Countable partition of a probability space I am trying to prove the following statement. Even though it seems almost obvious that it must be true, I am having trouble with making my arguments precise.
Let $\{D_i: i\in \mathbb{N}\}$ be a countable partition of $\Omega$, i.e. $\cup_ {i\in \mathbb{N}} D_i =  \Omega$ and $D_i \cap D_j =\emptyset$ whenever $i\neq j$. Define $\mathcal{D}$ as the $\sigma$-algebra generated by $\{D_i\}$. Show that $G \in \mathcal{D}$ if and only if $G = \cup_{i\in \mathbb{I}}D_i$ for some $\mathbb{I} \subseteq \mathbb{N}$.
The sufficiency part immediately follows from the definition of a $\sigma$-algebra. To show the necessity as I argue as follows. $G \in \mathcal{D} \Rightarrow G^\mathsf{c} \in \mathcal{D}$. Since $G \cup G^\mathsf{c} = \Omega = \cup_ {i\in \mathbb{N}} D_i$ and $D_i \cap D_j =\emptyset$, both $G$ and $G^\mathsf{c}$ must be the unions of some elements of $\{D_i\}$ since I cannot break down $D_i$s to smaller elements within $\mathcal{D}$. The last sentence is not good. But how do I make myself precise here?
Edit: (Based on Karolis Juodele's answer)
$$D_k^\mathsf{c} = \Omega\setminus D_k = \cup_ {i\in \mathbb{N}} D_i \setminus D_k = \cup_ {i\in \mathbb{N}\setminus \{k\}} D_i$$
I do the same for the countable unions of $\{D_k\}$.
 A: From wikipedia:

A subset $Σ ⊂ 2^X$ is called a σ-algebra if it satisfies the following three properties:

*

*X is in Σ.

*Σ is closed under complementation: If A is in Σ, then so is its complement, X\A.

*Σ is closed under countable unions: If A1, A2, A3, ... are in Σ, then so is A = A1 ∪ A2 ∪ A3 ∪ … .


You need to show that the complement of every $D_i$ is a union of the rest  of them and that the complement of every union of $D_i$ is another such union. Then we have that property 2 adds no new sets and that your sigma algebra only contains unions.
$$(\bigcup_{j \in J} D_j)^c = (\bigcup_{i \in I} D_i) \setminus (\bigcup_{j \in J} D_j) = ((\bigcup_{i \in I \setminus J} D_i) \setminus (\bigcup_{j \in J} D_j)) \cup ((\bigcup_{i \in J} D_i) \setminus (\bigcup_{j \in J} D_j)) =\\= (\bigcup_{i \in I \setminus J} D_i) \setminus (\bigcup_{j \in J} D_j) = \bigcup_{i \in I \setminus J} (D_i \setminus (\bigcup_{j \in J} D_j)) = \bigcup_{i \in I \setminus J} (D_i \setminus (\bigcup_{j \in J} D_j \cap D_i)) =\\= \bigcup_{i \in I \setminus J} (D_i \setminus (\bigcup_{j \in J\cap \{i\}} D_j \cap D_i)) = \bigcup_{i \in I \setminus J} (D_i \setminus \emptyset) = \bigcup_{i \in I \setminus J} D_i$$
Where the property disjointness of $D_i$ is used in the = going from line 2 to line 3.
