Definition 5.6.5: Lebesgue-Stieltjes Measure

Definition 5.6.5 (Lebesgue-Stieltjes Measure).label Let $\fraks{S}_{x}=\fraks{S}_{x}(\R,\R)$ and $\cali{M}_{lf}=\cali{M}_{lf}(\R)$ be as in Definition 5.6.1 and Definition 5.1.6 respectively. Let $\cali{A}$ be the ring from Lemma 5.6.4 then $\cali{B}_{\R}=\s(\cali{A})$. There exists a unique Borel measure $\mu_{F}\in\cali{M}_{lf}$ satisfying

\begin{align*}\mu_{F}((a,b])=F(b)-F(a)&&\fall-\infty<a<b<\infty\end{align*}

Moreover for any fixed $x\in\R$, this gives an ordered convex cone isomorphism

\begin{align*}\p:\fraks{S}_{x}&\bij \cali{M}_{lf}\\ F&\mto\mu_{F}\end{align*}

whose inverse is

\begin{align*}\mu\mto F_{\mu}&&F_{\mu}(t)=\begin{cases}\mu((x,t])&t\geq x\\ -\mu((t,x])&t\leq x\end{cases}\end{align*}

The measure $\mu_{F}$ is called the Lebesgue-Stieltjes measure associated with $F$.

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

\begin{align*}\mu_{0}:\cali{A}\to [0,\infty)&&\dcups{n}{j=1}(a_{j},b_{j}]\mto \sums{n}{j=1}\braks{F(b_j)-F(a_j)}\end{align*}

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

\begin{align*}F(b)-F(a)=F(b_{n})+\sums{n-1}{j=1}\braks{F(b_j)-F(a_{j+1})}-F(a_{1})=\sums{n}{j=1}\braks{F(b_j)-F(a_j)}\end{align*}

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 $N\in\N$

\begin{align*}\sums{}{n\leq N}\mu_{0}((a_{n},b_{n}])\leq\mu_{0}((a,b])\end{align*}

thus

\begin{align*}\sums{}{n\in\N}\mu_{0}((a_{n},b_{n}])\leq\mu_{0}((a,b])\end{align*}

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

\begin{align*}F(a+\de)-F(a)<\frac{\e}{2}&&F(b_{n}+\de_{n})-F(b_{n})<\frac{\e}{2^{n+1}}\end{align*}

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

\begin{align*}[a+\de,b]\suf \cups{}{n\leq N}(a_{n},b_{n}+\de)\suf \cups{}{n\leq N}(a_{n},b_{n}+\de_{n}]\end{align*}

Thus

\begin{align*}\mu_{0}((a,b])&\leq F(a+\de)-F(a)+\mu_{0}([a+\de,b])\\&\leq \frac{\e}{2}+\sums{}{n\leq N}\mu_{0}((a_{n},b_{n}+\de_{n}])\\&\leq \frac{\e}{2}+\sums{}{n\leq N}\mu_{0}((a_{n},b_{n}])+\sums{}{n\leq N}\braks{F(b_n+\de_n)-F(b_n)}\\&\leq \e+\sums{}{n\leq N}\mu_{0}((a_{n},b_{n}])\end{align*}

Since $\e$ was arbitrary

\begin{align*}\mu_{0}((a,b])\leq \sums{}{n\leq N}\mu_{0}((a_{n},b_{n}])\leq \sums{}{n\in\N}\mu_{0}((a_{n},b_{n}])\end{align*}

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 $-\infty<a<b<\infty$

\begin{align*}\mu_{F}((a,b])=F(b)-F(a)=G(b)-G(a)=\mu_{G}((a,b])\end{align*}

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 $a<b\in\R$

\begin{align*}F(b)-F(a)=G(b)-G(a)\end{align*}

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 $-\infty<a<b<\infty$

\begin{align*}\mu((a,b])=\begin{cases}\mu((0,b])-\mu((0,a])&a,b\geq0\\ \mu((a,0])-\mu((0,b])&a\leq 0,b\geq0\\ \mu((a,0])-\mu((b,0])&a,b\leq0\end{cases}\end{align*}

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

\begin{align*}\p(\lam F+G)((a,b])&=\mu_{\lam F+G}((a,b])\\&=(\lam F+G)(b)-(\lam F+G)(a)\\&=\lam (F(b)-F(a))+ (G(b)-G(a))\\&=\lam\mu_{F}((a,b])+\mu_{G}((a,b])\\&=\lam \p(F)((a,b])+\p(G)((a,b])\end{align*}

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

\begin{align*}\p(F)((a,b])=\mu_{F}((a,b])=F(b)-F(a)\leq G(b)-G(a)=\mu_{G}((a,b])=\p(G)((a,b])\end{align*}

$\square$