11.6 The Lebesgue-Stieltjes Integral

Definition 11.6.1 (Lebesgue-Stieltjes Measure). Let $[a, b] \subset \real$, $E$ be a Banach space, and $G \in BV([a, b]; E^{*})$, then there exists a unique $\mu_{G} \in M_{R}([a, b]; E^{*})$ such that:

For any $f \in \text{Reg}([a, b], G; E)$,
\[\int_{a}^{b} f(t) G(dt) = \int_{[a, b]}\dpn{f(t), \mu_G(dt)}{E}\]
For any $a \le c < d < b$,
\[\mu_{G}((c, d]) = G(d+) - G(c+) = \lim_{z \downto d}G(z) - \lim_{z \downto c}G(z)\]

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

Proof. By Proposition 11.4.3, the mapping $f \mapsto \int f(t)G(dt)$ is a continuous linear functional on $C_{0}([a, b]; E) = C([a, b]; E)$. By Singer’s Representation Theorem, there exists $\mu_{G} \in M_{R}([a, b]; E^{*})$ such that (1) holds for $C([a, b], G; E)$.

(2): Fix $x \in [a, b)$. For each $u \in (x, b)$, there exists a non-increasing function $\phi_{u} \in C_{c}([a, b))$ such that $\phi_{u}|_{[a, x]}= 1$ and $\supp{\phi_u}\subset [a, u]$. By outer regularity of $\mu_{G}$,

\[\mu_{G}([a, x]) = \lim_{u \downto x}\mu_{G}([a, u]) = \lim_{u \downto x}\int_{[a, b]}\phi_{u} d\mu_{G} = \lim_{u \downto x}\int_{a}^{b} \phi_{u} dG\]

For each $u > x$, using integration by parts,

\begin{align*}\int_{a}^{b} \phi_{u} dG&= \phi(b)G(b) - \phi(a)G(a) - \int_{a}^{b} G d\phi_{u} \\&= - G(a) - \int_{a}^{b} G d\phi_{u}\end{align*}

Let $v \in (x, u)$ and $(P = \bracs{x_j}_{0}^{n}, c = \bracs{c_j}_{1}^{n})$ be a tagged partition containing $\bracs{x, v, u}$, then since $\phi_{u}$ is non-increasing,

\begin{align*}\int_{a}^{b} Gd\phi_{u}&= \sum_{x < x_j \le v}G(c_{j})[\phi(x_{j}) - \phi(x_{j - 1})] + \sum_{v < x_j \le u}G(c_{j})[\phi(x_{j}) - \phi(x_{j - 1})] \\&\in [\phi(v) - \phi(x)]G([x, u]) + [\phi(u) - \phi(v)]G((x, u]) \\&\subset [\phi(v) - \phi(x)]G([x, u]) - G((x, u])\end{align*}

Since the above holds for all $v \in (x, u)$, $\int_{a}^{b} Gd\phi_{u} \in -\overline{G((a, u])}$, and

\[\int_{a}^{b} \phi_{u} dG \in \overline{G(x, u]}- G(a)\]

As $E^{*}$ is a Banach space, $G(x+)$ exists by Proposition 11.2.4, so

\[\mu_{G}([a, x]) = \lim_{u \downto x}\int_{a}^{b} \phi_{u} dG = \bigcap_{u > x}\overline{G((x, u])}- G(a) = G(x+) - G(a)\]

(1): Let $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]}\in RS([a, b], G)$ with

\[\int_{a}^{b} x \cdot \one_{[c, d]}dG = x[G(d) - G(c)] = \int_{[a, b]}x \cdot \one_{[c, d]}d\mu_{G}\]

By linearity, for any regulated step map $f: [a, b] \to E$, $\int_{a}^{b} f(t) G(dt) = \int_{[a, b]}\dpn{f(t), \mu_G(dt)}{E}$. Therefore (1) holds for $\text{Reg}([a, b], G; E)$ by the Linear Extension Theorem.$\square$

Proposition 11.6.2. Let $[a, b] \subset \real$, $E$ be a normed space, $G \in C^{1}([a, b]; E^{*})$, and $\mu_{G} \in M_{R}([a, b]; E^{*})$ be the associated Lebesgue-Stieltjes measure, then

\[\mu_{G}(dt) = DG(t)dt\]

Proof. By the Fundamental Theorem of Calculus, for any $[c, d] \subset [a, b]$,

\[\int_{a}^{b} \one_{[c, d]}dG = \int_{a}^{b} \one_{[c, d]}DG(t)dt\]

By Definition 11.6.1, applied to both $G$ and $\text{Id}$,

\[\int_{[a, b]}\one_{[c, d]}d\mu_{G} = \int_{a}^{b} \one_{[c, d]}dG = \int_{a}^{b} \one_{[c, d]}DG(t)dt = \int_{[a, b]}\one_{[c, d]}DG(t)dt\]

Thus by linearity, for any regulated step map $f: [a, b] \to E$,

\[\int_{[a, b]}f(t) \mu_{G}(dt) = \int_{[a, b]}\dpn{f(t), DG(t)}{E}dt\]

By Definition 11.5.3, the regulated step maps are dense in $\text{Reg}([a, b], G; E)$. Therefore by the Linear Extension Theorem, the above holds for all $f \in \text{Reg}([a, b], G; E)$. By uniqueness of the Lebesgue-Stieltjes measure, $\mu_{G}(dt) = DG(t)dt$.$\square$

Jerry's Digital Garden

11.6 The Lebesgue-Stieltjes Integral

Direct References

Jerry's Digital Garden

Direct References