Let $X = \mathbb{Q}\cap (0,1], \mathcal{E}= \{ (a,b]\cap\mathbb{Q}: 0 \leq a <b\leq 1\}.$
Let $\mu : \mathcal{A}(\mathcal{E}) \rightarrow [0, \infty]$ be given by
$$ \mu(A) = \begin{cases} \infty, & \text{if $|A| = \aleph_0$ } \\ 0, & \text{if $A = \emptyset$ } \end{cases}$$ a) Prove that $\mu$ is well defined on $ \mathcal{A}(\mathcal{E})$
b) $\mu$ is a $\sigma - $additive pre-measure
c) $\sigma (\mathcal{E}) = \mathcal{P}(X)$
I need help with c) : How can I prove that $ \mathcal{P}(X) \subset \sigma (\mathcal{E})$?
$\sigma (\mathcal{E}) = \{ S \subseteq \mathcal{P}(X) : S \quad \text{is a} \ \sigma \text{- algebra and} \quad \mathcal{E} \subseteq S\},$ so $ \sigma (\mathcal{E}) \subseteq \mathcal{P}(X) .$
Let $x \in \mathcal{P}(X) .$ Then $ x \subseteq X.$ Maybe I should see that $x=S,$ with $S$ being a $\sigma-$ algebra which contains $\mathcal{E},$ but this is not clear to me.
First we have that $\{q\}\in\sigma(\mathcal{E})$ for all $q\in X$. This is because for a strictly increasing sequence $a_n$ in $X$ which converges to $q$, we have $(a_n, q]\cap\mathbb{Q} \in \sigma(\mathcal{E}) \> \forall n\in\mathbb{N}$ and $\bigcap_{n}(a_n, q]\cap\mathbb{Q}=\{q\}$. Thus clearly all finite subset of $\mathcal{P}(X)$ is in $\sigma(\mathcal{E})$. For the case that it is not a finite set, since $X$ is countable, any element of $\mathcal{P}(X)$ is a countable set. Thus it can be written as $\{q_n:n\in\mathbb{N}\}=\bigcup_n\{q_n\}$ for some sequence $q_n$ in $X$, and with $\{q_n\}\in\sigma(\mathcal{E}) \> \forall n \in \mathbb{N}$ we conclude that $\mathcal{P}(X) \subset \sigma(\mathcal{E})$.