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.


1 Answer 1


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.

  • $\begingroup$ 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, $\mathfrak{S}(\mathcal{B})\subset\mathfrak{S}(\mathcal{A})$. $\endgroup$
    – antonio
    Dec 23, 2021 at 20:03
  • $\begingroup$ 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 $\{A\}$. $\endgroup$
    – antonio
    Dec 27, 2021 at 20:11
  • $\begingroup$ $\{\emptyset, A, A^c, X\}$ 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 $\min\{f(x):x\in B\}\le \min \{f(x):x\in A\}$). $\endgroup$
    – user140541
    Dec 27, 2021 at 20:25
  • $\begingroup$ BTW, your first comment is correct. $\endgroup$
    – user140541
    Dec 27, 2021 at 20:33
  • 1
    $\begingroup$ 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. $\endgroup$
    – antonio
    Dec 27, 2021 at 21:07

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