I have a proposition in the textbook: Let $\Pi$ be a partition of a set $\Omega$. Then $\sigma(\Pi)$ consists of all countable unions of sets in $\Pi$ (including the empty set, which is the union of an empty set of sets) and all complements of such countable unions.
The task is to prove this proposition. The proof provided by the answer key is as follows:
Let $\mathcal{G}$ denote the set of all countable unions of sets in $\Pi$ and all complements of such countable unions. We need to show that $\sigma(\Pi) = \mathcal{G}$. It suffices to show that $\mathcal{G}$ is a $\sigma$-algebra.
If $A \in \mathcal{G}$, then clearly $A^c \in \mathcal{G}$. The empty set is in $\mathcal{G}$, because it is the union of an empty set of sets in $\Pi$. It follows that $\Omega$ also is in $\mathcal{G}$.
Suppose $A_n$ is a sequence of sets in $\mathcal{G}$, and let us show that their union is in $\mathcal{G}$. If all of the sets $A_n$ are countable unions of sets in $\Pi$, then their union is a countable union of sets in $\Pi$ and so it is in $\mathcal{G}$. Alternatively, if one of the sets $A_n$ is not a countable union of sets in $\Pi$, then its complement $A_n^c$ is a countable union of sets in $\Pi$, $A_n^c = \cup_m B_m$, where $B_m \in \Pi$ for $m \in \mathbb{N}$, and therefore $$A_n = (\cup_m B_m)^c = \cap_m B_m^c \in \mathcal{G}$$
This shows that $\mathcal{G}$ is a $\sigma$-algebra. QED.
I have 2 questions regarding this answer key:
- Why does it suffice to show that $\mathcal{G}$ is a $\sigma$-algebra? I believe showing $\mathcal{G}$ is a $\sigma$-algebra merely shows that $\sigma(\Pi) \subset \mathcal{G}$, while we still need further steps to show $\mathcal{G} \subset \sigma(\Pi)$, before we can claim their equality.
- In order to prove $\sigma$-algebra, we need to show that $\cup A_n \in \mathcal{G}$. How does the condition that $A_n \in \mathcal{G}$ help us with the proof? Why do we even need to prove $A_n \in \mathcal{G}$, as it is part of our assumption for the sequence of sets?
I am very new to measure theory. Can anyone tell me what the answer key is intending to show? Or if the answer key is indeed problematic, can anyone give me a correct and complete proof for the proposition?
Thank you so much!