# Pre-Measure of Finite Stieltjes Function is Pre-Measure

## Theorem

Let $\mathcal J_{ho}$ denote the collection of half-open intervals in $\R$.

Let $f: \R \to \R$ be a finite Stieltjes function.

Then the pre-measure of $f$, $\mu_f: \mathcal{J}_{ho} \to \overline \R_{\ge 0}$ 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 $\mu_f \left({\varnothing}\right) = 0$.

Now suppose that for some half-open interval $\left[{a \,.\,.\, b}\right)$ one has:

$\left[{a \,.\,.\, b}\right) = \displaystyle \bigcup_{n \mathop \in \N} \left[{b_n \,.\,.\, b_{n+1}}\right)$

where $b_0 = a$ and $\displaystyle \lim_{n \mathop \to \infty} b_n = b$.

Then we compute:

 $\displaystyle \sum_{n \mathop \in \N} \mu_f \left({\left[{b_n \,.\,.\, b_{n + 1} }\right)}\right)$ $=$ $\displaystyle \sum_{n \mathop \in \N} f \left({b_{n + 1} }\right) - f \left({b_n}\right)$ Definition of $\mu_f$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} f \left({b_{n + 1} }\right) - f \left({b_0}\right)$ Telescoping Series $\displaystyle$ $=$ $\displaystyle f \left({b}\right) - f \left({a}\right)$ Definition of $\left({b_n}\right)_n$ $\displaystyle$ $=$ $\displaystyle \mu_f \left({\left[{a \,.\,.\, b}\right)}\right)$ Definition of $\mu_f$

which verifies the second condition for a pre-measure.

Hence $\mu_f$ is indeed a pre-measure.

$\blacksquare$