Measure theory: Countable mayhem

[Galileo]

Homework Statement

Given is the measure space $(A,\mathcal{P}(A),\mu)$ where $\mu$ is the counting measure on the powerset $\mathcal{P}(A)$ of $A$, i.e. $\mu(E)=\#E$
I have to show that if $\int_A f d\mu <\infty$, then the set $A_+=\{x\in A| f(x)>0\}$ is countable.

2. Relevant theorems
I wish I knew. Results I have so far (don't know if they are of any use):
(1) $\int_A f d\mu =\sup\{\sum_{x\in E}f(x) | E\subset A, E \mbox{ finite}\}$
(2) I know that $\{x\in A| f(x)=\infty\}$ has measure zero, which in this case means that f is finite everywhere.
(3) There's a general theorem which states that if $f:A\to[0,\infty]$ is measurable and $\int_A f d\mu<\infty$ then the set $\{x\in A| f(x)>0\}$ is sigma-finite. If I can prove this, then the result follows easily.

The Attempt at a Solution

I tried many things, the closest I got was by writing:
$$A_+=\bigcup_{n=0}^\infty \{x\in A| f(x)\in (n,n+1]\}=\bigcup_{n=0}^\infty C_n$$ where $C_n:=\{x\in A| f(x)\in (n,n+1]\}$,
and showing that $\mu(C_n)$ is finite for n=1,2,3,..., since
$$\int_A fd\mu \geq \int_A f\chi_{C_n}d\mu \geq \int_A \chi_{C_n} =\mu(C_n)=\#C_n$$
Because f>=1 for n>=1. I can't handle the case n=0 though...

I'd rather prove the theorem (3) in general instead for this special case, but any help is appreciated.

 

[StatusX]

It's equivalent to show that for any uncountable set of positive numbers, there is a countable subset whose infinite sum diverges. Do you see why this is equivalent, and if so, is it any easier for you to prove?

 

[Galileo]

Yes, I realized that. I can prove A_+ is uncountable => the integral is infinite, which means (by (1)) that the supremum of $\{\sum_{x\in E} f(x) | E \subset A, \mu(E)< \infty\}$ is infinite. So there should be a sequence of finite subsets E_i such that the sequence {sum f(x) with x in E_j} diverges.
That's what you mean right?

 

[StatusX]

For clarity, define the following properties:

A) "A_+ is uncountable"
B) "the integral is infinite",
C) there exists "a sequence of finite subsets E_i such that the sequence {sum f(x) with x in E_j} diverges"

What you're asked to show is that ~B implies ~A, which is clearly equivalent to A implies B. You seem to be saying you've shown A implies B. If so, then your done. I was suggesting showing A implies C, and then using the equivalence of B and C to get A implies B.

 

[Galileo]

Then I expressed myself poorly. I was trying to say the same as you did in your last post and I pointed out the equivalence of B) and C).

However, I've tried A => C and didn't get far. Any hints?

 

[StatusX]

Form a sequence of nested sets whose union is A+. Then if A+ is uncountable, you know one of the sets in the sequence must be infinite, since the countable union of finite sets is countable. If you pick these sets right, then you should be able to show that if any of them is infinite, the sum over the elements in the set diverges.

 

[Hurkyl]

The Attempt at a Solution

I tried many things, the closest I got was by writing:
$$A_+=\bigcup_{n=0}^\infty \{x\in A| f(x)\in (n,n+1]\}=\bigcup_{n=0}^\infty C_n$$ where $C_n:=\{x\in A| f(x)\in (n,n+1]\}$,
and showing that $\mu(C_n)$ is finite for n=1,2,3,..., since
$$\int_A fd\mu \geq \int_A f\chi_{C_n}d\mu \geq \int_A \chi_{C_n} =\mu(C_n)=\#C_n$$
Because f>=1 for n>=1. I can't handle the case n=0 though...

I'd rather prove the theorem (3) in general instead for this special case, but any help is appreciated.

The only reason your proof fails is because the interval (0, 1] doesn't have a positive lower bound, right? Doesn't that suggest something else you could try?

 

[Galileo]

It popped into my head when jumped into bed yesterday :P
But thanks! (Both of you).

Just write $A_+=B \cup C$, where

$$B=\cup_{n=1}^\infty \{x|f(x)\in [\frac{1}{n+1},\frac{1}{n}]\}$$
and
$$C=\{x|f(x)\geq1\}$$
Then the intervals all have a lower bound.

Now that we know A_+ is sigma finite we know it is a countable union of finite sets, and thus countable.