# (Countable) partition generated $\sigma$-algebra

I am working on exercise 1.9 b) in the book "Probability and Stochastics" by Erhan Çınlar to practice my understanding of $\sigma$-algebras (although this might possibly be future homework in my classes).

Anyway, here's the exercise:

Let $\mathcal{C}$ be a (countable) partition of E. Show that every element of $\sigma \mathcal{C}$ is a countable union of elements taken from $\mathcal{C}$.

So, $\mathcal{C} = \{ C_n: n \in \mathbb{N}\}$ is a partition of $E$. To prove the exercise, I have introduced $\mathcal{E} = \{\cup_{i \in I} C_i: I \subseteq \mathbb{N}\}$ and am trying to show that $\mathcal{E}$ is a $\sigma$-algebra.

For this, I have to show the following three parts/properties (correct me if I'm wrong):

1. E $\in \mathcal{C}$: $\mathcal{C}$ is a partition of E, so $\cup \mathcal{C}=E.$ Since $\mathcal{C}$ is countable, $\cup \mathcal{C}$ is the union of countably many members of $\mathcal{C}$.

2. Show that $\mathcal{E}$ is closed under countable unions: Since $\mathcal{E}$ consists only of sets which are countable unions of elements of $\mathcal{C}$, $\mathcal{E}$ is closed under countable unions.

3. Show that $\mathcal{E}$ is closed under complement: Let $A \in \mathcal{C}$, so there exists a countable $\mathcal{C}_A \subseteq \mathcal{C}$ such that $A = \cup \mathcal{C}_A.$ Let $\mathcal{D} = \mathcal{C} \setminus \mathcal{C}_A$; then $\mathcal{D}$ is a countable subset of $\mathcal{C}$, and if $D =\cup \mathcal{D},$ then $D \in \mathcal{C}.$ Since $\mathcal{C}$ is a partition on E, $D = E \setminus A = A^c$, and so $A^c\in \mathcal{C}$.

Now, in the last part I want to show that $\mathcal{E} = \sigma\mathcal{C}$, but I'm not sure how exactly this is done.

• what does $I \leq \mathbb{N}$ mean ? Just precise it (ok thanks). Otherwise, you seem to have it all. You showed that $\mathcal{E}$ is a sigma-algebra, obviously contained in $\sigma \mathcal{C}$. What could you do to show this is an equality ? Mar 3 '15 at 7:44
• Ah that is a typo, it is supposed to be $I \subseteq \mathbb{N}$, I've changed it in the post just now. Mar 3 '15 at 7:50
• Well I guess then next I would show that $\sigma \mathcal{C}$ is contained in $\mathcal{E}$? I'm also not completely sure if my proof for the "countable under countable union" property is 100% correct. Mar 3 '15 at 7:53
• Your proof for pt 2 is right. If you want it more rigorous, just write it with the index : let $(A_i)$ be a countable family of sets in $\mathcal{E}$. Then, for every $i$, there is a countable set $J_i$ such that $A_i = \cup_{j \in I_i} C_j$. Therefore, $\cup_{i \in I} A_i = \cup_{j \in K} C_i$, where $K$ is the union of all the $J_i$ over $i \in I$. As a countable union of countable sets is clearly countable, $K$ is countable and $\cup_{i \in I} A_i$ is in $\mathcal{E}$. Mar 3 '15 at 7:56
• To prove that $\mathcal{E} = \sigma (\mathcal{C})$, you could maybe use a property of $\sigma(\mathcal{C})$... Mar 3 '15 at 7:58

Here's a proof using only the material covered in Çınlar before your exercise. As you showed, $$\mathcal{E}$$ is a $$\sigma$$-algebra containing $$\mathcal{C}$$. It is obviously contained in $$\sigma(\mathcal{C})$$ : to see that, all we have to say is that $$\sigma(\mathcal{C}$$ must contain all the countable unions of elements in $$\mathcal{C}$$, so it contains all the elements of $$\mathcal{E}$$. Therefore, $$\mathcal{E} \subset \sigma (\mathcal{C})$$.

But, $$\sigma(\mathcal{C})$$ is defined as the intersection of all the $$\sigma$$-algebras containing $$\mathcal{C}$$. Therefore, as $$\mathcal{E}$$ is a $$\sigma$$-algebra containing $$\mathcal{C}$$, it is clear that $$\sigma(\mathcal{C})$$ is contained in $$\mathcal{E}$$.

So, $$\mathcal{E} = \sigma (\mathcal{C})$$.

This kind of reasonning is very very common in measure theory, analysis, probability. Make sure you understand all the things here. It works like this : to prove that two $$\sigma$$-algebras are equal, you write one as the $$\sigma$$-algebra generated by a collection, and you show that the other is a $$\sigma$$-algebra containing this collection and contained in the first one.

Have a nice time with Çınlar's excellent textbook :)

• Thanks for the very intuitive summary! Mar 3 '15 at 8:20
• (btw, you made a small typo: "reasonning" is written with only one n: reasoning) Mar 3 '15 at 8:29
• You mean, "reasoning" is written with two $n$, not three. Damn, my English is not perfect ^^ Mar 3 '15 at 8:30
• @TlönUqbarOrbisTertius Kind of late to the game here, but thank you for this answer and explanation! I really appreciate it. Apr 10 at 16:01