A hyperrectangle in $\mathbb{R}^d$ has the form $H = I_1 \times \ldots \times I_d$, where $I_1, \ldots, I_d$ are intervals. The natural volume $|H|$ is defined as $|H| : = |I_1| \cdot \ldots \cdot |I_d|$.
A simple set $M \subset \mathbb{R}^d$ can be written as finite union $H_1 \cup \ldots \cup H_n$ of disjoint hyperrectangles and the natural measure of $M$ is defined as $|M| : = |H_1| + \ldots + |H_n|$.
Now I want to prove the following theorem:
Theorem. For disjoint simple sets $\{M_j\}_{j=1}^{\infty}$, where $\bigcup_{j=1}^{\infty} M_j$ is a simple set again, it holds that $$ \left| \bigcup_{j=1}^{\infty} M_j \right| = \sum_{j=1}^{\infty} |M_j|.$$
By trying to prove this theorem I realised that the function $|\cdot|$ on simple sets has properties we are used from measures. For example we have $M_1 \subset M_2 \Rightarrow |M_1| \leq |M_2|$. It follows from the definition of $|\cdot|$ that we have
$$ \left| \bigcup_{j=1}^{n} M_j \right| =\left| \bigcup_{j=1}^{n} \bigcup_{i=1}^{k_j}H_{i,j} \right|= \sum_{j=1}^{n} \sum_{i=1}^{k_j}|H_{i,j}| = \sum_{j=1}^{n} |M_j|.$$
At this point I have no idea how to extend this property to countable unions.
I will use the following lemma, which is somewhat annoying to prove, but here is a sketch.
Lemma: The union of two (not necessarily disjoint) rectangles $R_1$ and $R_2$ is a simple set, and $|R_1 \cup R_2| \leq |R_1| + |R_2|$.
Sketch of proof: We have $R_1 = I_1 \times I_2 \times\cdots\times I_d$ and $R_2 = J_1 \times J_2 \times\cdots\times J_d$, where the $I_k$ and $J_k$ are intervals. Suppose that the combined set of endpoints of $I_k$ and $J_k$ is written in increasing order as $a,b,c,d$. Then $I_k \cup J_k$ can be partitioned canonically as a disjoint union of some or all of the intervals $\{a\}, (a,b), \{b\}, (b,c), \{c\}, (c,d), \{d\}$. We do this for each $k$. Then consider the finite set $\mathcal L$ of all rectangles of the form $L = L_1\times L_2 \times \cdots \times L_d$, where each $L_k$ is one of the intervals in the canonical partition of $I_k \cup J_k$. It's easy to see that these $L$'s are pairwise disjoint. Each $L$ is either contained in $R_1 \cup R_2$ or is disjoint from $R_1 \cup R_2$. If we include only those $L$ which are contained in $R_1 \cup R_2$, then the they form a finite disjoint partition of $R_1 \cup R_2$. Therefore, $R_1 \cup R_2$ is simple as claimed.
Similarly, we can express $R_2 \setminus R_1$ as a finite disjoint union of elements of $\mathcal L$, which shows that $R_2 \setminus R_1$ is also simple.
Finally, we observe that $R_1$ and $R_2 \setminus R_1$ are disjoint, and $R_1 \cup R_2 = R_1 \cup (R_2 \setminus R_1)$, so $|R_1 \cup R_2| = |R_1| + |R_2 \setminus R_1| \leq |R_1| + |R_2|$, where the inequality is due to monotonicity.
Note that by induction, the lemma implies that the union of any finite number of (not necessarily disjoint) rectangles $R_1,R_2,\ldots,R_n$ is a simple set, and $|\cup_{n=1}^{N} R_n| \leq \sum_{n=1}^{N}|R_n|$.
Also by induction, this further implies that the union of any finite number of (not necessarily disjoint) simple sets $M_1,M_2,\ldots,M_n$ is a simple set, and $|\cup_{n=1}^{N} M_n| \leq \sum_{n=1}^{N}|M_n|$.
Now we prove the main result.
Let $A = \cup_{j=1}^{\infty}M_j$ where the $M_j$ are disjoint simple sets. For finite $J$, $$A_J = \bigcup_{j=1}^{J}M_j$$ is a finite disjoint union of simple sets, each of which is a finite disjoint union of rectangles, so $A_j$ is simple, and $$|A_J| = \sum_{j=1}^{J}|M_j|$$ As $A_J \subset A$, and you have already established monotonicity for simple sets, it follows that $$\sum_{j=1}^{J}|M_j| = |A_J| \leq |A|$$ This holds for any positive integer $J$, so $$\sum_{j=1}^{\infty}|M_j| = \lim_{J \to \infty}\sum_{j=1}^{J}|M_j| \leq |A|$$ For the reverse inequality, first, note that since $A$ itself is simple, we can write $$A = \bigcup_{n=1}^{N}R_n$$ where the $R_n$ are disjoint rectangles, and so $|A| = \sum_{n=1}^{N}|R_n|$.
Let $\epsilon > 0$.
For each $n=1,2,\ldots,N$, we can find a (possibly empty) closed rectangle $K_n \subset R_n$ with volume $|K_n| \geq |R_n| - \epsilon/N$.
Also, $M = \cup_{j=1}^{\infty}M_j$ is the countable disjoint union of simple sets $M_j$, each of which is a finite disjoint union of rectangles. By increasing the size of these rectangles, we see that $M_j$ is contained in set $U_j$ which is the finite (not necessarily disjoint) union of open rectangles. By the lemma, $U_j$ is simple, and $|U_j|$ is no larger than the sum of the volumes of the open rectangles. We can choose the open rectangles small enough that $|U_j| \leq |M_j| + \epsilon/2^j$.
Then $K = \cup_{n=1}^{N}K_n$ is covered by $\{U_j\}_{j=1}^{\infty}$, a sequence of open sets. As $K$ is compact, only finitely many of the $U_j$'s are needed, say $$K \subset \bigcup_{j=1}^{L}U_j$$ Therefore, $$\bigcup_{n=1}^{N}K_n = K \subset \bigcup_{j=1}^{L}U_j$$ The sets on the left and right hand sides are simple (the latter by the lemma), so again by monotonicity, we have $$\sum_{n=1}^{N}|K_n| = \left|\bigcup_{n=1}^{N}K_n\right| \leq \left|\bigcup_{j=1}^{L}U_j\right| \leq \sum_{j=1}^{L}|U_j|$$ where the last inequality is again by the lemma.
So, $$\sum_{n=1}^{N}|R_n| - \epsilon \leq \sum_{n=1}^{N}|K_n| \leq \sum_{j=1}^{L}|U_j| \leq \sum_{j=1}^{\infty}|U_j| \leq \sum_{j=1}^{\infty}|M_j| + \epsilon$$ Therefore, $$|A| = \sum_{n=1}^{N}|R_n| < \sum_{j=1}^{\infty}|M_j| + 2\epsilon$$ As $\epsilon > 0$ is arbitrary, we conclude that $$|A| \leq \sum_{j=1}^{\infty}|M_j|$$