Durrett Theorem 1.3.1 I'm having difficulties with understanding the last point of the proof for this thoerem.

Theorem 1.3.1. If $ \{ \omega : X(\omega) \in A \} \in \mathcal{F} \quad \forall A \in \mathcal{A} $ and $ \sigma(\mathcal{A}) = \mathcal{S} $, then $X$ is measurable.

The full proof can be found here in the book. X is a random variable $\omega \rightarrow S $ and our sets and sigma algebras as $ (\Omega, \mathcal{F} ) \rightarrow (S, \mathcal{S}) $, B is defined as
$$\mathcal{B} = \{ B : \{ \omega : X(\omega) \in B \} \in \mathcal{F} \}$$
How I see, the entire idea behind the proof is the following:


*

*define a collection $ \mathcal{B} $ which is the set of all sets where the measurable map exists

*prove that the set of all sets is contained in $ \mathcal{S} $,  so X is measurable

*this should be done by proving that $\mathcal{A}$ is contained in $\mathcal{B}$ and $\sigma(\mathcal{A}) \subset \mathcal{B}$
The second part of (3) follows I believe, because $\mathcal{B}$ is a $\sigma$-field, and so IF a collection $ \mathcal{A} $ is contained, then at most it will be the $ \mathcal{B} $, but I don't see here why there cannot be equality.
But more importantly, I'm having problem understanding the first part of (3), why $ \mathcal{A} \subset \mathcal{B} $?  I assume it is true, because it is a collection of sets where the measurable map exists, but not necessarily all the sets, which is $ \mathcal{B}$. But I guess equality could still hold there? 
Another view of my problem
We wish to establish that $\mathcal{A} \subset \mathcal{B}$, the latter is defined as
$$\mathcal{B} = \{ B : \{ \omega : X(\omega) \in B \} \in \mathcal{F} \}$$
The proposition of the proof could be rewritten in similar form,
$$\mathcal{A} = \{ A \in \mathcal{A}: \{ \omega : X(\omega) \in A \} \in \mathcal{F} \}$$
which proves in my opinion indeed that $\mathcal{A} \subseteq \mathcal{B}$ because I could have choosen $\mathcal{B}$ as my $\mathcal{A}$, but the answer is still $ \mathcal{A} \subset \mathcal{B} $.
 A: The question is not really clear. I guess you define $$\mathcal B=\{S\in \sigma (\mathcal A)\mid X^{-1}(S)\in \mathcal F\},$$
right ? 


*

*The fact that $\mathcal B\subset \sigma (\mathcal A)$ is by definition.

*By hypothesis, $X^{-1}(A)\in \mathcal F$ for all $A\in \mathcal A$. Therefore, $\mathcal A\subset \mathcal B$.

*Moreover, $\mathcal B$ is a $\sigma -$algebra (prove it !). Since $\mathcal A\subset \mathcal B$, then $\sigma (\mathcal A)\subset \mathcal B$. This because $\sigma (\mathcal A)$ is the smallest $\sigma -$algebra that contain $\mathcal A$, and since $\mathcal B$ is also a $\sigma -$algebra that contain $\mathcal A$, we must have $\sigma (\mathcal A)\subset \mathcal B$.

*At the end, you indeed get $\mathcal B=\sigma (\mathcal A)$. Nevertheless, notice that the only interesting inequality is $\sigma (\mathcal A)\subset \mathcal B$. This because what we want to prove is :

If $X^{-1}(A)$ is measurable for all $A\in \mathcal A$, then
  $X^{-1}(A)$ is measurable for all $A\in \sigma (\mathcal A)$,

which is a very powerful result. 
