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

  • 2
    $\begingroup$ Note that the definition of $\mu$ makes no sense if $X$ is countable. This because then every set $A\subseteq X$ is countable (so $\mu(A)=0$) and also the complement of that set is countable (so $\mu(A)=1$). This makes your effort to prove "not countably additive" senseless. $\endgroup$
    – drhab
    Commented Sep 27, 2017 at 11:13
  • $\begingroup$ Hi drhab, thanks a lot for this comment. I think it is true. But could you look at the Example 5.2.2 and look at my comment below. Because it seems there is something wrong or a typo in his example, but he also wants to show the function is NOT countably additive. $\endgroup$
    – john_w
    Commented Sep 28, 2017 at 3:07
  • $\begingroup$ In 5.2.2. $\mu(A)=1$ if $A$ is countable. You accident switched the values $0,1$. $\endgroup$
    – drhab
    Commented Sep 28, 2017 at 14:39
  • $\begingroup$ But I thought in 5.2.2, $\mu(A)=1$ if $A$ is countable would not make the function finitely additive. I thought there is a mistake here in this function. Because say $A_1, ...,A_{10}$ disjoint, then $\mu(\cup^{i=10}_{i=1}A_i ) =1$, but $\sum_{i=1}^{10} \mu(A_i) = 10$ , so finitely additive won't hold if we have $\mu(A) = 1$. $\endgroup$
    – john_w
    Commented Sep 29, 2017 at 2:38
  • $\begingroup$ You are right. It might be that the problem is underestimated. Also have a look here. $\endgroup$
    – drhab
    Commented Sep 29, 2017 at 14:21

1 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.

  • $\begingroup$ Hi drhab, thanks a lot for your comment. The question I asked is from here: stat.colostate.edu/~estep/assets/chapter5.pdf specifically Example 5.2.2 . The author actually wants to show the function is NOT countably additive but only finitely additive. But I think there is a mistake in his definition of the function, so I change it to $\mu(A) = 0$ when $A$ is countable, and $\mu(A)=1$ when $A^c$ is countable and I used this in my question when I posted here. I think it is true that it is finitely additive. But the author says it is NOT countably additive. Could you comment more? $\endgroup$
    – john_w
    Commented Sep 28, 2017 at 3:02
  • $\begingroup$ I think the author wants to show it is not countably additive based on the correct definition of $\mu(A)$ (i.e. the one that I used here) because I think there is something wrong in the definition he used (he used $\mu(A) = 1$ when $A$ is countable, and $\mu(A)=0$ when $A^c$ is countable which I think there maybe a typo, because it is not even finitely additive if I used his definition of the function). $\endgroup$
    – john_w
    Commented Sep 28, 2017 at 3:05

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