Pre-Measure of Finite Stieltjes Function is Pre-Measure
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Theorem
Let $\JJ_{ho}$ denote the collection of half-open intervals in $\R$.
Let $f: \R \to \R$ be a finite Stieltjes function.
Then the pre-measure of $f$, $\mu_f: \JJ_{ho} \to \overline \R_{\ge 0}$ is a pre-measure.
Here, $\overline \R_{\ge 0}$ denotes the set of positive extended real numbers.
Proof
It is immediate from the definition of $\mu_f$ that:
- $\map {\mu_f} \O = 0$
Now suppose that for some half-open interval $\hointr a b$ one has:
- $\ds \hointr a b = \bigcup_{n \mathop \in \N} \hointr {b_n} {b_{n + 1} }$
where $b_0 = a$ and $\ds \lim_{n \mathop \to \infty} b_n = b$.
Then we compute:
\(\ds \sum_{n \mathop \in \N} \map {\mu_f} {\hointr {b_n} {b_{n + 1} } }\) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map f {b_{n + 1} } - \map f {b_n}\) | Definition of Pre-Measure of Finite Stieltjes Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map f {b_{n + 1} } - \map f {b_0}\) | Telescoping Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f b - \map f a\) | Definition of $\sequence {b_n}_{n \mathop \in \N}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mu_f} {\hointr a b}\) | Definition of Pre-Measure of Finite Stieltjes Function |
which verifies the second condition for a pre-measure.
Hence $\mu_f$ is indeed a pre-measure.
$\blacksquare$