Subsets of Equidecomposable Subsets are Equidecomposable

Theorem
Let $$A, B \subseteq \R^n \ $$ be equidecomposable and let $$S \subseteq A \ $$. Then there exists $$T \subseteq B \ $$ such that $$S \ $$ and $$T \ $$ are equidecomposable.

Proof
Let $$X_1, \dots, X_m \ $$ be a decomposition of $$A, B \ $$ together with isometries $$\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m: \R^n \to \R^n \ $$ such that

$$A = \bigcup_{i=1}^m \mu_i(X_i) \ $$

and

$$B = \bigcup_{i=1}^m \nu_i(X_i) \ $$.

Define

$$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i) ) \ $$

Then

$$\bigcup_{i=1}^m \mu_i(Y_i) = \bigcup_{i=1}^m ( S \cap \mu_i(X_i) ) = S \cap \bigcup_{i=1}^m \mu_i(X_i) = S \cap A = S \ $$

and so $$\left\{{Y_i}\right\}_{i=1}^m \ $$ forms a decomposition of $$S \ $$. But for each $$i \ $$,

$$( S \cap \mu_i(X_i) ) \subseteq \mu_i(X_i) \ $$,

and so,

$$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i)) \subseteq \mu_i^{-1}(\mu_i(X_i)) = X_i \ $$

Hence

$$\nu_i(Y_i) \subseteq \nu_i(X_i) \ $$,

and so

$$\bigcup_{i=1}^m \nu_i(Y_i) \subseteq \bigcup_{i=1}^m \nu_i(X_i) = B \ $$.

Define $$\bigcup_{i=1}^m \nu_i(Y_i) = T \ $$.