# Volume function is additive set-function

I have included some definitions right before the theorem, for convenience.


Definition. Let $\cA$ be a set of subsets of $\bR^n$.
Then $\cA$ is a field of subsets of $\bR^n$ when it has the following properties:

(FS-1) The empty set $\emptyset$ belongs to $\cA$.

(FS-2) If $A,B \in \cA$, then the set-difference $A \setminus B$ still belongs to $\cA$.

(FS-3) If $A_1, \ldots , A_k \in \cA$, then the union $A_1 \cup \cdots \cup A_k$ and the intersection $A_1 \cap \cdots \cap A_k$

Definition. Volume function $\mbox{vol}_o : \cP_n \to [0, \infty )$, where for $P = ( a_1, b_1 ] \times \cdots \times ( a_n, b_n ] \in \cP_n$, $\mbox{vol}_o (P) := (b_1 -a_1) \cdots (b_n -a_n)$.

Definition. Half-open rectangle in $\bR^n$ is a set of the form $P = ( a_1, b_1 ] \times \cdots \times ( a_n, b_n ] \subseteq \bR^n$, with $a_1 < b_1, \ldots , a_n < b_n$ in $\bR$. The collection of all half-open rectangles in $\bR^n$ is denoted by $\cP_n$.

Definition. Let $A \subseteq \bR^n$ be a Jordan measurable set. Consider the real non-negative numbers $V^{-} (A)$ and $V^{+} (A)$ defined as follows: $V^{-}(A) := \sup \Bigl\{ \sum_{i=1}^r \mbox{vol}_o (P_i) \ \begin{array}{cc} \lvert & P_1, \ldots , P_r \in \cP_n, \mbox{ with } \cup_{i=1}^r P_i \subseteq A \\ \lvert & \mbox{and with$P_i \cap P_j = \emptyset$for$i \neq j$} \end{array} \Bigr\} .$

$V^{+}(A) := \inf \Bigl\{ \sum_{j=1}^s \mbox{vol}_o (Q_j) \mid Q_1, \ldots , Q_s \in \cP_n, \mbox{ with } \cup_{j=1}^s Q_j \supseteq A \Bigr\} .$

Then $V^{-}(A) = V^{+}(A)$. The volume of $A$ is defined as $\mbox{vol} (A) := V^{-}(A) = V^{+}(A).$

Definition. Let $\cA$ be a field of subsets of $\bR^n$. A function $\mu : \cA \to [0, \infty )$ is said to be an additive set-function when it has the property that: $\left( \begin{array}{c} A_1, \ldots , A_k \in \cA \\ \mbox{with$A_i \cap A_j = \emptyset$for$i \neq j$} \end{array} \right) \ \Longrightarrow \ \Bigl( \, \mu (A_1 \cup \cdots \cup A_k) = \sum_{i=1}^k \mu (A_i) \, \Bigr) .$

Theorem. Consider the function $\mbox{vol} : \cM_n \to [0, \infty )$, as defined above, where $\cM_n$ is the collection of all Jordan measurable sets on $\bR^n$. Prove that $\mbox{vol} : \cM_n \to [0, \infty )$ is an additive set-function.

I want to prove the theorem stated above. Please let me know if my approach is OK. I'm concerned that it is not rigorous enough.

Let $A_1, \dots, A_k\in \cM_n$, with $A_i \cap A_j = \emptyset$ if $i\ne j$. Then $V^-(A_1\cup \dots \cup A_k)=\sup\left\{ \sum\limits_{i=1}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^r P_i \subset A_1\cup \dots\cup A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}$.

Since $P_i$ are mutually disjoint and $A_m$ are mutually disjoint, if $x\in P_i$ for some $1\le i \le r$, $x\in A_m$ for some $1\le m\le k$. Reorder the indices of the sets $P_i$ in such a way as to have $\bigcup\limits_{i=1}^{r_1} P_{i} \subset A_1$, $\bigcup\limits_{i=r_1+1}^{r_2} P_{i} \subset A_2$, $\dots$, $\bigcup\limits_{i=r_s}^{r} P_{i} \subset A_k$ and consider the following:

$\sup\left\{ \sum\limits_{i=1}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^r P_i \subset A_1\cup \dots\cup A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\ = \sup\left\{ \sum\limits_{i=1}^{r_1}\mbox{vol}_0(P_i)+\sum\limits_{i=r_1+1}^{r_2}\mbox{vol}_0(P_i) + \dots + \sum\limits_{i=r_s}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^{r_1}\subset A_1,\dots, \bigcup\limits_{i=r_s}^{r}\subset A_k, \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\ =\sup\left\{ \sum\limits_{i=1}^{r_1}\mbox{vol}_0(P_i): P_1, \dots, P_{r_1}\in \cP_n, \text{ with } \bigcup\limits_{i=1}^{r_1} P_i \subset A_1 \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\ +\dots+\sup\left\{ \sum\limits_{i=r_s}^{r}\mbox{vol}_0(P_i): P_{r_s}, \dots, P_{r}\in \cP_n, \text{ with } \bigcup\limits_{i=r_s}^{r} P_i \subset A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\ =\mbox{vol}(A_1)+\dots+\mbox{vol}(A_k)$.

Q.E.D.

• What is a "field of subsets" ? – Duchamp Gérard H. E. Jul 17 '17 at 6:22
• Do you mean "closed under (finite) intersections, unions and complements" ? – Duchamp Gérard H. E. Jul 17 '17 at 6:25
• @DuchampGérardH.E. I've just added this definition at the top of the post. – sequence Jul 17 '17 at 6:38
• Do you suppose that $\mathbb{R}^n$ is in $\mathcal{A}$ (that was inside my previous question). – Duchamp Gérard H. E. Jul 17 '17 at 6:41
• No, that is not supposed. – sequence Jul 17 '17 at 7:01


Take $A_1,\ldots A_k \in \M$ pairwise disjoint and let $A = \cup_i A_i$.
Suppose $\varepsilon>0$. Then for each fixed $i$ ($1 \le i \le k$) we have some finite pairwise disjoint family $\mathcal{P}_i = \{P^i_1, P^i_2, \ldots, P^i_{n_i}\} \subseteq \P$, such that

$$\sum_{j=1}^{n_i} \vol_0(P^i_j) > \vol^-(A_i) - \frac{\varepsilon}{k}$$

which can be done by the definition of $\vol^-(A_i)$ as a supremum of such sums.

Then note that $\P' = \cup_{i=1}^k \P_i$ is also a pairwise disjoint subfamily of $\P$ (as the $A_i$ are) and

$$\sum_{P \in \P'} \vol_0(P) = \sum_{i=1}^k \sum_{j=1}^{n_i} vol_0(P^i_j) > \sum_{i=1}^k vol^-(A_i) - \varepsilon$$.

As this holds for all $\varepsilon >0$, and $\P'$ is a valid family in the definition of $\vol^-(A)$ we have

$$\vol^-(A) \ge \sum_{i=1}^k vol^{-}(A_i)$$ and the other inequality should be clear as well:

Suppose $\P'= \{P_1, \ldots P_N\}$ is a pairwise disjoint subfamily of $\P$ such that $\cup \P' \subseteq A$. Then we can split ( gap here!) each $P \in \P'$ into a part inside $A_i$ for each $i$ and use the $\vol^-(A_i)$ upperbounds to have a bound for $\vol^-(A)$... I think the gap can be filled with ideas from Tao's post on these matters, working with properties of what he calls elementary sets, i.e. unions of half-open boxes, which form the smallest algebra of sets containing $\P$, and first extend $\vol_0$ to these...

Just some ideas to get you started, you see it's all quite subtle.