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Let $(X,M,\mu)$ be a measure space. Assume that $\mu$ is a non-atomic finite measure, namely for every $E \in M$ with $\mu(E)>0$, there exists $F \subset E$ such that $0<\mu(F)<\mu(E)$. Suppose that $\mu(X)=1$. Show that for each $\alpha \in [0,1]$, there exists $E \in M$ with $\mu(E)=\alpha$.

Thoughts on Question: I've thought about this question for a very long time but I can't seem to figure it out.

Here is my initial partial attempt.

Proof (first following a hint given to me):

Let $E \in M$ with $\mu(E)>0$. Also let $\epsilon>0$. Since $\mu$ is a non-atomic finite measure, there exists $F\subset E$ so that $0<\mu(F)<\mu(E)$. We know that $\mu(F)+\mu(E \cap F^c)=\mu(E)$ by additivity. We know that one of $F$, $E\cap F^c$ must satisfy $0<\mu(F_1)\leq 1/2*\mu(E)$. Let $F_1$ represent the element that does. Let's continue this process again. Let $F_2 \subset F_1$. Then we have: $\mu(F_2) \leq 1/2* \mu(F1) \leq 1/4*\mu(E)$. For every n, there exists $F_n$ so that $0<F_n\leq 1/2^n*\mu(E)$. Take n big enough so that $1/(2^n) \leq \epsilon$. Thus we know that for $\epsilon>0$ we can find $G\subset E$ so that $0<\mu(G)<\epsilon$.

Concerns: So I have now proved that for $\epsilon>0$ there exists $0<\mu(G)<\epsilon$,as per a hint but how do I use this to prove the the result the question is asking:for each $\alpha \in [0,1]$, there exists $E \in M$ with $\mu(E)=\alpha$? I have seen some things online but I can't seem to follow any of them. Also is proof correct for the part above? Did I say anything, even something small inaccurately? Could someone, if possible, help me clean up my proof above because I want to precise as possible? Thank you very much. I just need help utilizing this to prove the result.

kemb
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  • @MikeEarnest I saw that but I don't really understand what they are doing, in the second part. – kemb Sep 11 '18 at 01:29
  • Your proof is good. The proof linked uses Zorn's Lemma, which is counter-intuitive if you are not familiar with it. The intuitive proof would be this: start with $A=\varnothing$, and as long as $\mu(A)<\alpha$, add set with size less than $\alpha-\mu(A)$ to $A$. Since this process cannot go on forever, eventually $\mu(A)=\alpha$. Proving that the process must stop is surprisingly tricky, and uses the axiom of choice in some form. – Mike Earnest Sep 11 '18 at 01:40
  • @MikeEarnest is there a way to do it without the axiom of choice, I'm not too familiar with it. Also in the proof of the link you posted they say this fact: If $A=\cup_{n=1}^\infty A_n$, then $\mu(A)=\sum_{n=1}^\infty\mu(A_n)\leq\epsilon$. Where does this come from? I'm not seeing how they know this. thanks. – kemb Sep 11 '18 at 01:43
  • See this answer for a proof in line with my last comment. It requires transfinite induction. But you should research other examples of using Zorn's lemma if you want to get comfortable with the first proof I linked. I believe the axiom of choice is required. – Mike Earnest Sep 11 '18 at 01:45
  • @MikeEarnest does your proof in that link require what I proved above^ that we can find $G \subset E$ so that $0<\mu(G)<\epsilon$. – kemb Sep 11 '18 at 01:48
  • @MikeEarnest: I don't get two things in the accepted answer of the proof: If $A=\cup_{n=1}^\infty A_n$, then $\mu(A)=\sum_{n=1}^\infty\mu(A_n)\leq\epsilon$. How does he know this sum is less than or equal to $\epsilon$. Also: how does he come up with/rational for the statement: "so by construction necessarily we have $\mu(A_n)\geq\frac1n$ for all $n\geq N$," – kemb Sep 11 '18 at 01:56

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