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I wanted to check my understanding of this concept. The sigma algebra $M(\psi)$ generated by a set $\psi$ is the intersection of all sigma-algebras that contain $\psi$.

As an example if $X=\{1,2,\dots,6\}, \psi=\{\{2,4\},\{6\}\}$ then $M(\psi)$ would be the sigma algebra that is the intersection of all sigma algebras that contain $\psi$. For a sigma algebra to contain $\psi$ it must (if I've calculated correctly) contain $\{\psi, \emptyset, \{2,4,6\},\}$ and then also the complements of these and the complements of $\{2,4\}$ and $\{6\}$.

Other than direct computation is there a faster way to calculate this?

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    $\begingroup$ The $\sigma$-algebra's involved contain $\psi$ as a subcollection. Not as an element. $\endgroup$
    – drhab
    Commented Aug 22, 2019 at 14:11

3 Answers 3

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Here is a method to compute the $\sigma$-algebra generated by two non-empty disjoint sets say $S_1$ and $S_2$ on a set $S$.

  1. Let $S_3:=S\setminus (S_1\cup S_2)$. Then $\{S_1,S_2,S_3\}$ is a partition of $S$.
  2. Let $E$ a subset of $S$. Define $E^0=\emptyset$ and $E^1=E$. Let $$A_{i_1,i_2,i_3}=S_1^{i_1}\cup S_2^{i_2}\cup S_3^{i_3}, i_1,i_2,i_3\in \{0,1\}. $$ Then $\sigma(S_1,S_2)=\{A_{i_1,i_2,i_3},i_1,i_2,i_3\in \{0,1\}\}$.

This extends readily to the $\sigma$-algebra generated by a finite collection of non-empty pairwise disjoint subsets $S_1,\dots,S_n$: let $S_{n+1}:= S\setminus \bigcup_{i=1}^nS_i$: then $$ \sigma(S_i,1\leqslant i\leqslant n)=\left\{ \bigcup_{q=1}^{n+1} S_q^{i_q},i_q\in \{0,1\} \right\}. $$

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  • $\begingroup$ Nitpick: a non-empty $S_i$ might exists, so I would rather call it a "pseudo partition". $\endgroup$
    – drhab
    Commented Aug 22, 2019 at 14:31
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    $\begingroup$ In your last equation, I think it should be $\bigcup_{q=1}^{n+1} S_q^{i_q}$. As in the case of $S_1$ and $S_2$, we have $S_1^{i_1} \cup S_2^{i_2} \cup S_3^{i_3}$. $\endgroup$
    – MAOC
    Commented Feb 19, 2023 at 21:15
  • $\begingroup$ @MAOC You are right. This has been fixed. $\endgroup$ Commented Feb 20, 2023 at 9:58
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Actually $\psi$ must be a subset (I think this is what you meant) since both are by definitions subsets of $2^X$. My approach would be the following:

I know that $\emptyset, X, \lbrace 2,4\rbrace $ and $\lbrace 6 \rbrace $ must be contained. You must add $\lbrace 2,4,6 \rbrace$, otherwise not every union would be contained. Do we now have a $\sigma$-algebra? No, because we need the complements. So we have to add the complement of each set to obtain $$\lbrace \emptyset, \lbrace 2,4 \rbrace, \lbrace 6\rbrace,\lbrace 2,4,6\rbrace,\lbrace 1,3,5,6 \rbrace, \lbrace 1,2,3,4,5\rbrace ,\lbrace 1,3,5\rbrace, X \rbrace.$$

This set is constructed such that it must be contained in every $\sigma$-algebra which contains $\psi$, so all you have to do now ist to verify that this is a $\sigma$-algebra. Then, it would also be the smallest.

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  • $\begingroup$ In fact I also don't know any faster way, but usually it is not necessary to construct such $\sigma$-algebras explicitly. You may think of the Borel-sigma-algebra, which is the smallest such that all open sets of a metric space are contained. That is a very common example and it is very useful to know ways of construction, but I don't know a way to explicitly characterise the contained sets. $\endgroup$
    – user592521
    Commented Aug 22, 2019 at 14:31
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At first hand - if $\psi=\{A_1,\cdots,A_n\}$ (so is finite) - go for sets of the form $E_1\cap\cdots\cap E_n$ where $E_i\in \{A_i,A_i^{\complement}\}$ for $i\in\{1,\dots,n\}$.

In your case that leads to the sets:

  • $\{2,4\}\cap\{6\}=\varnothing$
  • $\{2,4\}\cap\{6\}^{\complement}=\{2,4\}$
  • $\{2,4\}^{\complement}\cap\{6\}=\{6\}$
  • $\{2,4\}^{\complement}\cap\{6\}^{\complement}=\{1,3,5\}$

At second hand go for the sets that can be written as a union of these sets.

They together form the $\sigma$-algebra generated by $\psi$.

(Actually the algebra but in finite case it coincides with the $\sigma$-algebra.)

There are $2^3=8$ of such unions in your case.

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