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

  1. Nov 24, 2005 #1
    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.

    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.
  2. jcsd
  3. Nov 25, 2005 #2


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    If T is a cube, how is Vol(T) = 0? Perhaps T is a cube in 4-dimensional space? I think you need to clarify the question.
    Where are you getting this from? This is false. If you're assuming that the continuity of g gives you this, then you're wrong. Let g be a real-valued function of one real variable, specifically, the exponential function. Then your claim above fails. Continuity says that for epsilon' > 0, there is an S > 0 such that |x - y| < S implies |g(x) - g(y)| < epsilon'.

    For Vol(T) = 0 to be true, you must be able to cover T by some number of cubes (not necessarily finite) such that the total volume of all the cubes can be made arbitrarily small. If we're talking about the real line, for instance, then let's say T is some one-point set. We can cover it by a cube (open interval) of radius e/2 (for any e > 0 we want) centered at the element of T. Any set with a countable number of points can be covered by cubes in the following way:

    Since the set is countable, order the elements t1, t2, ... (i.e. index the elements by the natural numbers). Around ti, place a cube Ci of volume e/2i. Then certainly, these cubes cover T and the sum of their volumes is:

    [tex]\sum _{k=1} ^{\infty }\frac{e}{2^i} = e[/tex]

    So for your problem, you are given the fact that for any e > 0, you can cover T by some number of cubes such that their total volume is < e. What can you say then about cubes that cover g? Also, what theorems do you know that involve det Dg <> 0?
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