Volume of Set A: Proposition and Proof for Continuous Functions f:A→ℝ

[alex.]

Homework Statement

Let $A\subset E^n$ be a set with volume and $f:A\to\mathbb{R}$ a continuous function. Show that if the set $\{x\in A:f(x)=0\}$ has volume zero, then the set $\{x\in A:f(x)>0\}$ has volume.

Homework Equations

None

The Attempt at a Solution

A proposition in my book states, if $A\subset E^n$ has volume zero and the set $B\subset E^n$ has volume, then $\text{vol}(B\cup A)=\text{vol}(B-A)=\text{vol}(B).$

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.

Since the set $\{x\in A:f(x)=0\}$ 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 $\epsilon.$

The phrase "$A$ has zero volume" means "$A$ has volume and that volume is zero".

 

[Dick]

Homework Statement

Let $A\subset E^n$ be a set with volume and $f:A\to\mathbb{R}$ a continuous function. Show that if the set $\{x\in A:f(x)=0\}$ has volume zero, then the set $\{x\in A:f(x)>0\}$ has volume.

Homework Equations

None

The Attempt at a Solution

A proposition in my book states, if $A\subset E^n$ has volume zero and the set $B\subset E^n$ has volume, then $\text{vol}(B\cup A)=\text{vol}(B-A)=\text{vol}(B).$

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.

Since the set $\{x\in A:f(x)=0\}$ 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 $\epsilon.$

The phrase "$A$ has zero volume" means "$A$ has volume and that volume is zero".

Does the problem mean to say $\{x\in A:f(x) \ne 0\}$ has volume or did you misstake something? Otherwise it's false. f could be zero on a set of volume 0 and negative otherwise.

 

[alex.]

Does the problem mean to say $\{x\in A:f(x) \ne 0\}$ has volume or did you misstake something? Otherwise it's false. f could be zero on a set of volume 0 and negative otherwise.

By volume, it is okay to have volume of zero in which case our set $f(x)>0$ can be an empty set. By empty set I mean, has zero volume. Suppose A is empty. Fix any box I. Then I contains A. The volume is the integral of the function that equals 1 on A and 0 outside A. Thus the function is zero everywhere, and its integral over I is zero

 

[haruspex]

By volume, it is okay to have volume of zero in which case our set $f(x)>0$ can be an empty set. By empty set I mean, has zero volume. Suppose A is empty. Fix any box I. Then I contains A. The volume is the integral of the function that equals 1 on A and 0 outside A. Thus the function is zero everywhere, and its integral over I is zero

I think you missed Dick's point. Consider f(x)=-1 for all x.

 

[alex.]

I am not sure I understand? I copied the question exactly like in the book.

 

[haruspex]

I am not sure I understand? I copied the question exactly like in the book.

OK, but you see that the question must be wrong, yes? I agree with the substitution Dick suggested: assume the question should read f(x)≠0 rather than f(x)>0.

 

[micromass]

No, the question is alright. The wording "the set has volume" should be interpreted as "the set is measurable" and not as "the set has nonzero volume".

 

[alex.]

The book is kinda old, so maybe they used different terms. But, hopefully micromass comment cleared the issue.

 

[haruspex]

No, the question is alright. The wording "the set has volume" should be interpreted as "the set is measurable" and not as "the set has nonzero volume".

Hmm.. then how should one read "the set $\{x\in A:f(x)=0\}$ has volume zero"? Is that "the set $\{x\in A:f(x)=0\}$ is measurable and has measure zero"? If so, I don't see that its having measure zero would be of interest.

 

[micromass]

Hmm.. then how should one read "the set $\{x\in A:f(x)=0\}$ has volume zero"? Is that "the set $\{x\in A:f(x)=0\}$ is measurable and has measure zero"? If so, I don't see that its having measure zero would be of interest.

Yes, I think that's how we should read it. I don't really see how it would be interesting to know it has measure 0 either. But the OP doesn't work with measure theory yet. He defined "volume" and "has volume" as the existence of some kind of integral. This integral likely isn't very general and requires some very restrictive conditions to exist.

So to the OP, how did you define that integral? For what kind of functions does that integral exist? Do you know theorems about this?

Also, which book are you reading?