The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition

Let $(Q_k)$ be the cells of the grid generated by the concatenated list, indexed by the multi-indices $k$ with $k_i<N_i$ for $i<n$; they are nonempty, pairwise disjoint, cover ${\mathbb{R}}^n$, and each of them is either contained in or disjoint from each $B_j$ and each $C_l$.

If one of the boxes in either decomposition has infinite volume, then both sums are $+\mathrm{\infty }$ and there is nothing left to prove. Indeed, by [L3] a nonempty box has infinite volume exactly when some endpoint is infinite, and such a box is unbounded. Conversely, a finite union of boxes all of whose endpoints are real is bounded: for each such box $B(a,b)$ every coordinate of every point of $B$ lies between the real endpoints $a_i$ and $b_i$, so choosing one real bound for each box and then taking the maximum over the finite list bounds the whole union. Therefore, if $A$ contains an unbounded box from one decomposition, the other decomposition cannot consist entirely of finite-volume boxes, since that would make $A$ bounded. So it remains only to treat the case in which every $B_j$ and every $C_l$ has finite volume; from now on all the volumes that appear are real numbers and the finite-sum laws of [F1] apply to them.

For each $j<m$ let $w_k^j:=\mathrm{vol}(Q_k)$ when $Q_k\subseteq B_j$ and $w_k^j:=0$ otherwise; then $\mathrm{vol}(B_j)={\sum }_{k_0<N_0}\cdots {\sum }_{k_{n-1}<N_{n-1}}{w}_k^j$, because for $B_j=\mathrm{\varnothing }$ no cell is contained in it and both sides are $0$, while for $B_j=B(a^j,b^j)$ nonempty the parameters $a_i^j$ and $b_i^j$ occur among the grid points, say $a_i^j=c_{i,{\alpha }_i}$ and $b_i^j=c_{i,{\beta }_i}$ with ${\alpha }_i<{\beta }_i$, the cells contained in $B_j$ are exactly those with ${\alpha }_i\le k_i<{\beta }_i$ in every coordinate, the lists $c_{i,{\alpha }_i}<\cdots <c_{i,{\beta }_i}$ subdivide $B_j$ so that [L2] applies, and widening each summation range from ${\alpha }_i\le k_i<{\beta }_i$ to $k_i<N_i$ only inserts zero terms.

A cell is contained in $A$ exactly when it is contained in exactly one $B_j$, since a cell contained in $A$ meets $A$, hence meets some $B_j$ and is contained in it, while a cell contained in two of the pairwise disjoint boxes would be empty; so, writing $v_k:=\mathrm{vol}(Q_k)$ when $Q_k\subseteq A$ and $v_k:=0$ otherwise, one has ${\sum }_{j<m}{w}_k^j=v_k$ for every $k$.

Summing the identities of step 2.1 over $j<m$ and regrouping the resulting finite real sums by repeated use of the additivity law in [F1] gives ${\sum }_{j<m}\mathrm{vol}(B_j)={\sum }_{k_0<N_0}\cdots {\sum }_{k_{n-1}<N_{n-1}}{\sum }_{j<m}{w}_k^j$. Since step 2.2 identifies the inner sum with $v_k$, this is ${\sum }_{k_0<N_0}\cdots {\sum }_{k_{n-1}<N_{n-1}}{v}_k$.

The right-hand side of step 3.1 is built from $A$ and the grid alone, and the same computation applied to the list $C_0,\dots ,C_{p-1}$, whose parameters also generate the same grid, gives ${\sum }_{l<p}\mathrm{vol}(C_l)$ for the same value; hence the two sums agree, and since every elementary set has at least one presentation by a finite list of pairwise disjoint half-open boxes, the assignment ${\mu }_0$ is a well-defined function on ${\mathcal{E}}_n$ with ${\mu }_0(\mathrm{\varnothing })=0$ and ${\mu }_0(B)=\mathrm{vol}(B)$ for a single box.