# $\sigma$-algebra generated by a subcollection

1. 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 $$\mathcal{A} \subset \Sigma$$.

2. 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:

1. Any element of $$\mathcal{A}$$ is both in $$\sigma(\mathcal{A})$$ and $$\sigma(\mathcal{B})$$, because $$\mathcal{A} \subset \mathcal{B} \subset \sigma(\mathcal{B})$$.

2. 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.

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.
• Further, while counterintuitive, $\mathfrak{S}(\mathcal{A})$ is larger than $\mathfrak{S}(\mathcal{B})$. In fact, every $\sigma$-algebra $\Sigma_\mathcal{B}$ in $\mathfrak{S}(\mathcal{B})$ includes $\mathcal{B}$, which includes $\mathcal{A}$ (i.e. $\mathcal{A}\subset\mathcal{B}\subset\Sigma_\mathcal{B}$), so $\Sigma_\mathcal{B} \in \mathfrak{S}(\mathcal{A})$ too. However, for some $\Sigma_\mathcal{A}$ in $\mathfrak{S}(\mathcal{A})$, $\mathcal{A}\subset\Sigma_\mathcal{A}$, but $\mathcal{B}\not\subset\Sigma_\mathcal{A}$. Hence, $\mathfrak{S}(\mathcal{B})\subset\mathfrak{S}(\mathcal{A})$. Dec 23, 2021 at 20:03
• Sorry, is this true also using the collection of the SAs including the family $\mathcal{F}$, that is $\mathfrak{S}(\mathcal{F})$, not $\sigma({\mathcal{F}})$? Given the families $\mathcal{A}=\{A\}$ and $\mathcal{B}=\{A,B\}$, it seems the SA $\{\emptyset,X,A,A^c\}$ belongs to both $\mathfrak{S}(\mathcal{A})$ and $\mathfrak{S}(\mathcal{B})$, while the SA $\{\emptyset,X,A,A^c,B,B^c,A\cup B,A\cap B,etc.\}$ belongs only to $\mathfrak{S}(\mathcal{B})$, so the SA collection $\mathfrak{S}(\mathcal{B})$ is smaller than $\mathfrak{S}(\mathcal{A})$, while not the smallest SA including $\{A\}$. Dec 27, 2021 at 20:11
• $\{\emptyset, A, A^c, X\}$ does not belong to $\mathfrak{S}(\mathcal{B})$ because it does not contain $\mathcal{B}$. However, $\mathfrak{S}(\mathcal{B})$ is indeed smaller than $\mathfrak{S}(\mathcal{A})$ because every member of $\mathfrak{S}(\mathcal{B})$ belongs to $\mathfrak{S}(\mathcal{A})$. Thus, the intersection of $\mathfrak{S}(\mathcal{A})$ is smaller. (It is like finding the minimum of a function over restricted domains; if $A\subset B$, then $\min\{f(x):x\in B\}\le \min \{f(x):x\in A\}$).
• My bad, when typing, I swapped the SAs. With respect to your example, I meant the SA $\{\emptyset,X,A,A^c,B,B^c,A\cup B,A\cap B,etc.\}$, based on $\{A,B\}$, belongs to both $\mathfrak{S}(\mathcal{A})$ and $\mathfrak{S}(\mathcal{B})$, while the SA $\{\emptyset,X,A,A^c\}$ belongs only to $\mathfrak{S}(\mathcal{A})$, therefore $\mathfrak{S}(\mathcal{B})$ is smaller. Dec 27, 2021 at 21:07