Equidecomposable Nested Sets

Theorem
Let $$A, B, C \ $$ be sets such that $$A \ $$ and $$C \ $$ are equidecomposable and $$A \subseteq B \subseteq C \ $$.

Then $$B \ $$ and $$C \ $$ are equidecomposable.

Proof
Let $$\left\{{X_k}\right\}_{k=1}^n \ $$ be a decomposition of $$A \ $$ and $$C \ $$, so that there are isometries $$\left\{{\phi_k}\right\}_{k=1}^n \ $$ and $$\left\{{\psi_k}\right\}_{k=1}^n \ $$ such that

$$A = \bigcup_{k=1}^n \phi_k(X_k) \ $$

$$C = \bigcup_{k=1}^n \psi_k(X_k) \ $$

Let $$Y_k = \psi_k^{-1} \left({ B \cap \psi_k(X_k) }\right) \ $$. Then $$\bigcup_{k=1}^n \psi_k (Y_k) = B \cap \bigcup_{k=1}^n \psi_k(X_k) = B \cap C = B \ $$.

Let $$Z_k = \phi^{-1} \left({ A \cap \phi_k(X_k) }\right) \ $$. Then $$\bigcup_{k=1}^n \phi_k (Z_k) = A \cap \bigcup_{k=1}^n \phi_k(X_k) = A \cap C = A \ $$.