# sigma algebra generated by a set example

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?

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

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 $$\sigma(S_i,1\leqslant i\leqslant n)=\left\{ \bigcup_{q=1}^{n+1} S_q^{i_q},i_q\in \{0,1\} \right\}.$$

• Nitpick: a non-empty $S_i$ might exists, so I would rather call it a "pseudo partition". Commented Aug 22, 2019 at 14:31
• 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 $S_1^{i_1} \cup S_2^{i_2} \cup S_3^{i_3}$.
– MAOC
Commented Feb 19, 2023 at 21:15
• @MAOC You are right. This has been fixed. Commented Feb 20, 2023 at 9:58

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.

• 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.
– user592521
Commented Aug 22, 2019 at 14:31

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 $$i\in\{1,\dots,n\}$$.

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