Below is the description of my problem.
Definition: An interval is a subset of $\mathbb{R}$ of the form $(a,b]$, $[a,b]$, $(a,b)$, $[a,b)$, $(a,\infty)$, $[a,\infty)$, $(-\infty,b)$, $(-\infty,b]$ or $(-\infty,\infty)=\mathbb{R}$ in which $a< b$ are real numbers.
Proposition: Let $$\mathbb{A}:=\big\{\cup_{k=1}^nI_k\in 2^\mathbb{R}:I_k=\emptyset \vee I_k\text{ is an interval}\vee (\exists a\in\mathbb{R})(I_k=\{a\})\big\}.$$ Then $\mathbb{A}$ is an algebra of subsets of $\mathbb{R}$. Define $\rho:\mathbb{A}\to\overline{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,\infty\}$ by
- $\rho(\emptyset ):=0$
- If $I\neq\emptyset $ is a bounded interval, then $\rho (I):=\sup I-\inf I$;
- If $I$ is a unbounded interval, then $\rho(I):=\infty$;
- If $I$ and $J$ are two disjoint intervals, then $\rho (I\cup J):=\rho (I)+\rho (J)$.
Then $\rho$ is a premeasure.
I'm trying to demonstrate that $\rho:\mathbb{A}\to\overline{\mathbb{R}}$ is a premeasure. The only thing I haven't been able to demonstrate is this: if $\{A_n\}_{n\in\mathbb{N}}$ is a collection of pairwise disjoint sets of $\mathbb{A}$, then $\rho (\cup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}\rho (A_n)$.
I already demonstrated that $\sum_{n\in\mathbb{N}}\rho (A_n)\leq \rho(\cup_{n\in\mathbb{N}}A_n)$. Now I need to prove the following inequality: $ \rho(\cup_{n\in\mathbb{N}}A_n)\leq \sum_{n\in\mathbb{N}}\rho (A_n)$.
My question: How can I prove that if $\{A_n\}_{n\in\mathbb{N}}$ is a collection of pairwise disjoint sets of $\mathbb{A}$, then $\rho (\cup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}\rho (A_n)$?
Below is a lemma that I'm trying to use to prove the inequality $ \rho(\cup_{n\in\mathbb{N}}A_n)\leq \sum_{n\in\mathbb{N}}\rho (A_n)$.
Lemma: Let $\varepsilon >0$ and $n\in\mathbb{N}$ be any elements. If $I\subset \mathbb{R}$ is a bounded interval, then there's an open interval $J\subset\mathbb{R}$ such that $I\subseteq J$ and $\rho (J)= \rho (I)+\varepsilon /2^n$