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There is an uncountable set $X$ given. There is also a $\sigma$-algebra given:
$$F = \{A \subseteq X: |A| = \aleph_0 \vee |A^c| \le \aleph_0 \}$$

I decided that something like that would be good:
$$\forall A \in F \mu(A) = \begin{cases} 0 &\text{if } |A|\le\aleph_0\\1 &\text{if } |A^c|\le\aleph_0 \end{cases}$$ However there is a problem (I think) with this point in axioms:
$$\mu \bigg(\bigcup_{n=1}^{\infty} A_n \bigg) = \sum_{n=1}^{\infty}\mu(A_n)$$ How can I solve this problem? Maybe the measure is wrong?

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    $\begingroup$ What is the problem? Aren't the $A_n$ supposed to be pairwise disjoint? -- Wait, the main problem is that $F$ is not a $\sigma$ algebra in the first place because $X\notin F$ $\endgroup$ Commented Nov 2, 2017 at 21:10
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    $\begingroup$ As you've defined it, $F$ is not a $\sigma$-algebra (nor an algebra, for that matter). There ought to be "$\le\aleph_0$" every time you've written "$=\aleph_0$". $\endgroup$
    – user228113
    Commented Nov 2, 2017 at 21:12
  • $\begingroup$ Thanks! That's right. I have corrected that mistake. $\endgroup$
    – Hendrra
    Commented Nov 2, 2017 at 21:27

1 Answer 1

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If you make your $=\aleph_0$s into $\le \aleph_0$s you do have a $\sigma-$algebra. The measure is fine. The requirement for additivity is only countable unions and only when the $A_n$ are disjoint. If all your $A_n$ are countable, the union of countably many is countable as well, so it has measure zero just like the sum on the right. If one of the $A_n$s is cocountable, the measure is $1$, as is the right hand side. No other one can be cocountable and disjoint from that one, so there will be only one $1$ on the right.

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  • $\begingroup$ Why no other can be cocountable and disjoint from that one? $\endgroup$
    – Hendrra
    Commented Jan 27, 2018 at 20:07
  • $\begingroup$ To have two sets be disjoint one has to be a subset of the complement of the other. If the first set is cocountable, its complement is countable. The second, being a subset of that countable set, is at most countably infinite. It therefore cannot be cocountable because all cocountable sets are uncountable. $\endgroup$ Commented Jan 27, 2018 at 20:42
  • $\begingroup$ I think it's a bit clearer now. So that means that if we would like fo find another disjoint set with the cocountable one it must be countable? $\endgroup$
    – Hendrra
    Commented Jan 27, 2018 at 20:47
  • $\begingroup$ That is right. It works just like cofinite sets in $\Bbb N$. Two cofinite sets have to have infinite intersection. Similarly in any uncountable set, two cocountable subsets have to intersect in uncountably many points because the points not in their intersection are the union of their complements, which are each countable so their union is as well. $\endgroup$ Commented Jan 27, 2018 at 21:34

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