Sigma algebra from partiton

Proposition Let $\mathcal{P}=\{E_1,E_2,\dots \}$ be a countable partition of a set $E$, that is

$$\bigcup _k{E}_k=E,E_i\cap E_j=\varnothing \text{ if }i\ne j$$

Then $\sigma (\mathcal{P})=\mathcal{E}:=\{{\cup }_{i\in I}{E}_i|I\subseteq \mathbb{N}\}$, i.e. the Sigma algebra is simply all the possible unions (even contables) of the base sets.

Proof First we show that $\{{\cup }_{i\in I}{E}_i|I\subseteq \mathbb{N}\}$ is a $\sigma$-algebra. By definition is closed under countalbe unions, and contains the empty set (the union of zero sets), and it’s closed under complement, since

$$A=\bigcup _{i\in I_A}{E}_i$$

then

$$A^c=\bigcap _{i\in I_A}{E}_i^c=\bigcap _{i\in I_A}\bigcup _{j\ne i}{E}_j=\bigcup _{i\notin I_A}{E}_i$$

so indeed $\mathcal{E}$ is a $\sigma$-algebra. Clearly $\mathcal{P}\subset \mathcal{E}$, so $\sigma (\mathcal{P})\subseteq \sigma (\mathcal{E})=\mathcal{E}$ But $\mathcal{E}\subseteq \sigma (\mathcal{P})$, since the sigma algebra that contains $\mathcal{P}$ also need to contain every countable union of its elements, so that

$$\sigma (\mathcal{P})=\mathcal{E}\square$$