Let $\Omega$ be a nonempty set and let $\mathcal{A} := \{A_{i}:i\in\mathbb{N}\}$ be a partition of $\Omega$. Let $\mathcal{F} = \left\{\bigcup_{i\in J}A_{i}:J\subset\mathbb{N}\right\}$ where, for $J = \varnothing$, $\bigcup_{i\in J}A_{i} = \varnothing$. Show that $\mathcal{F}$ is a $\sigma$-algebra.
MY ATTEMPT
To begin with, let us notice that $\Omega\in\mathcal{F}$: it suffices to take $J = \mathbb{N}$.
Moreover, we do also have that \begin{align*} A\in\mathcal{F} \Rightarrow A = \bigcup_{i\in J}A_{i} \Rightarrow A^{c} = \bigcup_{i\in\mathbb{N}\backslash J}S_{i}\in\mathcal{F},\,\, S_{i}\in\mathcal{A}. \end{align*}
Now, if $A\in\mathcal{F}$and $B\in\mathcal{F}$, we conclude that \begin{align*} A = \bigcup_{i\in J}A_{i}\,\,\text{and}\,\,B = \bigcup_{i\in J'}B_{i} \Rightarrow A\cup B = \bigcup_{i\in J\cup J'}S_{i},\,\,S_{i}\in\mathcal{A}.\ \end{align*}
Then $\mathcal{F}$ is an algebra. To prove the countable union property, we shall proceed as follows \begin{align*} A_{n}\in\mathcal{F} \Rightarrow A_{n} = \bigcup_{i\in J_{n}}A_{i} \Rightarrow \bigcup_{i\in\mathbb{N}}A_{n} = \bigcup_{i\in J}A_{i}\in\mathcal{F},\,\,\text{where}\,\,J = \bigcup_{j\in\mathbb{N}}J_{n} \end{align*}
Is the wording of my proof correct? Can it be improved? Any contribution is appreciated.
It’s fine apart from a fairly serious notational error in the last line. When you write $A_{ni}$, you’re talking about the member of $\mathcal{A}$ whose index is the product $ni$. The problem began because at the beginning of the line when you used $A_n$, which is already the name of a particular member of $\mathcal{A}$, as the name of an arbitrary member of $\mathcal{F}$. What you really meant is that if $F_n\in\mathcal{F}$ for each $n\in\Bbb N$, then for each $n\in\Bbb N$ there is a $J_n\subseteq\Bbb N$ such that $F_n=\bigcup_{i\in J_n}A_i$, and therefore
$$\bigcup_{n\in\Bbb N}F_n=\bigcup_{n\in\Bbb N}\bigcup_{i\in J_n}A_i=\bigcup_{i\in\bigcup_{n\in\Bbb N}}A_i\in\mathcal{F}\;.$$