The problem I'm trying to solve is the following:

Consider Ω uncountable and F = {A ⊆ Ω: A is countable or Ac is countable}.

1) Show that F is a σ-algebra of Ω

2) Consider P: F → [0,1]. If A is uncountable P(A) = 1 and if A is countable P(A) = 0. Is P a probability measure?

I think that for this problem I just have to prove that the conditions for a σ-algebra is valid for 1) and Kolmogorov axioms are valid for 2).

However, I'm having a lot of trouble with countable (and uncountable) sets properties. For example, if A is countable, Ac may be countable or uncountable, right?

At the moment I have the following:


i) F is non-empty. This is true, since A is in F

ii) F is Closed under complementation. I'm not sure on how to work A and Ac in this

iii) F is Closed under countable unions.


i) P>= 0. This is true, it can only be 0 or 1.

ii) P(Ω) = 1. Also true, do I need to prove this?

iii) P(Union of all the sets) = ∑ P(each set) if they are disjoint. I'm not sure on how to prove this one.

Sorry for all the questions. I'm not a mathematician and moving areas has been very hard for me.

  • $\begingroup$ If $A$ is countable, then $A^c$ must be uncountable. First try to prove this (it should be short). $\endgroup$ – Shalop Mar 18 '15 at 22:57
  • $\begingroup$ Are you sure that if A is countable Ac must be uncountable? For example, A = Even numbers then Ac = Odd numbers and both are countable. Or is there something wrong with my logic? $\endgroup$ – mechanical_fan Mar 19 '15 at 0:42
  • $\begingroup$ But in this example, the set $\Omega$ is uncountable. That is why $A^c$ is uncountable whenever $A$ is countable. If the overlying set $\Omega$ was countable (as in your even-odd example), then you're correct that it's not true. Thus, the cardinality of the universal set $\Omega$ is critical here. $\endgroup$ – Shalop Mar 19 '15 at 0:46
  • $\begingroup$ So it must be because A + Ac = Ω, so one of them must be uncountable (because Ω is uncountable), right? What if A is uncountable, must Ac then be countable in this case? $\endgroup$ – mechanical_fan Mar 19 '15 at 20:56
  • $\begingroup$ If $A \subset \Omega$ is uncountable, then it is not necessarily true that $A^c$ is countable. For example, if $\Omega=[0,2]$ and $A=[0,1]$, then $A^c=(1,2]$, and both $A$ and $A^c$ are uncountable. However, if $A$ is uncountable and $A \in \mathcal{F}$, then $A^c$ must be countable, by definition of $\mathcal{F}$. $\endgroup$ – Shalop Mar 19 '15 at 21:01

Hints for statement (ii) in Part 1:

If $A \in \mathcal{F}$ then $A$ is countable or $A^c$ is countable. If $A^c$ is countable, then $A^c \in \mathcal{F}$. If $A = (A^c)^c$ is countable, then $A^c \in \mathcal{F}$. Thus, in any case $A^c \in \mathcal{F}$.

Hints for statement (iii) in Part 1:

Any countable union of countable sets is necessarily countable (assuming the Axiom of Choice). If $\{A_n \}_{n \in \mathbb{N}}$ is any countable subcollection of $\mathcal{F}$, consider the case where there exists an uncountable $A_n$ separately from the case where all $A_n$ are countable.

Hints for statement (iii) in Part 2:

If $ \{ A_n \}_{n \in \mathbb{N}}$ is any countable collection of pairwise disjoint sets in $\mathcal{F}$, then it can contain at most one uncountable set (because if $A_n, A_m$ are two disjoint uncountable sets, then $A_n^c, A_m^c$ are countable, so $A_n^c \cup A_m^c$ is countable, and thus $A_n \cap A_m = (A_n^c \cup A_m^c)^c$ is uncountable and thus non-empty, which is a contradiction).


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