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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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    @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