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Let A be an n by n matrix. Let h:R^n->R^n be the linear transformation h(x)=A x. Let S be a rectifiable set (the boundary of S BdS has measure 0) in R^n. Then v(h(S))=|detA|v(S) (v=volume).

The author starts his proof by considering tha case of A being a non-singular matrix (invertible).

I think I understand his steps in that case (I basically had to prove that h(intS)=int h(S) and h(S) is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).

He proceeds by considering the case where A is singular, so detA=0. He tries to show now that v(T)=0. He states that since S is bounded so is h(S) (I think thats true because |h(x)-h(a)|<=n|A|*|x-a| for each x in S and fixed a in S, if there is again a better explanation please tell me).

Then he says that h(R^n)=V with dimV=p<n and that V has measure 0 (for each ε>0 it can be covered by countably many open rectangles of total volume less than ε), a statemant that I have no clue how to prove. Then he says that the closure of h(S)=cl h(S) is closed and bounded and has neasure 0 (of course cl h(S) is closed but why is it bounded with measure 0?). Then makes an addition step (which I understand) and proves the theorem for that case too.

Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!