Let $X$ be a non empty set. Define set function $\mu$ on $E$ subset of $X$ by $\mu (E)=0$ if $E$ is finite and $\mu (E)=1$ if $X\backslash E$ finite. (a) If $X$ is countably infinite, show that for $\mu$ is not countably additive. (b) If $X$ is uncountable, show that for $\mu$ is countably additive.

My answer: (a) Since $X$ is countable, we can write $X$ as $X=\{x_1,x_2,...,x_n\}$. Let $E_i=\{x_i\}$ for every $i \in \mathbb{N}$. Since $E_i$ is finite, we have $\mu(E_i)=0$ hence $\sum_{i=1}^{\infty} \mu(E_i)=0$. Since, $\bigcup_{i=1}^{\infty} E_i=X$, we have $\mu (\bigcup_{i=1}^{\infty} E_i)=1 \neq \sum_{i=1}^{\infty} \mu(E_i)$.

(b) Let $\{E_n\}_{n=1}^{\infty}$ be a disjoint collection of subsets of $X$ that is either finite or cofinite. We will prove that $\mu (\bigcup_{i=1}^{\infty} E_i)= \sum_{i=1}^{\infty} \mu(E_i)$

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    $\begingroup$ It looks like you stopped short for part (b)... $\endgroup$ – anon271828 Feb 4 '13 at 23:31
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    $\begingroup$ $X$ is countably infinite, so we certainly can't say $X=\{x_1,x_2,...,x_n\}$. You mean perhaps that $X=\{x_n:n\in\Bbb N\}$, where the $x_n$ are distinct, yes? Also, I think you mean "Since $E_i$ is finite, we have $\mu(E_i)=0$" in your part (a) work. $\endgroup$ – Cameron Buie Feb 4 '13 at 23:31
  • $\begingroup$ @anon271828, : Part (a), i just wanna make sure that is correct, but part (b) i have no idea. Any Hint? $\endgroup$ – beginner Feb 4 '13 at 23:50
  • $\begingroup$ Currently $(b)$ does not make sense. If $E_{n}$ is finite for each $E_{n}$, then $\bigcup E_{n}$ is countably infinite, and $X\setminus \bigcup E_{n}$ is uncountable. Hence $\mu(\bigcup E_{n})$ does not exist. $\endgroup$ – T. Eskin Feb 5 '13 at 7:58
  • $\begingroup$ In fact, in $(b)$, if every $\mu(E_{n})$ is defined, then by going through all the possible cases you will notice that $\mu(\bigcup E_{n})$ being defined implies that $E_{n}=\emptyset$ all but finitely many $n$. $\endgroup$ – T. Eskin Feb 5 '13 at 8:23

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