This is my first time posting here. I am working on exercise 1.9 from Probability and Stochastics by Erhan Çinlar. The problem is formulated as follows:

"Let E = $\mathbb{R}$, the set of all real numbers. Let $\mathcal{C}$ be the collection of all singleton subsets of $\mathbb{R}$, that is, each element of $\mathcal{C}$ is a set that consists of exactly one point in $\mathbb{R}$. Show that every element of $\sigma \mathcal{C}$ is either a countable set or the complement of a countable set." (Here $\sigma \mathcal{C}$ is the sigma algebra generated by the collection $\mathcal{C}$)

I did find questions here, here, and here that are very closely related to mine. I tried working through them, but I got stuck understanding the proofs. Below I detail what I misunderstood.

Let $\mathcal{A}$ be the collection of all subsets of $\mathbb{R}$ that are either countable or the complement of a countable set. By showing that $\sigma \mathcal{C}$ = $\mathcal{A}$, I would have proven the question, if I understand correctly. Then this breaks down to showing that (1) $\sigma\mathcal{C} \subseteq \mathcal{A}$ and (2) $\mathcal{A} \subseteq \sigma\mathcal{C}$.

Regarding (1), I confirmed that $\mathcal{A}$ is a $\sigma$-algebra. The collection $\mathcal{C}$ from the question is clearly a subcollection of $\mathcal{A}$. What I don't understand, is that this seems to lead directly to the conclusion that the $\sigma\mathcal{C} \subseteq \mathcal{A}$. Why is that so? I tried writing out examples with very small $\sigma$-algebras explicitly, but I did not get much further doing this. Regarding (2), I am even more confused. If we take $A \in \mathcal{A}$, then $A$ is either a countable or co-countable subset of $\mathbb{R}$. How do I move from this fact to showing that $A$ is in $\sigma\mathcal{C}$? In the proofs I read, this was also taken as an obvious step, but I have trouble seeing the logic.

Apparently this type of proof is very common in showing that $\sigma$-algebras are equal. As another question, what are resources on proof techniques that will be helpful in working through the book by Çinlar? Especially the early part on measure theoretic probability, that is. Some more background knowledge would be useful, instead of going empty-handed into these exercises.


1 Answer 1


I think the idea that you might be missing is that given a collection of sets $\mathcal{C}$, the $\sigma$-algebra generated by $\mathcal{C}$ is the smallest $\sigma$-algebra containing every set in $\mathcal{C}$.

In your case, $\mathcal{C}$ is the collection of singleton sets in $\mathbb{R}$. Let $\sigma \mathcal{C}$ denote the $\sigma$-algebra generated by $\mathcal{C}$, and $\mathcal{A}$ denote the collection of sets which are countable or cocountable. Suppose we have shown part (2). Then we would have:

  1. $\mathcal{A} \subseteq \sigma \mathcal{C}$
  2. $\mathcal{A}$ is a $\sigma$-algebra
  3. $\mathcal{C} \subseteq \mathcal{A}$

If $\mathcal{A} \neq \sigma(\mathcal{C})$, then $\mathcal{A}$ is a smaller $\sigma$-algebra containing $\mathcal{C}$. This cannot exist by the definition of $\sigma(\mathcal{C})$. Therefore they are equal. This should address (1).

To address (2), if $A$ is a countable set, then $A$ is a countable union of singletons. If $A$ is cocountable, then $A$ is the set difference of $\mathbb{R}$ and a countable union of singletons. Since $\sigma$-algebras are closed under countable unions and differences, it follows that these are in $\sigma \mathcal{C}$.

  • $\begingroup$ Fantastic. That fills in the details I was misunderstanding. Accepted as the answer. Thank you. One last question, is it common to use the one inclusion in proving the other? I hadn't thought of looking at the problem through that lens. $\endgroup$
    – BasMts
    Commented Nov 11, 2022 at 19:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .