If $\mathcal A$ generates $\mathcal S$ then $\sigma (X )=\sigma (X ^{-1 } ( \mathcal A ))$ Show that if $\mathcal A$  generates $\mathcal S$ then $X ^{-1 } ( \mathcal A )= \{\{X \in A \}  : A \in \mathcal A  \} $ generates $\sigma (X) =\{\{X \in B \}  : B \in \mathcal S  \}$
First I interpret this as $\sigma (X )=\sigma (X ^{-1 } ( \mathcal A ))$
Edit $\mathcal S $ is a $\sigma $-algebra .

I suppose we verify this by showing inclusion both ways.
Since $\mathcal A \subset \mathcal S$, for any $A \in \mathcal A $, $X^{-1 } (A) \in \sigma (X)$ implying $\sigma (X ^{-1 } ( \mathcal A )) \subset \sigma (X)$ Correct?
How about the other inclusion?
Thanks in advance! 
 A: It comes to proving that: $$X^{-1}\left(\sigma\left(\mathcal{C}\right)\right)=\sigma\left(X^{-1}\left(\mathcal{C}\right)\right)$$
where in general $\sigma\left(\mathcal{D}\right)$ denotes the smallest
$\sigma$-algebra that contains $\mathcal{D}$.
On base of the fact that $\sigma\left(\mathcal{C}\right)$ is a $\sigma$-algebra
it is straightforward to prove that $X^{-1}\left(\sigma\left(\mathcal{C}\right)\right)$
is also a $\sigma$-algebra, and this with $X^{-1}\left(\mathcal{C}\right)\subseteq X^{-1}\left(\sigma\left(\mathcal{C}\right)\right)$. This leads to $\sigma\left(X^{-1}\left(\mathcal{C}\right)\right)\subseteq X^{-1}\left(\sigma\left(\mathcal{C}\right)\right)$
wich is the part that you allready worked on in your question.
To prove the other side it must be verified that any collection of
the form $\left\{ Z\mid X^{-1}\left(Z\right)\in\mathcal{D}\right\} $
is a $\sigma$-algebra whenever $\mathcal{D}$ is a $\sigma$-algebra. This also is straightforward and applying this result on $\mathcal{D=}\sigma\left(X^{-1}\left(\mathcal{C}\right)\right)$
we find that $\mathcal{E}:=\left\{ Z\mid X^{-1}\left(Z\right)\in\sigma\left(X^{-1}\left(\mathcal{C}\right)\right)\right\} $
is a $\sigma$-algebra. It is evident that $\mathcal{C}\subseteq\mathcal{E}$ which allows the conclusion that $\sigma\left(\mathcal{C}\right)\subseteq\mathcal{E}$
or equivalently $X^{-1}\left(\sigma\left(\mathcal{C}\right)\right)\subseteq\sigma\left(X^{-1}\left(\mathcal{C}\right)\right)$.
