Equidecomposability Unaffected by Union

Theorem
Let $\left\{{S_1, \ldots, S_m}\right\}, \left\{{T_1, \ldots, T_m }\right\}$ be sets of sets in $\R^n$ such that:
 * for each $k \in \left\{{1, \dots, m}\right\}, S_k$ and $T_k$ are equidecomposable.

Then the set $\displaystyle S = \bigcup_{i \mathop = 1}^m S_i$ is equidecomposable with $\displaystyle T = \bigcup_{i \mathop = 1}^m T_i$.

Proof
We have for each $k \in \left\{{1, \dots, m}\right\}$ a decomposition $\left\{{S_{k, 1}, \cdots, S_{k, l_k}}\right\}$ and set of isometries $\phi_{i, j}: \R^n \to \R^n$ such that:


 * $\displaystyle S_k = \bigcup_{a \mathop = 1}^{l_k} \phi_{k, a} \left({S_{k, a} }\right)$

and similarly for $T_k$ and some isometries $\theta_{i, j}: \R^n \to \R^n$.

Thus:


 * $\displaystyle S = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \phi_{k, i} \left({S_{k, i} }\right)$

and:


 * $\displaystyle T = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \theta_{k, i} \left({T_{k, i} }\right)$

but since by definition, each of the $S_{a, b}, T_{a, b}$ are congruent, they yield equivalent decompositions of $S$ and $T$.