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The general set theory claims that there exists a set which has no members. From the point of view of $\mathbb{R}$ what are null sets? More specifically when we define structures such as algebras or subsets of $\mathbb{R}$ etc.

I am reading Measure theory and in there one is talking about the algebra of sets of all finite unions of the form $(a, b]$, $(-\infty, b]$, $(a, \infty)$, $(-\infty, \infty)$ how do you show that $\emptyset$ belongs to the family F of all unions of sets as mentioned above?

One way to look at it is to say sets of measure zero are null sets, but we have not defined what lebesgue measure is yet.

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  1. The set which has no members is called the empty set. Some people mean this when they say ‘null set’, but this is confusing in the context of measure theory.

  2. The empty set is the union of no sets, just the same way that zero is the sum of no numbers.

  3. Sets of measure zero are precisely the null sets, but this is by definition. But there is another way of characterising sets of Lebesgue measure zero: a set $Y$ is a null set if and only if for all positive $\epsilon$, there exists a cover of $Y$ by countably many open intervals $U_i$ such that the sum of the lengths of all the intervals $U_i$ is less than $\epsilon$.

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    Probably Zhen's answer is best, zero is finite, so finite unions include unions of zero sets. But in this case we might allow $(a,a] = \emptyset$ to get the emptyset.2011-08-22
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    Thanks for the response. Using 2, I suppose the fact that the null set or empty set is union of no sets or 0 sets implies it is finite. One can claim that the $\emptyset$ therefore belongs automatically to the algebra. Using definition 3 implies that measure has already been defined. I am uncomfortable either way but I guess it is what it is.2011-08-22
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    @Ramesh: In case it wasn't clear, there are non-empty null sets. For example, $\mathbb{Q}$ is an null set in $\mathbb{R}$.2011-08-22
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    @Zhen: I understand that there are non-empty null sets and it has to be defined according to the context. In defining the algebra one needs to already define what the measure is and the algebra seems to be interconnected. In that sense one cannot say what a null set is with out defining the measure, at least in the context of measure theory. $\mathbb{Q}$ is a null set in $\mathbb{R}$ under lebesgue measure. I think I get it.2011-08-22
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    @Ramesh: You seem to be confused about something, but I'm not sure what. $\emptyset$ is the _empty_ set, which is _a_ null set. The definition of an algebra of sets does _not_ require the measure to be already defined; null sets are irrelevant to the definition.2011-08-22
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    @Zhen: If you were asked to prove that the family of sets F obtained by all finite unions of the sets of the form $(a,b] , (-\infty, b], (a, \infty), (-\infty, \infty)$is an algebra of sets in R. How do u justify that $\emptyset \in F$.2011-08-22
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    @Ramesh: That's (2) in my answer above.2011-08-23
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    @Zhen: Thanks again, I figured that as in my response earlier. I appreciate your help.2011-08-23
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    @Zen: I understand the subtle difference between null sets (i.e. includes sets of measure zero) and empty set. Thanks again, i think that was the source of the confusion.2011-08-23
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    The empty set can be easily constructed as $(-\infty,a)\cap(a,\infty)$; complement of $(-\infty,\infty)$ and so on. It makes more sense than "union of zero many sets" if you are not used to think about trivial sums, unions, and so on.2011-09-21
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    @Asaf Karaglia: Intersection of non-empty sets yes can produce a non-empty set. However not unions. The most logical answer though not convincing is that the empty set (not null) is a union of no zero sets. I will leave it at that.2015-12-22