Inverse image of $\sigma$-algebra Is this proof correct?

Let $f$ be a function mapping $\Omega$ to $E$ with $\mathcal E$ a $\sigma$-algebra on $E$. Show that $\mathcal A=\{f^{-1}(B):B\in \mathcal E\}$ is a $\sigma$-algebra on $\Omega$.

Pf. 


*

*Let $A\in\mathcal A$; then $A=f^{-1}(B)$ for some $B\in\mathcal E$. $A^c=(f^{-1}(B))^c=f^{-1}(B^c)$ which is $\in \mathcal A$ because $B^c\in \mathcal E$.

*Let $(A_i)_{i\in I}$ be a countable family of elements of $\mathcal A$ and consider the union $\bigcup A_i$. $A_i=f^{-1}(B_i)$ for all $i\in I$. So: $$\bigcup A_i=\bigcup f^{-1}(B_i)=f^{-1}(\bigcup B_i)$$
which is $\in \mathcal A$ because $\bigcup B_i\in \mathcal E$.

*It remains to show that $\Omega \in \mathcal A$.Since $E\in \mathcal E$, we have: $$\Omega=f^{-1}(E)\in\mathcal A$$


Yes, it's mostly correct. Note that in 2 you want to consider only countable families. 
