Does Mapping with Bounded Distortion Preserve Zero Volume in Higher Dimensions?

[alex.]

Homework Statement

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 $\text{vol}(f(A))=0.$

Homework 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.

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 $\text{vol}(f(A))=0.$

 

[gopher_p]

Homework Statement

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 $\text{vol}(f(A))=0.$

Homework 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.

Some questions regarding the problem statement:

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 $\mathbb{R}^n$?

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.

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

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 $\epsilon.$

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)?

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 $\epsilon.$

$\text{vol}(A)$ 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?

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 $\text{vol}(f(A))=0.$

What does a step function $h:A\rightarrow E^m$ look like?

 

[alex.]

Some questions regarding the problem statement:

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 $\mathbb{R}^n$?

1) Yes, it is a subset of $\mathbb{R}.$

2) That's correct

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)?

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

$\text{vol}(A)$ 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?

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 $\text{vol}(A)=\int_I f$.

What does a step function $h:A\rightarrow E^m$ look like?

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