14.8 Lebesgue-Stieltjes Measures
Definition 14.8.1 (Stieltjes Function). Let $F: \real \to \real$, then $F$ is a Stieltjes function if $F$ is right-continuous and non-decreasing.
Definition 14.8.2 (h-interval). A h-interval is an interval of the form $(a, b] \subset \real$ where $-\infty < a \le b < \infty$.
Lemma 14.8.3. Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the collection of all $h$-intervals, then:
$\ci$ is an elementary family.
The collection
\[\alg = \bracs{\bigsqcup_{j = 1}^n I_j \bigg | \seqf{I_j} \subset \ci, n \in \natp}\]is a ring.
Proof. (1): Let $(a, b], (c, d] \in \ci$ and assume without loss of generality that $a < b$ and $c < d$, then
$(a, b] \cap (c, d] = (\max(a, c), \min(b, d)] \in \ci$.
$(a, b] \setminus (c, d] = (a, \min(b, c)] \sqcup (\max(a, d), b]$.
(2): By Proposition 13.3.2, $\alg$ is a ring.$\square$
Definition 14.8.4 (Lebesgue-Stieltjes Measure, [Theorem 1.16, Fol99]). Let $\mu: \cb_{\real} \to [0, \infty]$ be a Borel measure on $\real$ such that for any $K \subset \real$ compact, $\mu(K) < \infty$, then
There exists a Stieltjes function $F: \real \to \real$ such that for all $-\infty < a < b < \infty$,
\[\mu((a, b]) = F(b) - F(a)\]For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_{F}: \cb_{\real} \to [0, \infty]$ satisfying (1), and $\mu_{F}$ is the Lebesgue-Stieltjes measure associated with $F$.
Proof. (1): Let
then $F$ is a Stieltjes function by monotonicity and continuity from above (Proposition 14.1.5). For any $-\infty < a < b < \infty$,
In all three cases, $\mu((a, b]) = F(b) - F(a)$.
(U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
Converse: Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
Define
Let $(a, b] \in \ci$ with $(a, b] = \bigsqcup_{j = 1}^{n} (a_{j}, b_{j}]$. Assume without loss of generality that $\seqf{a_j}$ is non-decreasing, then $b_{j} = a_{j+1}$ for each $1 \le j \le n - 1$, $b = b_{n}$, and $a = a_{1}$. Thus
and $\mu_{0}$ is well-defined and finitely-additive on $\alg$.
Let $\seq{(a_n, b_n]}\subset \ci$ such that $\bigcup_{n \in \natp}(a_{n}, b_{n}] = (a, b]$. By monotonicity,
Let $\eps > 0$. By right-continuity of $F$, there exists $\delta > 0$ such that $F(a + \delta) - F(a) < \eps/2$, and $\seq{\delta_n}\subset \real_{> 0}$ such that $F(b_{n} + \delta_{n}) - F(b_{n}) < \eps/2^{n+1}$ for all $n \in \natp$. The intervals $\seq{(a_n, b_n + \delta_n)}$ forms an open cover for $[a + \delta, b]$. By compactness, there exists $N \in \natp$ such that
Thus
Since such an $N$ exists for each $\eps > 0$,
Therefore $\mu_{0}$ is a premeasure on $\alg$. By Carathéodory’s Extension Theorem, $\mu_{0}$ extends uniquely to a Borel measure on $\real$, which satisfies (1).$\square$