$\sigma$-algebra of $\mathbb{R}$ generated by $\mathcal{P}(\mathbb{N})$ What is the $\sigma$-algebra of $\mathbb{R}$ generated by $\mathcal{P}(\mathbb{N})$?
I thought it is $$\Sigma = \{\emptyset, \mathbb{N}, \mathcal{P}(\mathbb{N}), \mathbb{R}, \mathbb{R}-\mathbb{N}, \mathbb{R}-\mathcal{P}(\mathbb{N}),(\mathbb{R}-\mathcal{P}(\mathbb{N})) \cup \mathbb{N}, (\mathbb{R}-\mathbb{N}) \cup (\mathbb{R}-\mathcal{P}(\mathbb{N})) \}$$
Is that correct?
 A: So the sigma algebra generated by $\mathcal{P}(\mathbb{N})$ is the smallest $\sigma$-algebra containing $\mathcal{P}(\mathbb{N})$.  So what's going to have to be in it?  Well, we'll need $\mathbb{R}$ and $\emptyset$, and we'll certainly need all of $\mathcal{P}(\mathbb{N})$.  Notice that $\mathcal{P}(\mathbb{N})$ is already closed under countable unions, so we won't need to worry so much about that.  But it's not closed under complements.  So we'll need to include every set of the form $\mathbb{R} \setminus A$ for $A \subset \mathbb{N}$.
Anything else?  Well, we just included new sets, so we need to check if the new sets is compatible with our old ones, or if we're going to need to add more. We can of course take complements of sets of the form $\mathbb{R} \setminus A$ and just get $A$, which is in $\mathcal{P}(\mathbb{N})$ so we're fine there.  If we take unions, $\bigcup_{\alpha} (\mathbb{R} \setminus A_{\alpha}) = \mathbb{R} \setminus (\bigcup_{\alpha} A_{\alpha})$, so we're also fine there.  
So I believe the answer to your question ought to be $\sigma(\mathcal{P}(\mathbb{N})) = \{ \mathbb{R}, \emptyset \} \cup \mathcal{P}(\mathbb{N}) \cup \{\mathbb{R} \setminus A: A \subset \mathbb{N} \}$
A: This indeed can be more general. Let $X$ be any nonempty set and $Y\subset X$ be any nonempty subset. Let $\sigma_Y(Y)$ be any $\sigma$-algebra on $Y$.
Then $$\sigma_X(\sigma_Y(Y))=\sigma_Y(Y)\bigcup\{A^c|A\in\sigma_Y(Y)\}.$$ Because any $\sigma$-algebra that contains $\mathcal{P}(Y)$ must contain the right hand side, it is enough to show the right hand side is a $\sigma$-algebra. 
For any sequence $\{B_n\}$ from the right hand side, let $I_1=\{n|B_n \in\sigma_Y(Y)\}$ and $I_2=\{n|B_n^c\in\sigma_Y(Y)\}$. Then 
$$\bigcup_n B_n=\bigcup_{n\in I_1} B_n\cup \bigcup_{n\in I_2} B_n=C\cup D^c=(D\backslash C)^c$$ where $C=\bigcup_{n\in I_1} B_n$ and $D=\bigcap_{n\in I_2} B_n^c$. The claim follows by observing that $C,D, D\backslash C\in\sigma_Y(Y)$.
In your example, $X=\mathbb{R}$, $Y=\mathbb{N}$ and $\sigma_Y(Y)=\mathcal{P}(\mathbb{N})$.
