How can I show that the following sigma algebras are equal? 
Let $\Omega$ be a set and $(B_n;~n\in \Bbb{N})$ it's partition. We define $G=\sigma(B_n)=\{\bigsqcup_{n\in L} B_n: L\subset \Bbb{N}\}$. Consider a random variable $V=\sum_{n\in \Bbb{N}}n1_{B_n}$ where $V:(\Omega,G)\rightarrow (E,\epsilon)$. I need to show that $\sigma(V)=G$.

I know that by definition $\sigma(V)=\{\Lambda \subset \Omega: V^{-1}(B)=\Lambda~~\text{for some}~~B\in \epsilon\}$. Now I thought that I can take $\Lambda\in \sigma(V)$ which means that $V^{-1}(B)=\Lambda$. But then clearly $\Lambda=\bigsqcup_{n\in L} B_n$ for some $L\subset \Bbb{N}$. Now let me take $C\in G$ which means that $G=\bigsqcup_{n\in L} B_n$. But then I somehow do not see how to continue to show that $C\in \sigma(V)$. I think in this direction we need to use the definition of $V$ but I somehow do not see how I should do it. Maybe my idea is also totally wrong.
Can maybe someone help me further?
 A: $(\subseteq)$. Note $V$ is real valued. Also
$$\{\omega\in \Omega:V(\omega)\leq x\}=\bigcup_{k\leq \lfloor x\rfloor }B_k\in G,\,\quad \forall x\geq 1$$
$$\{\omega\in \Omega:V(\omega)\leq x\}=\emptyset \in G,\,\quad \forall x <1$$
By using the generator of the Borel sets $\{(-\infty,x],x \in \mathbb{R}\}$ we get
$$\sigma(V)=\sigma(\{\{\omega\in \Omega:V(\omega)\leq x\},x \in \mathbb{R}\})\subseteq G$$
$(\supseteq)$. Note that
$$C \in G\implies C =\bigcup_{k \in I}B_k,\,I \subseteq \mathbb{N}$$
for some $I$, and that
$$V^{-1}(\{k\})=\{\omega \in \Omega:V(\omega)=k\}=B_k\implies C=\bigcup_{k \in I}V^{-1}(\{k\})\in \sigma(V)$$
But this means that $C \in \sigma(V),\forall C \in G$. So we conclude $G=\sigma(V)$.
A: The statement is true if a random variable is a measurable function $V:\Omega\rightarrow E$ to $E\subseteq\mathbb R$ equipped with the canonical $\sigma$-algebra (the Borel algebra induced by one of the equivalent norms).
In this case (or any case where $\{n\}\in\epsilon$ for all $n\in\mathbb N$) and for any $B\in\epsilon$ we have $V^{-1}(B)=V^{-1}(B\cap\mathbb N)=\bigsqcup_{n\in B\cap\mathbb N}B_n$ by the very definition of $V$, so $\sigma(V)\subseteq G$.
Clearly, for any $B\subseteq\mathbb N$ and $\bigsqcup_{n\in B}B_n\in G$ we have $B\in\epsilon$ and $V^{-1}(B)=\bigsqcup_{n\in B}B_n$, so $G\subseteq\sigma(V)$.
For the general case, recall that $\epsilon_\circ=\{\emptyset,E\}$ is a $\sigma$-algebra. For $\epsilon=\epsilon_\circ$ we have $\sigma(V)=\{\emptyset,\Omega\}\subseteq G$, so $V$ is a random variable. But usually we don't have $\sigma(V)=G$ (say for $\Omega=\mathbb N$ with $B_n=\{n\}$).
