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I'm confused because the proof of Hahn's Lemma in Royden requires enumerating sets of negative measure. (We're working on signed measures) If you exclude the counterexample of using subsets of measure zero, say singletons, will you now be forced to say that there are only countably many subsets?

Edit: I added the condition of disjoint.

user54358
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  • Well $\lambda([0,1]) = 1$ and $\lambda([0,x]) = x$ with $x \in (0,1]$. This gives you an uncountable number of nonzero measure subsets. – Stefan Mar 21 '13 at 13:10
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    See this page for your answer. – Clayton Mar 21 '13 at 13:29
  • Yes, Clayton thank you. But in what I'm studying we have slightly relaxed the condition to allow negative measures. I also feel it really boils down to talking about an uncountable sum becoming finite. – user54358 Mar 21 '13 at 13:34

2 Answers2

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No. Consider the interval $[a,b]$ and its subsets of the form $[c,b]$ with $a<c<b$.

Edit: Since you're now adding disjointness as a condition, then the answer is "yes." Observe in particular that there are infinitely-many positive-measured subsets or infinitely-many negative-measured subsets, so Quiaochu's (linked) argument works or GEdgar's (below) works. Either way, you've your answer.

Community
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Cameron Buie
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  • Ok thanks. I understand now. I mean, I understand parts of the proof I'm reading better. Will adding the condition of disjoint make it true? – user54358 Mar 21 '13 at 13:11
  • Do you mind if I edit the question to add the condition of disjoint because I sort of have it at the back of my head? – user54358 Mar 21 '13 at 13:16
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Suppose sets $A_i$ with $i \in I$, an uncountable index set, satisfy: $A_i$ are pairwise disjoint and $\mu(A_i)<0$. If $I_n = \{i \in I: \mu(A_i) < -1/n\}$, then $I$ is the union of countably many sets $I_n$. We know a countable union of countable sets is countable, so $I_n$ is uncountable for some $n$. Fix such an $n$. Choose $i_1,i_2,\cdots \in I_n$ all different. Then $$ \mu\left(\bigcup_{k=1}^\infty A_{i_k}\right) = \sum_{k=1}^\infty \mu\left(A_{i_k}\right) = -\infty $$ (an infinite series where every term is ${}\le -1/n\;$). Measure $-\infty$ is not allowed for a signed measure.

GEdgar
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