# Partition generates sigma algebra

I was reading (Countable) partition generated $\sigma$-algebra but I can't understand few parts.

First, we know that in order to show that a partion $$\mathcal{C}=\left\{C_{1},C_{2},...,C_{n}\right\}$$ of a set E, generates a sigma algebra that contains all the unions of elements of C, we have to do the following steps:

1. We define the sigma algebra $$\sigma(\mathcal{C})$$

2. We also define $$\mathcal{E}$$ that we want to show that is generated from $$\mathcal{C}$$ as $$\mathcal{E}=\left\{\bigcup_{i\in I}C_{i}:I\subset \mathbb{N}\right\}$$ and we show that is sigma algebra

3. And the last part is to show that $$\sigma(C)=\mathcal{E}$$

Back to the proof, at first step, they show that

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 $$\mathcal{C}$$.

But we shouldn't do that $$E\in\mathcal{E}$$ and not $$E\in \mathcal{C}$$? As we want to show that $$\mathcal{E}$$ is a sigma algebra.

And on the third part of the proof, they 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 $$A^c\in \mathcal{C}$$.

But how do we conclude that $$\mathcal{E}$$ is closed under complements? Here we showed that C is closed under complements. And also what is this $$\mathcal{C}_{A}$$? Is it a new partition that contains A?

Any advice would be helpful.

• I think there are a lot of notational issues here, you have a $C$ disappearing, a $\mathcal{C}$ appearing, etc. Perhaps you should read again and correct these notations to understand it better Commented Nov 8, 2017 at 7:01
• You are correct, I fixed it but I'm still confused Commented Nov 8, 2017 at 8:16
• I am confused trying to read your question. You said: "But we shouldn't do that $E\in\mathcal{E}$ and not $E\in \mathcal{E}$??" What's the difference between $E\in\mathcal E$ and $E\in\mathcal E$????
– bof
Commented Nov 8, 2017 at 9:37

## 1 Answer

The high level structure of the proof is as follows.

1. Show $\mathcal E$ is a sigma algebra.
2. Show $\mathcal C\subseteq\mathcal E$.
3. Therefore, $\sigma(\mathcal C)\subseteq\mathcal E$.
4. Show that $\mathcal E\subseteq\sigma(\mathcal C)$.
5. Therefore $\sigma(\mathcal C)=\mathcal E$.

The proof in the question you linked to is purely trying to prove point (1). However, there are many typos in the question, which are understandably causing confusion.

(1) $E\in\mathcal C$ is false. The author is in fact showing $E\in\mathcal E$, by expressing $E$ as a countable union of elements of $\mathcal C$.

(2) There are no typos in part two, but the author is implicitly using the fact that a countable union of countable families of sets is again a countable union of those sets.

(3) We want to show $\mathcal E$ is closed under complements, so we should take $A\in\mathcal E$ and show $A^c\in\mathcal E$. This is what the author meant, $A\in\mathcal C$ is a typo. If $A\in\mathcal E$, then by definition $A$ is a union of some family of sets in $\mathcal C$, the author calls that family $\mathcal C_A$. We then show that $A^c$ is in fact the union of $(C_A)^c$, and thus is an element of $\mathcal E$.