Restricting the generating set of a $\sigma$-algebra Given a collection of sets $I$, define
$$ I|_A := \{B \in I | B \subseteq A\}.$$
It is then the case that $\sigma(J)|_A$ is a $\sigma$-algebra for any collection $J$ containing $A$.
My question is: does $\sigma(J|_A) = \sigma(J)|_A$ when $A \subseteq J$? That the left-hand side is a subset of the right is straightforward, but I am having trouble showing the reverse inclusion.

We suppose that if $F$ is a class of subsets of the base space $\Omega$, then under $\sigma(F)$ we understand a minimal $\sigma$-algebra of subsets of the space $\Omega$ generated by the class $F$.     
Then,  the  equality $\sigma(J|_A) = \sigma(J)|_A$, in general,  is not valid.
Indeed, let $A=\{1\},$ $J=\{ \{1,2\},\{1,2,3\}\}$. $\Omega=\{1,2,3\}$. Then
$J|_A=\emptyset$. Hence $\sigma(J|_A)=\{\emptyset, \Omega\}$.  
$\sigma(J)=\{ \{1,2\},\{3\},\emptyset, \Omega\}$.
Hence $\sigma(J)|_A=\{ \emptyset \}$. 
