Prove that if Vol(T)=0, and g is differentiable with det Dg not equal to 0, then Vol(g(T))=0. T is a cube.(adsbygoogle = window.adsbygoogle || []).push({});

Here's my idea, although it doesn't incorporate the fact that det Dg <> 0.

Since Vol(T)=0, then T is a subset of a the union of sub cubes T1, T2, ..., Tn. Furthermore the sum of all volumes of Ti, i=1,2,...,n, < epsilon. [at this point, im not sure how to choose my epsilon, but I suppose I can't know 'til later.]

since g is differentiable, g is continuous [Do i have enough to say g is uniformly continuous? b/c that may help]. So for S>0, there is an epsilon' >0, such that |x-y|< S implies |g(x)- g(y)|< epsilon'.

Here is where I'm stuck. My next intuition is to take g of tiny intervals which make each Ti, and then show that the sum of all volumes g(Ti) < epsilon'-usingi the fact that g is continous. I'm just unsure how to go about doing so. Any help would be much appreciated.

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# Vol(T)=0 proof about Vol(g(T))

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