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Let $X$ be a nonempty set and let $\mathcal{M}$ be the $\sigma$-algebra of countable subsets and cocountable subsets of $X$. Let $\mu(A) = 0$ if $A$ is countable and 1 if $A$ is cocountable. I want to prove that $\mu$ is a measure, but I'm worried that it might not be well defined since $X$ can be countable, and thus any subset of $X$ is both countable and cocountable. Any ideas?

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    $\begingroup$ You will have to assume that $X$ is uncountable. $\endgroup$ – PhoemueX Jul 12 '15 at 21:55
  • $\begingroup$ That's what I figured. The question I'm reading from didn't require that, but it was from my instructor and not a book, so she probably left out that detail on accident. $\endgroup$ – change_picture Jul 12 '15 at 22:01
  • $\begingroup$ duplicate of : math.stackexchange.com/questions/509394/… $\endgroup$ – zoli Jul 12 '15 at 23:02
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Suppose that $X$ is uncountable.

The first part of showing that this measure is well-defined is to verify that $\mathcal M$ is indeed a $\sigma$-algebra. It is clearly closed under complements. To show that it is closed under countable intersections, there are two cases to consider: either the intersection involves only cocountable sets (in which case it is also cocountable), or it involves some countable set (in which case it is countable). Thus $\mathcal M$ is indeed a $\sigma$-algebra.

To show the measure is well-defined, we must show that $\mu$ assigns a unique value to every $A\in \mathcal M$. Then either $A$ or $A^c$ is countable, by definition of $\mathcal M$. Moreover $A$ and $A^c$ cannot be simultaneously countable, since otherwise $X$ would be countable. Hence the measure is well-defined.

Lastly, to verify that $\mu$ is a measure, one must check that $\mu(\varnothing)=0$ (clear) and that $$\mu(\bigcup_{n\in \mathbb N} A_n)=\sum_{n\in \mathbb N}\mu(A_n)$$ for any disjoint countable family of sets $\{A_n\}_{n\in \mathbb N}\subset \mathcal M$. First, observe that there is at most one cocountable set among the $\{A_n\}$ (which follows by disjointness and the fact that $X$ is uncountable). Next if there is precisely one cocountable set, observe that both sides are $1$. In the remaining case when all sets are countable, both sides evaluate to $0$. Thus we have a well-defined measure.

Suppose that $X$ is countable.

In this case, the measure is not well-defined, because $0=\mu(X)=1$.

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  • $\begingroup$ Thanks. I already knew how to prove it but the way the question was presented to me did not make any assumptions about X being uncountable, so I was worried I was misinterpreting something. Indeed, this works only if X is uncountable. $\endgroup$ – change_picture Jul 13 '15 at 6:56
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One way to make sure that the measure is well-defined is to assign to cocountable sets infinity instead of $1$1. In other words, set $\mu(A) = \infty$ if $A$ is cocountable, and $\mu(A) = 0$ if $A$ is countable. This will only work if $X$ is uncountable to begin with, otherwise cocountable set will be countable by itself.

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  • $\begingroup$ Thanks. I already knew how to prove it but the way the question was presented to me did not make any assumptions about X being uncountable, so I was worried I was misinterpreting something. Indeed, this works only if X is uncountable. $\endgroup$ – change_picture Jul 13 '15 at 6:57

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