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Volume of zero; Real Analysis

  1. Apr 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Let ##A\subset E^n## and let ##f:A\to E^m.## Consider the condition that there exist some ##M\in\mathbb{R}## such that ##d(f(x),f(y))\le Md(x,y)## for all ##x,y\in A.##

    Show that if the condition is satisfied, if ##m=n##, and ##\text{vol}(A)=0##, then ##\text{vol}((f(A))=0.## Now suppose ##m>n## and ##A## is bounded then show that ##\text{vol}(f(A))=0.##


    2. Relevant equations

    For an arbitrary subset ##A\subset E^n,## we say that ##A## has volume, and define the volume of ##A## to be ##\text{vol}(A)=\int_A 1,## if this integral exists.

    3. The attempt at a solution


    I am not sure how to do the second part of the question and I am not sure that my outline for the first part of the proof is correct.

    For the first part, since the subset ##A## has volume zero then given any ##\epsilon>0## there exists a finite number of closed intervals in ##E^n## whose union contains ##A## and the sum of whose volumes is less that ##\epsilon.## So if a define ##\text{vol}(A)=\int_I f## for ##A\subset I## then for any ##\epsilon>0## there is a partition of ##I## such that any Riemann sum for ##f## corresponding to this partition has absolute value less that ##\epsilon.## So for ##x,y\in I,## let ##\delta=\frac{\epsilon}{M}## then ##|f(x)-f(y)|<\epsilon.## Then I can create step functions such that ##f## is sandwiched between the two step functions and since ##\text{vol}(A)=0## and ##m=n## then ##\text{vol}(f(A))=0.##
     
  2. jcsd
  3. Apr 26, 2014 #2
    Some questions regarding the problem statment:

    1) Is ##E## supposed to be some subset of ##\mathbb{R}##?

    2) Is the integral that you're using to define volume the Riemann integral on ##\mathbb{R}^n##?

    And some remarks on what you have done and questions for you to consider.

    What is a closed interval in ##E^n##? Are you sure this statement is even true in the case where ##E=\mathbb{R}## and ##n=1## (i.e. the simplest case)?

    ##\text{vol}(A)## is already defined per your relevant equations as ##\text{vol}(A)=\int_A 1##. Furthermore, considering that ##f## is a map with codomain, ##E^m##, what does ##\int_I f## even mean? What does a Riemann sum for ##f## look like?

    What does a step function ##h:A\rightarrow E^m## look like?
     
  4. Apr 26, 2014 #3
    1) Yes, it is a subset of ##\mathbb{R}.##

    2) That's correct

    Hmm, maybe I am not sure what you mean? The phrase I used was the definition the book gave me for volume zero.

    Hmm, I should start by saying what exactly is ##\text{vol}(A)=\int_A 1,## the book didn't clarify? But to answer your question, I defined ##\int_I f## to be the function ##f:I\to\mathbb{R}## by setting ##f(x)=1## if ##x\in A, \ f(x)=0## if ##x\in I-A,## so that ##\text{vol}(A)=\int_I f##.

    Well, I was going to define my step functions later since I didn't know if my outline was correct or not
     
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