Proving Compact Set Exists with m(E)=c

[the_dane]

Homework Statement

Suppose E1 and E2 are a pair of compact sets in R^d with E1 ⊆ E2, and let a = m(E1) and b=m(E2). Prove that for any c with a<c<b, there is a compact set E withE1 ⊆E⊆E2 and m(E) = c.

Homework Equations

m(E) is ofcourese referring to the outer measure of E

The Attempt at a Solution

I know that for d=1 measurable subset of [0,1]. Is it worth it to look at the measure m(E∩[0,t]) as function of t?
I really don't know how to tackle this one

 

[Ray Vickson]

Homework Statement

Suppose E1 and E2 are a pair of compact sets in R^d with E1 ⊆ E2, and let a = m(E1) and b=m(E2). Prove that for any c with a<c<b, there is a compact set E withE1 ⊆E⊆E2 and m(E) = c.

Homework Equations

m(E) is ofcourese referring to the outer measure of E

The Attempt at a Solution

I know that for d=1 measurable subset of [0,1]. Is it worth it to look at the measure m(E∩[0,t]) as function of t?
I really don't know how to tackle this one

The name of the subject is "measure" theory, not "measurement" theory.

 

[Samy_A]

I know that for d=1 measurable subset of [0,1]. Is it worth it to look at the measure m(E∩[0,t]) as function of t?

Yes, I think that is the way to do it.

 

[Mark44]

The name of the subject is "measure" theory, not "measurement" theory.

Fixed...

 

[fresh_42]

What does it mean for $E_1$ to be compact? Does anything change if you reduce the task to $E_2$ as topological space? What do you know about the closure of finitely many open sets of finite mass? (You probably won't need to regard $d$ at all.)

 

[the_dane]

Yes, I think that is the way to do it.

Could you help a bit along the way?

 

[S.G. Janssens]

I think you may need to exploit continuity of the measure (w.r.t. inclusion).

 

[Samy_A]

Instead of an interval you have to use a d-dimensional cube with side t. Let's call that cube C(t).
Now define the function $f(t)=m(E_1\cup(E_2\cap C(t))$ on a well chosen interval $[x,y]$ of $\mathbb R$ so that $f(x)=a$ and $f(y)=b$.

Remember that $E_1$ and $E_2$ are compact, and thus bounded.

EDIT: not to discourage you, but it does look like quite a difficult exercise. I think it should work with the function $f$ given above though.

EDIT2: maybe the way @fresh_42 suggests will be easier.

 

[the_dane]

EDIT: not to discourage you, but it does look like quite a difficult exercise. I think it should work with the function ff given above though.

I know what you mean, but I want to learn " the hard way" in order to understand it better. I am trying to teach myself measure theory so for me it's not "just homework".

Instead of an interval you have to use a d-dimensional cube with side t. Let's call that cube C(t).
Now define the function $f(t)=m(E_1\cup(E_2\cap C(t))$ on a well chosen interval $[x,y]$ of $\mathbb R$ so that $f(x)=a$ and $f(y)=b$.

Remember that $E_1$ and $E_2$ are compact, and thus bounded.

I can't see how to apply the last information?

 

[Samy_A]

I can't see how to apply the last information?

First thing to do is to find $x$ and $y$ satisfying $f(x)=a,f(y)=b$.
I think that $x$ is easy, but for a arbitrary $E_2$, it is not even sure that $y$ exists. But $E_2$ being compact (and thus bounded) makes it easy to find some $y$ large enough to satisfy $f(y)=b$.

Next challenge will be to prove that $f$ is continuous.
Note that $f$ is a monotone function: $s \leq t \Rightarrow f(s) \leq f(t)$

Set $E(t)=E_1\cup(E_2\cap C(t))$
Verify that if $s \leq t$, $E(t)\setminus E(s) \subset C(t) \setminus C(s)$

Also remember that if $A \subset B$, then $m(B \setminus A)=m(B)-m(A)$ (assuming that A and B are both measurable of course).

 

[the_dane]

Also remember that if $A \subset B$, then $m(B \setminus A)=m(B)-m(A)$ .

Can I use this information to conclude then?

 

[Samy_A]

Can I use this information to conclude then?

I'm not sure I understand your question. I used that in the final step proving that the function $f$ is continuous.