How to analytically prove that the area of a closed curve is conserved under "tearing-and-reattaching" transformation? In physics we have some quantities that are conserved. For example, charges, energy, momentum, volume of incompressible fluid, etc are conserved.
When proving Pythagorean theorem,  we assume that the area is conserved under "tearing-and-reattaching" transformation. How to analytically prove that the area of a closed curve is conserved under such a transformation?
 A: Depending on how are allowed to tear up the shape, it's not true that area is preserved: http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox
Suppose you are using the Lebesgue measure. Chopping your (measurable) shape X into two peices corresponds to finding creating two disjoint measurable subsets A, B of X such that $A \cup B = X$. It is then known that $m(X) = m(A) + m(B)$. 
Suppose that T is a rotation composed with a translation. If you want to show that $m(TA) = m(A)$, you could set up a correspondence between all coverings of TA by rectangles and all similar covering of A. For if $\{V_{\alpha} \}$ is a covering of A, then $\{T V_{\alpha} \}$ covers TA, and the total area of $\{V_{\alpha} \}$ is the same as the total area of $\{TV_{\alpha} \}$, if you show T preserves the area of rectangle. This gives $m(A) \geq m(TA)$, since for every volume greater than A there is a covering of TA with that volume. The other direction follows by similar reasoning to give equality.
To shows that $T$ preserves the area of a rectangle, consider it first as a translation and then as a rotation. Translations preserve area by using the area formula and rotations preserve area by (for instance) the multivariate change of variables formula, since the determinant of a rotation is 1. The composition of two area preserving functions will preserve area by transitivity of equality.
