in\fraks{S},x\in\R$. For any $F\in\fraks{S},x\in\R$ we have $\til{F}\define F-F(x)\in\fraks{S}_{x}$ so $\R F\cap\fraks{S}_{x}\neq \emp$. Let $G,H\in \R F\cap\fraks{S}_{x}$ then $G=H+c$ for some $a\in\R$ so

so $G=H$. By Lemma 5.6.2 $\fraks{S}_{x}\iso \fraks{S}/\sim$ as sets. Since $G-F+c$ is non-decreasing if and only if $G-F$ is non-decreasing for any $c\in\R$ we have by Lemma 5.6.2(1) $\fraks{S}_{x}\iso \fraks{S}/\sim$ as posets.

Let $F,G\in\fraks{S}$, $\al,\be\geq 0$ then

so $\fraks{S}$ is convex cone. Similarly so is $\fraks{S}_{x}$^[1]. By Lemma 5.6.2(2) we have an ordered convex cone isomorphism.$\square$

Proof. Consider the map $\p:Y\to X/G$ defined by $y\mto [y]$. Let $y_{1},y_{2}\in Y$ then $\p(y_{1})=\p(y_{2})\iff [y_{1}]=[y_{2}]\iff y_{1}=y_{2}$ since $\curl{y_1,y_2}\suf [y_{1}]\cap Y$. Let $E\in X/G$ then $E=[x]$ for some $x\in X$. Since $\abs{[x]\cap Y}=1$ there exists $y\in Y$ such that $y\in [x]$ so $[y]=[x]=\p(x)$.

1. (1)

   By hypothesis $[x_{1}]\leq_{X/G}[x_{2}]\iff x_{1}\leq_{X} x_{2}$ is a well-defined partial order on $X/G$. The restriction of $\leq_{Y}\define \leq_{X}|_{Y}$ to $Y$ is a partial order on $Y$. Suppose $y_{1},y_{2}\in Y$ then

   \begin{align*}y_{1}\leq_{Y} y_{2}\iff y_{1}\leq_{X} y_{2}\iff \braks{y_1}\leq_{X/G}\braks{y_2}\iff\p(y_{1})\leq_{X/G}\p(y_{2})\end{align*}

   
   
2. (2)

   By hypothesis $X/G$ is a convex cone with

   \begin{align*}\braks{x_1}+\braks{x_2}=\braks{x_1+x_2}&&\lam\braks{x}=\braks{\lam x_1}\end{align*}

   
   

   Suppose $y_{1},y_{2}\in Y,\lam\geq 0$ then

   \begin{align*}\p(y_{1}+y_{2})=\braks{y_1+y_2}=\braks{y_1}+\braks{y_2}=\p(y_{1})+\p(y_{2})\\ \p(\lam y_{1})=\braks{\lam y_1}=\lam \braks{y_1}=\lam\p(y_{1})\end{align*}

   
   

$\square$

Proof. (1): Let $(a,b],(c,d]\in\cali{I}$ and assume without loss of generality that $a<b$ and $c<d$ then

(2): By Proposition 4.4.2, $\cali{A}$ is a ring.$\square$

Proof. By Definition 5.6.1 it is equivalent to consider $\fraks{S}_{x}$ over $\fraks{S}/\sim$. By Proposition 4.3.4, $\cali{B}_{\R}=\s(\cali{I})=\s(\cali{A})$. We define

Let $(a,b]\in\cali{I}$ with $(a,b]=\dcups{n}{j=1}(a_{j},b_{j}]$. Assume without loss of generality that $\curl{a_j}_{1}^{n}$ is non-decreasing, then $b_{j}=a_{j+1}$ for each $1\leq j\leq n-1,b=b_{n}$ and $a=a_{1}$. Thus

so $\mu_{0}$ is well-defined and finitely-additive on $\cali{A}$. Suppose $(a,b]=\dcups{}{n\in\N}(a_{n},b_{n}]\in\cali{A}$ for $\curl{(a_n,b_n]}_{n\in\N}\suf\cali{I}$. By finite additivity, for each $N\in\N$

thus

Let $\e>0$. By right-continuity of $F$, pick $\de,\de_{n}>0$ for each $n\in\N$ such that

The collection $\curl{(a_n,b_n+\de_n)}_{n\in\N}$ forms an open cover for $[a+\de,b]$ so by compactness tehre exists $N\in\N$ such that

Thus

Since $\e$ was arbitrary

Thus $\mu_{0}$ is a premeasure onn $\cali{A}$. Since $\R=\cups{}{k\in\Z}(k,k+1]$ and $\mu_{0}((k,k+1])=F(k+1)-F(k)<\infty$ the premeasure is $\s$-finite. By Carathéodory Extension Theorem, there exists a unique Borel measure $\mu_{F}$ on $\s(\cali{A})=\cali{B}_{\R}$ extending $\mu_{0}$. Let $K\suf\R$ be compact then $K\suf [a,b]$ for some $-\infty<a<b<\infty$ then by monotonicity of measure $\mu_{F}(K)<\infty$.

If $F\sim G$ then for any $-\infty<a<b<\infty$

Since $\mu_{F}$ and $\mu_{G}$ are Borel measures and $\cali{B}_{\R}=\s(\cali{A})$, we have $\mu_{F}=\mu_{G}$ so the map is well-defined. Suppose $\mu_{F}=\mu_{G}$ then for all $a<b\in\R$

so $F-G$ is constant. Since $F(x)=G(x)=0$, the constant is $0$ so $F=G$. Therefore the map is injective. Let $\mu\in\cali{M}_{lf}$ then $F_{\mu}\in\fraks{S}$ by monotonicity and continuity from above. Observe $F_{\mu}(x)=0$ so $F_{\mu}\in\fraks{S}_{x}$. For $-\infty<a<b<\infty$

In all three cases, $\mu((a,b])=F(b)-F(a)$. By definition of $\mu_{F_\mu}$, $\mu_{F_\mu}|_{\cali{A}}=\mu|_{\cali{A}}$ so $\mu_{F_\mu}=\mu$. Therefore the map is surjective.

Suppose $F,G\in\fraks{S}_{x},\lam\geq 0$ then

Suppose $F\leq G$ then $G-F$ is non-decreasing so $G(a)-F(a)\leq G(b)-F(b)$ thus

$\square$