Is the countable union of measure-zero sets zero?

Question

Let $E=\cup_{i=1}^\infty A_i$ where the measure of $A_i$ is zero. How can I conclude that $E$ has measure zero?

You can change any union into a disjoint union: Replace $A_2$ with $A_2\setminus A_1$, $A_3$ with $A_3\setminus(A_1\cup A_2)$, etc. — Andrés E. Caicedo, May 12 '13 at 15:33

score 5 · Accepted Answer · answered May 12 '13 at 15:35

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If your measure $\mu$ is sub sigma additive you can surely say that \[ \mu(E)\leq \sum_{i=1}^\infty \mu (A_i)=0\] If it's not the statement doesn't need to be true.

answered May 12 '13 at 15:35

Dominic Michaelis

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Ah, ok, you are using "measure" to mean "finitely additive measure" rather than the usual countably additive requirement. – Andrés E. Caicedo May 12 '13 at 15:39
@AndresCaicedo Until now we only met the Lebesgue measure in Measure-theory. But as I know there are some more general concepts like the Haarmeasure. So I thought let's take the weakest argument which comes in my head and says that your measure needs it. # – Dominic Michaelis May 12 '13 at 15:45
By definition, measures are $\sigma$-additive. This implies sub-additivity, which is what you are using here. We use the word measure in other contexts, and then replace requirements accordingly. But the standard usage in measure theory has sigma-additivity as part of the definition. (Anyway, your answer is fine, of course.) – Andrés E. Caicedo May 12 '13 at 15:48
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@AndresCaicedo so the outer measure is not a measure? – Dominic Michaelis May 12 '13 at 15:50
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Not in general, but measures can be defined from outer measures by identifying what sets are measurable. (It is unfortunate notation. And yes, sometimes authors start by saying: " By "measure" I mean "outer measure" ", which does not help matters.) – Andrés E. Caicedo May 12 '13 at 15:53

score 4 · Answer 2 · answered May 12 '13 at 15:34

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Hint: Assuming the measure is Lebesgue measure, for each $A_i$, you can find an open set $O_i$ such that $A_i\subseteq O_i$ and the measure of $O_i$ is less than $\varepsilon/2^i$. If not, as long as the measure is subadditive, you can follow Dominic Michaelis' approach.

answered May 12 '13 at 15:34

Clayton

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If it is a measure, it is subadditive. Things that are not measures carry qualifiers, such as "finitely additive measure". – Andrés E. Caicedo May 12 '13 at 15:38

score 1 · Answer 3 · edited May 12 '13 at 15:39

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It follows by sigma-aditivity. $$m\bigg(\bigcup_{i\in\mathbb {N}}A_{i}\bigg)\leq\sum_{i\in\mathbb{N}}m(A_{i})=0$$

edited May 12 '13 at 15:39

Martin

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answered May 12 '13 at 15:31

Aleph

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@DominicMichaelis: Measures are by definition $\sigma$-additive. If $\sigma$-additivity is not assumed then the question is simply false (take the finitely additive probability measure associated with an ultrafilter on $\mathbb{N}$ which is disjoint from the ideal of finite sets. Then $\mathbb{N} = \bigcup_{n \in \mathbb{N}} {n}$ is a countable union of null-sets but has measure $1$.) – Martin May 12 '13 at 15:41

Is the countable union of measure-zero sets zero?

3 Answers3