(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given is the measure space [itex](A,\mathcal{P}(A),\mu)[/itex] where [itex]\mu[/itex] is the counting measure on the powerset [itex]\mathcal{P}(A)[/itex] of [itex]A[/itex], i.e. [itex]\mu(E)=\#E[/itex]

I have to show that if [itex]\int_A f d\mu <\infty[/itex], then the set [itex]A_+=\{x\in A| f(x)>0\}[/itex] is countable.

2. Relevant theorems

I wish I knew. Results I have so far (don't know if they are of any use):

(1) [itex]\int_A f d\mu =\sup\{\sum_{x\in E}f(x) | E\subset A, E \mbox{ finite}\}[/itex]

(2) I know that [itex]\{x\in A| f(x)=\infty\}[/itex] has measure zero, which in this case means that f is finite everywhere.

(3) There's a general theorem which states that if [itex]f:A\to[0,\infty][/itex] is measurable and [itex]\int_A f d\mu<\infty[/itex] then the set [itex]\{x\in A| f(x)>0\}[/itex] is sigma-finite. If I can prove this, then the result follows easily.

3. The attempt at a solution

I tried many things, the closest I got was by writing:

[tex]A_+=\bigcup_{n=0}^\infty \{x\in A| f(x)\in (n,n+1]\}=\bigcup_{n=0}^\infty C_n[/tex] where [itex]C_n:=\{x\in A| f(x)\in (n,n+1]\}[/itex],

and showing that [itex]\mu(C_n)[/itex] is finite for n=1,2,3,..., since

[tex]\int_A fd\mu \geq \int_A f\chi_{C_n}d\mu \geq \int_A \chi_{C_n} =\mu(C_n)=\#C_n[/tex]

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.

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# Measure theory: Countable mayhem

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