infinite set don't contain sigma algebra that have more than one countable complement

Question

I have a question regarding the complements of two disjoint set in a sigma algebra on an infinite set say $X$.

Basically this questions comes to my mind when I was trying to show the following:

Given $X$ an infinite set, and some collections of sets of $X$ say denoted as $M$ that satisfies the condition that for each $A \in M$, either $A$ is countable of $A^{c}$ is countable. The set that satisfies the condition can be shown to be a set of sigma algebra, i.e. $M$ is a sigma-algebra. Prove that the function defined below is finitely additive, but not countably additive:

$\mu(A) = 0$ if $A$ is countable , $\mu(A) = 1$ if $A^{c}$ is countable.

My attempt to show that it is not countably additive is because:

we have $X = \bigcup_{i=1} \{a_i\}$ where $\{a_i\}$ is an element in $X$ because of course $X$ can be written as union of each of its element even if $X$ is an infinite set.

Then I have $\mu(X) = 1$ because the complement of $X$ is the $\emptyset$, i.e. $X^{c} = \emptyset$. Since $\emptyset$ has nothing it in, so it is countable, and hence $\mu(X)=1$. But $\mu(\{a\}) = 0$ for each $\{a\}$ since the set $\{a\}$ is countable as it is a singleton. So $\sum_{i=1}^{\infty} \mu(\{a_i \})=0 \ne 1 = \mu(X)$. And hence it is not countably additive.

Is the above proof correct? could someone comment?

For the finitely additive part, basically I was able to show that in the sigma algebra $M$, given $A_i$ disjoint, i.e. $A_i \cap A_j =\emptyset$, there cannot exist two disjoint sets say $A_i$ and $A_j$ with $A_i^{c}$ being countable and $A_i^{c}$ being countable at the same time. i.e. say given $A_i$ is an infinite subset of $M$, so $A_i^{c}$ is countable, then there is no another set say $A_j$ in $M$ disjoint from $A_i$ that satisfies the above property, and because of this reason, finitely additive holds for the above function.

So here are my two questions: 1) My question is whether my proof on the Countable-additive part correct? (i.e. did I show that countably-additive DO NOT hold correctly?)

2) Given an infinite set say X=[0,1]. A sigma algebra I can think of is

$M = \{\{\}, [0,1], \{0.3,0.5\}, \{[0,0.3) \cup (0.3,0.5) \cup (0.5,1]\} \}$,

so say $A_1 =[0,1]$ and hence $A_1^{c} =\emptyset$, it seems it is true that there is no anther subset of $X$ that is disjoint from $A_1$ and have the complement being countable. But is it true that the set that satisfies the condition i.e. $A_i$ is uncountable, but $A_i^{c}$ is countable will always be the set $X$? i.e. in my example here, $A_1 = X$ because as we know a sigma algebra MUST always contain the set $X$ and the emptyset $\emptyset$, so if $X$ is itself infinite, then it always takes up a set $A_i$ in the sigma algebra $M$ that satisfies the above particular condition (i.e. $A$ is countable or $A^{c}$ is countable).

Could someone comment whether my proof on the Countable-additivity part is correct? and also comment on my Question 2?

Thank you

Answer 1

$X$ must be uncountably infinite to make $\mu$ well-defined (see my comment on your question) and in the sequel it is preassumed that $X$ is uncountably infinite.

Let $A_1,A_2,\dots\in\mathcal M$ be disjoint. We intend to prove the following statement: $$\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)\tag1$$

If all $A_i$ are countable then so is their union so in that case LHS and RHS of $(1)$ both take value $0$.

Now assume that e.g. $A_1$ is not countable.

Then $A_1\in\mathcal M$ implies that $A_1^c$ is countable.

Then also its subsets $(\bigcup_{n=1}^{\infty}A_n)^{\complement}, A_2,A_3,A_4,\dots$ are countable,

Then LHS and RHS of $(1)$ both will equal $1$.

We conclude that $\mu$ is countably additive and consequently that it is finitely additive.