Muhammad Haris Rao
Throughout, let $\mathcal{A}$ be the algebra of finite unions of half open intervals on $\mathbb{R}$ \begin{align*} \mathcal{A} &= \bigcup_{n \ge 0} \left\{ \bigcup_{k = 1}^n \left( a_k, b_k \right] \mid -\infty \le a_k < b_k \le \infty \text{ for all $k \le n$} \right\} \subseteq \mathcal{P}(\mathbb{R}) \end{align*}
Theorem: Let $F : \mathbb{R} \longrightarrow \mathbb{R}$ be non-decreasign and right-continuous. The function $\mu_0 : \mathcal{A} \longrightarrow \mathbb{R}_{\ge 0}$ defined by \begin{align*} \mu_0 \left( \bigsqcup_{k = 1}^n \left( a_k, b_k \right] \right) &= \sum_{k = 1}^n \left( F(b_k) - F(a_k) \right) \end{align*} is a premeasure on the algebra $\mathcal{A}$.
Proof. First, we have to show that this funciton is well-defined. For this, two things need to be demonstrated: that every element of $\mathcal{A}$ can be written as a disjoint union of half open intervals, and that the evaluation of the function in accordance with the rule does not depend on this representation. Any finite union of half open intervals is also a disjoint finite union because \begin{align*} \bigcup_{k = 1}^n \left( a_k , b_k \right] &= \bigsqcup_{S \subseteq \{ 1,2, \cdots, n \} \\ |S| \ge 1} \left( \bigcap_{k \in S} \left( a_k, b_k \right] \cap \bigcap_{k \notin S} \left( -\infty, a_k \right] \cap \bigcap_{k \notin S} \left( b_k, \infty \right] \right) \end{align*} Now to show that the expression defining $\mu_0$ is independent of representation. If we write for a single half open interval $(a, b] = \bigsqcup_{k = 1}^n ( a_k, b_k]$, then after relabelling we may assume that $a = a_1 \le b_1 \le a_2 \le b_2 \le \cdots \le a_n \le b_n = b$. Then, we have a telescoping sum \begin{align*} F(b) - F(a) = \sum_{k = 1}^n \left( F(b_k) - F(a_k) \right) \end{align*} This means that $F$ is well-defined when the argument is a half open interval. Then for finite unions of half open intervals with $\bigsqcup_{k = 1}^n (a_k, b_k] = \bigsqcup_{\ell = 1}^m (a_\ell^\prime, b_\ell^\prime]$ we will have \begin{align*} \sum_{k = 1}^n \left( F(b_k) - F(a_k) \right) &= \sum_{k = 1}^n \mu_0 \left( \left(a_k, b_k \right] \right) \\ &= \sum_{k = 1}^n \sum_{\ell = 1}^m \mu_0 \left( \left(a_k, b_k \right] \cap \left( a_k^\prime, b_\ell^\prime \right] \right) \\ &= \sum_{\ell = 1}^m \sum_{k = 1}^n \mu_0 \left( \left(a_k, b_k \right] \cap \left( a_k^\prime, b_\ell^\prime \right] \right) \\ &= \sum_{\ell = 1}^m \mu_0 \left( \left(a_\ell^\prime, b_\ell^\prime \right] \right) \\ &= \sum_{\ell = 1}^m \left( F(b_\ell) - F(a_\ell) \right) \end{align*} which proves that $\mu_0$ is well-defined on all of $\mathcal{A}$. Now to prove that it is a premeasure. Suppose $\bigsqcup_{k \ge 1} (a_k, b_k] = (a, b]$. Then, we have that \begin{align*} (a, b] &= \bigsqcup_{k = 1}^n (a_k, b_k] \sqcup \left( (a, b] \bigg\backslash \bigsqcup_{k = 1}^n (a_k, b_k] \ \right) \end{align*} The second expression in the brackets is a finite union of half open intervals as it is an element of the algebra $\mathcal{A}$. Thus, we can apply finite additivity of $\mu_0$ to get \begin{align*} \mu_0 \left( (a, b] \right) &= \mu_0 \left( \bigsqcup_{k = 1}^n (a_k, b_k] \right) + \mu_0 \left( (a, b] \bigg\backslash \bigsqcup_{k = 1}^n (a_k, b_k] \ \right) \ge \mu_0 \left( \bigsqcup_{k = 1}^n (a_k, b_k] \right) = \sum_{k = 1}^n \mu_0 \left( (a_k, b_k] \right) \end{align*} Taking $n \to \infty$ yields \begin{align*} \mu_0 \left( (a, b] \right) \ge \sum_{k = 1}^n \mu_0 \left( (a_k, b_k] \right) \end{align*} Now for the other inequality. Let $\varepsilon > 0$ be arbitrary, and let $\delta > 0$ be such that $F(a + \delta) - F(a) < \varepsilon/2$. For each $k \ge 1$, let $\delta_k > 0$ be such that \begin{align*} F(b_k + \delta_k) - F(b_k) < \frac{\varepsilon}{2^{k + 1}} \end{align*} The collection of open intervals $\{ (a_k, b_k + \delta_k) \}_{k \ge 1}$ then covers the compact set $[a + \delta, b]$. Hence, there is a finite subcover which we denote $\{ (a_k^\prime, b_k^\prime)\}_{k = 1}^N \subseteq \{ (a_k, b_k + \delta_k) \}_{k \ge 1}$. We may assume that $a_1 < a_2 < \cdots < a_N$ and that $b_k \in (a_{k + 1}, b_{k + 1} + \delta_k)$ for $k \in \{ 1, \cdots, N - 1 \}$. It follows that \begin{align*} \mu_0 \left( (a, b] \right) &= F(b) - F(a) \\ &< F(b) - F(a + \delta) + \frac{\varepsilon}{2} \\ &\le F\left( b_N + \delta_N \right) - F(a_1) + \frac{\varepsilon}{2} \\ &= F\left( b_N + \delta_N \right) - F(a_N) + \sum_{k = 1}^{N - 1} \left( F \left( a_{k + 1} \right) - F \left( a_k \right) \right) + \frac{\varepsilon}{2} \\ &< F\left( b_N + \delta_N \right) - F(a_N) + \sum_{k = 1}^{N - 1} \left( F \left( b_k + \delta_k \right) - F \left( a_k \right) \right) + \frac{\varepsilon}{2} \\ &= \sum_{k = 1}^N \left( F \left( b_k + \delta_k \right) - F \left( a_k \right) \right) + \frac{\varepsilon}{2} \\ &= \sum_{k = 1}^N \left( F \left( b_k + \delta_k \right) -F \left( b_k \right) \right) + \sum_{k = 1}^N \left( F \left( b_k \right) - F \left( a_k \right) \right) + \frac{\varepsilon}{2} \\ &= \varepsilon \sum_{k = 1}^\infty \frac{1}{2^{k + 1}} + \sum_{k = 1}^\infty \left( F \left( b_k \right) - F \left( a_k \right) \right) + \frac{\varepsilon}{2} \\ &= \sum_{k = 1}^\infty \mu_0 \left( (a_k, b_k] \right) + \varepsilon \end{align*}