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I was wondering, can the following theorem be true for finitely additive measures defined on algebras not $\sigma$-algebras. (Theorem is in Bogachev's Measure Theory Vol I).

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I was not sure about it, but I found a coubterexample for nonatomic measure, which is: $\lambda=\mu+2\nu$, where $\mu$ is the Lebesgue measure on the unit intrval $I$, and $\nu$ is a 0−1-valued finitely additive measure on $I$ such that $\nu(A)=0$ if $\mu(A)=0$. $\lambda$ is a nonatomic finitely additive measure, but has not partition for $0<\varepsilon<2$.

I really need a result like this: suppose $\mu$ is a finitely additive nonatomic measure on a countable Boolean algebra $B$. Then, for every $\varepsilon>0$, there is a finite partition $P$ of the unit element such that $\mu(p_i)<\varepsilon$" for each $p_i\in P$ and $i=1,\dots,n$.

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  • $\begingroup$ @EmilJeřábek do you mean for countable algebra ? or ... $\endgroup$
    – Ameen
    Jan 27, 2014 at 14:51
  • $\begingroup$ @Emil: presumably "nonatomic" means: any set with positive measure has a subset with strictly smaller, but still positive, measure. So ultrafilters are atomic... $\endgroup$ Jan 27, 2014 at 15:17
  • $\begingroup$ @GeraldEdgar: Ah, yes, of course. $\endgroup$ Jan 27, 2014 at 15:57

1 Answer 1

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The usual proof of this uses countable intersections. So for a counterexample, we need a case without countable intersections.

Consider the set $[0,1] \subseteq \mathbb R$. Let $U_n, n \in \mathbb N$ be a disjoint set of nonempty open intervals with total measure $1$. (For example, the intervals that make up the complement of the Cantor set.) Write $U = \bigcup_n U_n$. Let $\mathcal A$ be the collection of all finite unions of intervals $\subset U$ with rational endpoints. For $A \in \mathcal A$ let $\mu(A)$ be the Lebesgue measure of $A$. Note that $0 \le \mu(A) < 1$ for such $A$. Strictly less, because each such $A$ is contained in a finite union of the sets $U_n$.

Let $\mathcal B$ be the collection of complements $[0,1] \setminus A$ of elements of $\mathcal A$. If $B \in \mathcal B$, write $B = [0,1]\setminus A$ and define $\mu(B) = 10 - \mu(A)$. Note that $9 < \mu(B) \le 10$ for such $B$.

Now let $\mathcal F = \mathcal A \cup \mathcal B$. We should check that $\mathcal F$ is a countable algebra, $\mu$ is additive on $\mathcal F$, nonatomic on $\mathcal F$, and the whole space $[0,1]$ is not a disjoint union of sets with measure $< 9$.

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  • $\begingroup$ In the Cantor example, isn't is true that anything is a disjoint union of things with measure $\le 2^{-n}$? $\endgroup$ Jan 27, 2014 at 15:29
  • $\begingroup$ Sorry I do not get your point. Do you mean my example is also true? or .... Please explain more. $\endgroup$
    – Ameen
    Jan 27, 2014 at 15:33
  • $\begingroup$ I mean your example satisfies the "result you need", whereas mine doesn't. $\endgroup$ Jan 27, 2014 at 15:35

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