I have recently begun to learn about $\sigma$-algebra. Here, the set $\mathcal{F}\in P(\Omega)$ (where $P(\Omega)$ denotes the power set of $\Omega$) is a $\sigma$-algebra if $\mathcal{F}$ has the following properties: \begin{align} (1):& \ \ \emptyset\in\mathcal{F} \\ (2):& \ \ A\in\mathcal{F}\implies A^c\in\mathcal{F} \\ (3):& \ \ A_1,A_2,...\in\mathcal{F}\implies \bigcup_{i=1}^\infty A_i\in\mathcal{F} \end{align} I am struggling to understand the idea of the smallest $\sigma$-algebra. I have read that given a family $U$ of subsets of $\Omega$, the smallest $\sigma$-algebra is the intersection of all $\sigma$-algebras containing $U$. To help my understanding, I considered an example.

Let $\Omega=[0,1]$ and $U=\{[0,0.2],[0.8,1]\}$. Then the smallest $\sigma$-algebra, denoted by $\sigma_U$, is $$\sigma_U=\{\emptyset\ ,[0,0.2],[0.8,1]\}.$$ Is this correct? I am quite lost.

  • $\begingroup$ @KaviRamaMurthy Thanks, but what is the intuition here? I'm having difficulty understanding the definition of the smallest $\sigma$-algebra. Also, does $A$ \ $B$ denote elements that are in $A$ but not in $B$? $\endgroup$
    – M B
    Commented Feb 18, 2021 at 8:39
  • $\begingroup$ @littleO You mean $S$ is a family of subsets of $\Omega$, or $S$ is a subset of $\mathcal{P}(\Omega)$. $\endgroup$ Commented Feb 18, 2021 at 8:58
  • $\begingroup$ @littleO Can my example be visualised by using a Venn diagram? This will help my understanding. $\endgroup$
    – M B
    Commented Feb 18, 2021 at 9:03
  • $\begingroup$ @MB The intuition is similar to how you define any type of closure operations, e.g., closure in topology, group generated by a set of generators, transitive closure of relations, or algebraic closure of fields. You can either (1) concretely: take everything you can possibly get to by the operations, possibly transfinitely until it stabilises; or (2) abstractly: take the smallest set containing your starting point that is closed under the operations. $\endgroup$ Commented Feb 18, 2021 at 9:11
  • $\begingroup$ Oh, you're right, I think I should have said that the correct phrase is "the smallest $\sigma$-algebra containing each element of $U$", where $U$ is a collection of subsets of $\Omega$. For example, $U$ could be the collection of open subsets of $\mathbb R$. The smallest $\sigma$-algebra containing every open subset of $\mathbb R$ is (by definition) the Borel $\sigma$-algebra. $\endgroup$
    – littleO
    Commented Feb 18, 2021 at 9:12

1 Answer 1


I'll try to explain this from probability theory perspective.

  • Maybe someone will correct me here

In probability of discrete sample spaces we denote by $\Omega$ the sample space. For example by speaking of one dice rolling the sample space can be the possible results: $\Omega = {1,2,3,4,5,6}$ and then we can simply take $2^\Omega = \{ \emptyset,\{1\},\{2\},..,\{1,2\},\{1,3\},...,\{1,2,3,4,5,6\} \}$ to be every possible event.

But, when there is a need to speak of not countable sample spaces (which is the need for example when you investigate probability of events that occur on the interval $(0,1)$) we cant take $2^\Omega$ (There are going to be sets like Vitally sets that will cause problems)

Hence, we need a subset of $2^\Omega$ that maintains the important properties: empty event, complement of an event and union of events.

Now, the concept of minimal $\sigma$-algebra is important because intersection of $\sigma$-algebras is also $\sigma$-algebra and we can prevent ourselves from including problematic sets as Vitally sets and take for example the minimal sigma algebra that includes all the open subsets of $(0,1)$ and that is good enough for many applications.

(Of course we still have to prove that each set of this sigma algebra can have probability, but this can be done by Lebesgue measure).

  • 2
    $\begingroup$ Tips: To put things in italics, surround it with single asterisk *italic text* italic text. To put things in bold, surround it with double asterisks. Don't use $...$ for italicising text or otherwise you end up with this "sigma minus a l g e b r a". $\endgroup$ Commented Feb 18, 2021 at 9:16
  • $\begingroup$ Tried to make it better now, thanks $\endgroup$
    – e.ad
    Commented Feb 18, 2021 at 9:18

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