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A $\textbf{box}$ $Q\subset \mathbb{R}^n$ is defined as $Q:= (a_1,b_1) \times \ldots \times (a_n,b_n)$ and the same with closed or half opened intervals where $a_i < b_i \in \mathbb{R}$ and in case of open, half-opened intervals we allow $\pm \infty$. The $\textbf{volume}$ of a box $Q$, call it $\lambda(Q)$, is the product of the lengths of the intervals. An $\textbf{elementary set}$ $E$ is the finite disjoint union of boxes: $E:= \dot\bigcup_{i=1}^m Q_i$ with volume $\lambda(E)= \sum_{i=1}^m \lambda(Q_i)$.

The elementary sets form an algebra $\mathcal{A}$. This volume function $\lambda$ is a premeasure, but I can't prove that it is $\sigma$-additive, meaning for any disjoint sequence $\lbrace{E_i\rbrace}_{i\in \mathbb{N}} \subset \mathcal{A}$ with $\dot\bigcup_{i=1}^\infty E_i \in \mathcal{A}$ it follows that $\lambda(\dot\bigcup_{i=1}^\infty E_i)= \sum_{i=1}^\infty \lambda(E_i)$.

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  • $\begingroup$ You would first need to define what is meant by the volume of an infinite union of elementary sets. $\endgroup$
    – copper.hat
    Commented Jun 17 at 0:20
  • $\begingroup$ @copper.hat $\bigcup_{i=1}^\infty E_i$ belongs to $\mathcal{A}$, so $\lambda(\bigcup_{i=1}^\infty E_i)$ is well defined. $\endgroup$
    – azif00
    Commented Jun 17 at 2:08
  • $\begingroup$ @azif00 Thanks, I missed that. $\endgroup$
    – copper.hat
    Commented Jun 17 at 5:11
  • $\begingroup$ Look at math.stackexchange.com/a/2879192/27978 $\endgroup$
    – copper.hat
    Commented Jun 19 at 5:12

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