[Math] Partition generates sigma algebra

elementary-set-theorymeasure-theory

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.

Best 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 $\mathcal E$.

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