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Here is an old measure theory exercise I remember solving, but I'm now a bit fuzzy on the details.

Let $(X,\Sigma,\mu)$ be a finite measure space. Call $\mu$ nonatomic if for any $A\in\Sigma$ with $\mu(A)>0$ there exists $B\in\Sigma$ with $B\subset A$ and $0<\mu(B)<\mu(A)$. Call $\mu$ continuous if for any $A\in\Sigma$ with $\mu(A)>0$ and any $c\in\mathbb{R}$ with $0<c<\mu(A)$, there exists $B\in\Sigma$ with $B\subset A$ and $\mu(B)=c$.

Intuitively, if we think of measurable sets as rocks, a nonatomic measure allows us to chip off a small rock of some size from a larger one, while a continuous measure allows us to chip off a small rock of any size from a larger one. It follows that a continuous measure is nonatomic.

Show that the converse is also true: nonatomic measures are continuous.

I remember that my solution involved Zorn's lemma. Here are my questions: Can this exercise be solved without Zorn's lemma, and can we replace the finiteness hypothesis with something weaker?

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    $\begingroup$ It will involve dependent choice, but the full strength of Zorns lemma is overkill for this problem. $\endgroup$ – George Lowther Jun 11 '13 at 18:09
  • $\begingroup$ The proof is due to Sierpinsky. A short version of it is given in the wikipedia article: en.wikipedia.org/wiki/Atom_(measure_theory) $\endgroup$ – C-Star-W-Star Oct 21 '14 at 18:21
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The following proof makes ample use of dependent choice, but not more. This is line with the comment by George Lowther.

A measurable partition $\Pi$ is an $\epsilon$-partition if each cell has $\mu$-measure less than $\epsilon$ and there are only finitely many cells.

The proof of the following lemma was inspired by an answer by Brian M. Scott to one of my questions.

Lemma: For each $\epsilon>0$, there is an $\epsilon$-partition of $A$.

Proof: Let $l$ be the infimum over all $\epsilon>0$ such that an $\epsilon$-partition of $A$ exists. There exists a sequence $(\Pi_n)$ of partitions, such that for each $n$, $\Pi_n$ is an $l+1/n$-partition. By replacing $\Pi_n$ with $\bigwedge_{m=1}^n\Pi_m$, we can take $(\Pi_n)$ to have the property that $\Pi_{n+1}$ is a refinement of $\Pi_n$ for all $n$. Also, we can take $\Pi_1$ to be $\{A\}$. The set $\mathcal{T}=\bigcup_n\Pi_n$ is a tree when ordered by $\supseteq$. There is a chain $\mathcal{C}$ in $\mathcal{T}$ such that $\inf\{\mu(C):C\in\mathcal{C}\}=l$. The idea is to take $C_1=A$, $C_2$ to be a cell in $\Pi_2$ such that the infimum over all $\epsilon>0$ such that an $\epsilon$-partition of $C_2$ exist is $l$. Continue this way, and we get the desired chain $\mathcal{C}=\{C_1,C_2,\ldots\}$. Let $C^1=\bigcap_n C_n$. Then $\mu(C)=l$. Since we could break up $C^1$ into sets of smaller measure by nonatomicity, $l=\mu(C^1)<\mu(A)$.

We now repeat this approach to find a measurable set $C^2\subseteq A\backslash C^1$ such that $\mu(C^2)$ equals the infimum over all $\epsilon>0$ such that an $\epsilon$-partition of $A\backslash C_1$ exists. Since it is easier to find an $\epsilon$-partition of a smaller set, we have $\mu(C^2)\leq\mu(C^1)$. Continue this way to get the sequence $C^1,C^2,\ldots$. Since $\mu(A)<\infty$, there must be some index $n$ such that $\mu(C^{n+1})<\mu(C^n)=l$ and we take $n$ to be the first such index. So there is a measurable partition $\Pi$ of $A\backslash\bigcup_{m=1}^{n-1}C^m$ into finitely many sets of measure less than $l$. Split each $C^m$ with $m\leq n$ into two measurable sets of smaller measure $C^m_1$ and $C^m_2$. Then $\Pi\cup\{C^m_i:m\leq n, i=1,2\}$ is a partition of $A$ into finitely many measurable sets of measure less than $l$, which cannot be. $\square$

The main result is now easy to prove.

Theorem: Let $A$ be a measurable set with $\mu(A)>0$ and $0<\alpha<\mu(A)$. Then there exists a measurable set $B\subseteq A$ with $\mu(B)=\alpha$.

Proof: By taking finite unions of cells in an $\epsilon$-partition with $\epsilon$ small enough, we can find $B_1\subseteq A$ with $\alpha-1/2<\mu(B_1)<\alpha$. Given that $B_1,\ldots, B_n$ are already constructed, we can then find some $B_{n+1}\subseteq A\backslash\bigcup_{m=1}^n B_n$ with $\alpha-1/2^{n+1}<\sum_{m=1}^n \mu(B_m) + \mu(B_{n+1})<\alpha$. We can clearly take $B=\bigcup_n B_n$. $\square$

The only requirement for applying this proof to infinite measure spaces is that whenever $\mu(A)=\infty$ and $r$ is a number, there exists a measurable set $A'\subseteq A$ with $r<\mu(A')<\infty$. This is for example the case when the measure is $\sigma$-finite.

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