# Generalization of measure theory to uncountable unions

• I
Hi.

Is there some kind of measure theory generalized to uncountable unions? Of course one needs to take care how to make sense of sums over an uncountable index set. I was thinking about following formulation of the additivity property of the "measure":
$$\mu\left(\bigcup_{i\in\ I} A_i\right)=\sup_{J\subset\ I}\sum_{i\in\ J}\mu\left(A_i\right)$$
Here ##I## can be uncountable, but the supremum is taken over all countable subsets ##J##.

Obviously, if we allow all uncountable unions on ##P(\mathbb{R})##, only the trivial measure can satisfy this (because we can divide every set into subsets of single elements with "measure" 0). But can there be a (nontrivial) subset of uncountable unions that allows for a nontrivial "measure"?

## Answers and Replies

WWGD
Science Advisor
Gold Member
Hi.

Is there some kind of measure theory generalized to uncountable unions? Of course one needs to take care how to make sense of sums over an uncountable index set. I was thinking about following formulation of the additivity property of the "measure":
$$\mu\left(\bigcup_{i\in\ I} A_i\right)=\sup_{J\subset\ I}\sum_{i\in\ J}\mu\left(A_i\right)$$
Here ##I## can be uncountable, but the supremum is taken over all countable subsets ##J##.

Obviously, if we allow all uncountable unions on ##P(\mathbb{R})##, only the trivial measure can satisfy this (because we can divide every set into subsets of single elements with "measure" 0). But can there be a (nontrivial) subset of uncountable unions that allows for a nontrivial "measure"?

I am not sure, sorry, but I have thought of using hyperreal-valued measures, since this would allow for sums to assume infinite values. Still, it seems hard to interpret a set having a measure with non-zero non-standard part. I think I saw this dealt with somewhere; let me see if I can find a link to the source. Your taking limits over countable subsets kind of reminds me of Riemann sums.

Maybe I should have started with a more concrete question: Is there a partition of the unit interval into an uncountable number of subsets that all have nonzero Lebesgue measure?

WWGD
Science Advisor
Gold Member
Maybe I should have started with a more concrete question: Is there a partition of the unit interval into an uncountable number of subsets that all have nonzero Lebesgue measure?
If I understood correctly, no, if by partition you mean breaking down into pairwise-disjoint subsets.. For sum to converge ( meaning to be finite in this case) , support must be countable ; finite or infinite, of course.

Erland
Science Advisor
Maybe I should have started with a more concrete question: Is there a partition of the unit interval into an uncountable number of subsets that all have nonzero Lebesgue measure?
No, because then, there must be an infinite (in fact uncountable) set of sets from the partition which all have measure > 1/n for some specific n > 0 (otherwise the partition would be countable, since it would be a countable union of countable sets), and then the measure of the unit interval would be infinite, a contradiction.

I think this is what WWGD meant.

WWGD
WWGD
Science Advisor
Gold Member
No, because then, there must be an infinite (in fact uncountable) set of sets from the partition which all have measure > 1/n for some specific n > 0 (otherwise the partition would be countable, since it would be a countable union of countable sets), and then the measure of the unit interval would be infinite, a contradiction.

I think this is what WWGD meant.
Yes; exactly what I meant define ##S_n:=\{ s: s>1/n\} ## for ##s## in the collection . Then , by cardinality argument, at least one such ## S_n## is infinite...

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