I will flesh out the question here (Question taken from Rosenthal's Intro to probability theory book). (I omit parts (a) and (c))

Suppose that $\Omega = [0,1]$ is the unit interval, and $\mathcal{F}$ is the set of all subsets $A$ such that either $A$ or $A^c$ is countable (i.e., finite or countably infinite), and $P$ is defined by $P(A) = 0$ if $A$ is countable, and $P(A) = 1$ if $A^c$ is countable.

(b) Is $\mathcal{F}$ a $\sigma$-algebra?

(d) Is $P$ countably additive on $\mathcal{F}$? meaning that if $A1,A2,... \in \mathcal{F}$ are disjoint, and if it happens that $\cup_n A_n \in \mathcal{F}$, then $P(\cup_n A_n) = \sum_n P(A_n)$

I am wondering whether I am correct for b,d, and whether the (rough) proofs I provide look okay?


(b) Yes. $\emptyset, \Omega\in \mathcal{F}$ clearly, and $\mathcal{F}$ is clearly closed under complements (by def).

Is $\mathcal{F}$ closed under finite unions? Consider the following cases:

  1. Case: $ \cup_{i=1}^\infty A_i$ countable,$\ A_i \in \mathcal{F}$ and $A_i$ countable $\forall i$. Then $ \cup_{i=1}^\infty A_i \in \mathcal{F}$ because a countable union of countable sets is countable.
  2. Case: $ \cup_{i=1}^\infty A_i$, uncountable, $\ A_i \in \mathcal{F}$ and $A_i$ are countable $\forall i$ except one. WLOG assume $A_n$ is uncountable.Then $A_n^c$ is countable because $A_n \in \mathcal{F}$ and thus $ \cup_{i=1}^\infty A_i \in \mathcal{F}$ because $ (\cup_{i=1}^\infty A_i)^c= \cap_{i=1}^\infty A_i^c$ which cannot have more elements than the smallest set, which is the countable $A_n^c$ (because assumed that is the only uncountable set, so all other $A_i^c$ are uncountable). This argument can be expanded by induction to the case with more than one uncountable set.

(d) Yes. Let $A_1,A_2,\dots \in \mathcal{F}$ be disjoint such that $\cup_{i=1}^\infty A_i \in \mathcal{F}.$There are two cases to consider:

  1. Case: $\cup_{i=1}^\infty A_i$ is countable.

Then $A_i$ is countable $\forall i$ and then $$ P(\cup_{i=1}^\infty A_i) = 0 = \sum_{i=1}^\infty P(A_i) $$

  1. Case: $\cup_{i=1}^\infty A_i$ is uncountable.($(\cup_{i=1}^\infty A_i)^c$ is countable.).

Then at least one $A_i$ is uncountable, because if all $A_i$ we countable then the countable union would have to be countable.. Let $A_{n_0}$ be uncountable.

We will show that only one $A_i$ can be infinite: Consider $n\not = n_0$. Then, by assumption, $A_n \cap A_{n_0} = \emptyset \implies A_n \subseteq A_{n_0}^c$ but $A_{n_0}^c$ is countable, which implies that $A_n$ is countable. Thus, only $A_{n_0}$ is infinite. Therefore, $$ P(\cup_{i=1}^\infty A_i) = 1 +0 = \left(\sum_{n \not = n_0} 0\right) +P(A_{n_0}) = \left(\sum_{n \not = n_0} P(A_n)\right) +P(A_{n_0}) = \sum_{n =1}^\infty P(A_n) $$ Thus, $P$ is countably additive.


1 Answer 1


(b) Your cases 1. and 2. are not exaustive, since we could have $\bigcup A_i$ uncountable, and more than one of the $A_i$ being uncountable. The rest of the arguments are more-or-less ok, but it's quite confusing to read. I'll write a version here, which you will probably recognize as saying basically the same things as you did:

Case 2. $\bigcup_{i=1}^\infty A_i$ is uncountable.

Then at least one the $A_i$ is uncountable (because a countable union of countable sets is countable). Let $A_{n_0}$ be uncountable. Since $A_{n_0}\in\mathcal{F}$, then $A_{n_0}^c$ is countable. Since $(\bigcup_{i=1}^\infty A_i)^c=\bigcap_{i=1}^\infty A_i^c\subseteq A_{n_0}^c$, then $(\bigcup_{i=1}^\infty A_i)^c$ is countable, so $\bigcup_{i=1}^\infty A_i\in\mathcal{F}$.

(d) Again your second case is confusing, and you're using "infinite" to mean "uncountable", apparently, which is wrong. The arguments are ok, though, if you correct the usage of the words "infinite".

Also, be careful in the order of the elements in the equation you wrote: You should write "$0+1$" instead of "$1+0$" to have the other terms in the correct order.

  • $\begingroup$ Thank you very much for taking the time both to answer to to provide helpful comments. $\endgroup$
    – majmun
    Commented Aug 9, 2016 at 3:42
  • $\begingroup$ @Luiz. I think you meant to say "Since $(\bigcup_{i=1}^\infty A_i)^c=\bigcap_{i=1}^\infty A_i^c\subseteq A_{n_0}^c$," and not "Since $(\bigcup_{i=1}^\infty A_i)^c=\bigcap_{i=1}^\infty A_i^c\subseteq A_{n_0}$," $\endgroup$ Commented Mar 6, 2019 at 3:03
  • $\begingroup$ @TryingHardToBecomeAGoodPrSlvr Yes, I did. Corrected ;) $\endgroup$ Commented Mar 6, 2019 at 21:17

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