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Let $\mathcal{A}\subset \wp(X)$ and let $\mathfrak{S}(\mathcal{A})$ be the collection of all $\sigma$-algebras over $X$ including $\mathcal{A}$, therefore any $\sigma$-algebra in $\Sigma \in \mathfrak{S}(\mathcal{A})$ is such that $ \mathcal{A} \subset \Sigma$.
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Let $\sigma(\mathcal{A}) =\bigcap \mathfrak{S}(\mathcal{A})$, that is the smallest $\sigma$-algebra including $\mathcal{A}$. This, of course, assumes that we have proved that the intersection of $\sigma$-algebras is still a $\sigma$-algebra.
If $\mathcal{B}\subset \wp(X)$ is a superset of $\mathcal{A}$, the inclusion holds for $\sigma()$ too, that is:
$$ \mathcal{A} \subset \mathcal{B} \rightarrow
\sigma(\mathcal{A}) \subset \sigma(\mathcal{B}) $$
To most books, this is trivial and not dealt with.
I understand that most of the sets in $\sigma(\mathcal{A})$ are also in $ \sigma(\mathcal{B}) $, for example:
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Any element of $\mathcal{A}$ is both in $\sigma(\mathcal{A})$ and $ \sigma(\mathcal{B})$, because
$ \mathcal{A} \subset \mathcal{B} \subset \sigma(\mathcal{B})$. -
Countable sequences $A_1, \ldots, A_k \in \mathcal{A}$ have their intersections, unions, and complements, both in $\sigma(\mathcal{A})$ and $ \sigma(\mathcal{B})$, because
$A_1, \ldots, A_k \in \mathcal{B}$ too.
However, (1) and (2) do not exhaust all the possible sets in $\sigma(\mathcal{A})$, in fact, I can still reason about intersections, unions and complements of the sets built in (2). I can analyse specific cases. For example,
$(A_1 \cap A_2) \cap (A_3 \cap A_4) \in \sigma(\mathcal{A})$, and also
$$A_1, \ldots, A_4 \in \mathcal{B}
\rightarrow
(A_1 \cap A_2), (A_3 \cap A_4) \in \sigma(\mathcal{B})
\rightarrow
(A_1 \cap A_2) \cap (A_3 \cap A_4) \in \sigma(\mathcal{B})
$$
All elements of $\sigma(\mathcal{A})$ I can think of are in $ \sigma(\mathcal{B})$ too, but the possibilities are endless, and I cannot obtain a proof.
Note. This question is a generalisation of other questions, such as math.stackexchange.com/q/1667546/75616, which refer to specific collections $\mathcal{A},\mathcal{B}$ and is about the formulation of a formally correct proof.
Best Answer
You don't need to look at the elements of $\sigma(\mathcal{A})$. It suffices to note that any $\sigma$-algebra containing $\mathcal{B}$ contains $\mathcal{A}$. Thus, $\bigcap \mathfrak{S}(\mathcal{B})$ contains $\mathcal{A}$, but it need not be the smallest such a $\sigma$-algebra.