Abstract enunciation of the Good Set Principle in measure theory I am struggling with the Good Set Principle in Measure Theory, so is this rephrasing in the most abstract way ultimately correct?

Good Set Principle 
Let $(X, \Sigma)$ be a measurable space. We want to prove that 
$$(\ast) \hspace{1cm} \forall B \ ( B \in \Sigma \Longrightarrow P(B) ) ,$$
where $P(B)$ means that $B$ has the property $P$.
Let $\mathcal{G}:= \{ X \mid X \in \Sigma, P(X) \ \}$. Hence, $(\ast)$ is equivalent to
$$(\star) \hspace{1cm} \Sigma \subseteq \mathcal{G}. $$ 
Assume additionally that $\Sigma = \sigma (\mathcal{A})$, where $\mathcal{A} \subseteq \wp (X)$. Thus, $(\star)$ is equivalent to
$$ \mathcal{A} \subseteq \mathcal{G},$$
which ultimately is what we have to prove, along with the fact that $\mathcal{G}$ is a $\sigma$-algebra on $X$.

As always, thank you for you time.

Edit for reference (after the answer)
Just to clarify and make the overall question more perspicuous as a potential reference. As it is written above, the Principle is wrong. To make it correct simply:

Assume $\mathcal{G}$ is a $\sigma$-algebra, and simply prove that $\mathcal{A} \subseteq \mathcal{G}$.

 A: Sorry, your enunciation of the "Good Set Principle" in measure theory is not correct.
Let us look in details:

*

*Let $P(x)$ means that "$x$ has property $P$".
Let $\mathcal{G}:= \{ X \ | \ X \in \Sigma, P(X) \ \}$. Then it is true that
$$(\ast) \hspace{1cm} \forall B \ ( B \in \Sigma \Longrightarrow P(B) ) $$
is equivalent to
$$(\star) \hspace{1cm} \Sigma \subseteq \mathcal{G} $$
The proof is immediate.

So in this first step, your statement is correct.


*However, since there is no restiction on the property $P$, we may have $\mathcal{A} \subseteq \wp (X)$, such that  $ \mathcal{A} \subseteq \mathcal{G}$ and $\sigma (\mathcal{A})\nsubseteq \mathcal{G}$. In fact, $\mathcal{G}$ don't need to be a $\sigma$-algebra.

So the principle, as you have enounced it, does not work.
Example: Let $X$ be $\mathbb{N}$, $\Sigma$ be $\wp (\mathbb{N})$. Take $P$ to be the property of having finite cardinality. Let $\mathcal{A}$ be the set of finite subsets of $\mathbb{N}$. Then we have:
$(\ast)$ and $(\star)$ are both false (so equivalence is OK);
$ \mathcal{A} \subseteq \mathcal{G}$ but $\sigma (\mathcal{A})\nsubseteq \mathcal{G}$;
$\mathcal{G}$ is NOT a $\sigma$-algebra.
Remark: One way to correctly enunciate the Good Set Principle in measure theory is:

Let $(X,\Sigma)$ be a measurable space.  Suppose we want to prove that, for all $A\in \Sigma$, $A$ has property $P$ (which we will write $P(A)$). Then we can prove it by proving the three conditions:

*

*$\mathcal{G}:= \{ X \ | \ X \in \Sigma, P(X) \ \}$ is a $\sigma$-algebra. (Note that it is actually a condition on $P$);

*There is $\mathcal{A} \subseteq \wp (X)$, such that $\Sigma=\sigma(\mathcal{A})$;

*$\mathcal{A}\subseteq \mathcal{G}$.

(if the three conditions hold, then $\Sigma \subseteq \mathcal{G}$)

