There exists a minimal field, $ \sigma\text{-field}$, or monotone class generated by (or, containing) any specified class $\mathcal{C}$ of subsets of $\Omega$. We call $\mathcal{C}$ the generators. For example, $\sigma[\mathcal{C}]\equiv \bigcap\{ \mathcal{F}_\alpha:\mathcal{F}_\alpha$ is a $ \sigma\text{-field}$ of subsets of $\Omega$ for which $\mathcal{C \subset \mathcal{F}_\alpha}\}$ is the $\textit{minimal } \sigma\text{-field}$ generated by $\mathcal{C}$.

My attempt:

First, show that there exists at least one field, $ \sigma\text{-field}$, monotone class generated by $\mathcal{C}$. We know that $2^\Omega$ is the largest possible class as it is defined as all possible subset of $\Omega$ and it must also contain $\mathcal{C}$, hence we have proved that such field or $ \sigma\text{-field}$ or monotone class at least exists.

From the definition we know that, the minimal field (denote $g[\mathcal{C}]$) must satisfy the conditions of a field as well as containing the the specified call $\mathcal{C}$. Hence 1) $\Phi,\Omega \in g[\mathcal{C}]$; 2) $\forall A\in \mathcal{C}$, $A^c\in$ $\sigma[\mathcal{C}]$; 3) $\forall A,B\in\mathcal{C},A \cup B \in \sigma[\mathcal{C}]$. Taking intersection of all classes which satisfy these three conditions we will obtain the minimal field of subset of $\Omega$ which contain $\mathcal{C}$.

Following the same procedure, we can also find the $\sigma[\mathcal{C}]$ by intersecting all fields satisfied the conditions which contain $\mathcal{C}$. The $\sigma$-field that we select should satisfy :1) $\Phi,\Omega \in g[\mathcal{C}]$; 2) $\forall A\in \mathcal{C}$, $A^c\in$ $\sigma[\mathcal{C}]$; 3) $\forall$ monotone sequence $A_1, A_2,..\in \mathcal{C}$; $\bigcup^\infty A_n \in \sigma[\mathcal{C}]$ if the sequence is increasing, $\bigcap^\infty A_n \in \sigma[\mathcal{C}]$ if the sequence is decreasing.

To find $m[\mathcal{C}]$ follows the same logic; here we consider all monotone classes: $\forall$ monotone sequence $A_1, A_2,...\in \mathcal{C}$, $\bigcup^\infty A_n \in \sigma[\mathcal{C}]$ if the sequence is increasing, $\bigcap^\infty A_n \in \sigma[\mathcal{C}]$ if the sequence is decreasing and take the intersection.

My Question:

  1. Is this proof valid?
  2. I first try to show that there is at least one such field/$\sigma$-field/monotone class that can be generated from the $\mathcal{C}$ which is the subset of the collection of subset of $\Omega$. Then I tried to show that the minimal must exists and unique but for the second part, I don't know how to proceed. Actually, I am very sceptical about my step by listing all of them and take the intersection, as what if there are infinitely many such field/$\sigma$-field/monotone class? How can I prove that their limit is also a field/$\sigma$-field/monotone class? Once the "minimal must always exist" established the uniqueness seem to easy to establish as if it is not unique it will contradict the "minimal" property presupposed.
  • $\begingroup$ "Their limit" is not a precise description. There is no "limit" concept involved in this problem. $\endgroup$
    – Zhanxiong
    Commented Sep 4, 2020 at 16:30

1 Answer 1


Take $\sigma$-field for example. Your construction is correct, and you indicated that you tried to show $\sigma(\mathcal{C})$ is a $\sigma$-field. Regarding your concern about intersecting infinitely (possibly uncountably) many $\sigma$-fields, it is not as different as the scenario below: let $x_0$ be a real number, and consider the intersection of all the intervals that contains $x_0$. If you are comfortable with this scenario, the procedure of intersecting all the qualified $\sigma$-fields that contains $\mathcal{C}$ should not bother you much. Indeed, the existence is of our greatest interest, while how to implement them is not that important (as in many proofs in mathematics).

Back to the proof: What is missing is the step of showing the minimality of $\sigma(\mathcal{C})$, that is, for every $\sigma$-field that contains $\mathcal{C}$, it also contains $\sigma(\mathcal{C})$. This is trivial by construction of $\sigma(\mathcal{C})$: if $\mathcal{C} \subset \mathcal{G}$ and $\mathcal{G}$ is a $\sigma$-field, then $\mathcal{G}$ is one of the $\sigma$-fields in the intersection defining $\sigma(\mathcal{C})$, so that $\sigma(\mathcal{C}) \subset \mathcal{G}$.

In summary, $\sigma(\mathcal{C})$ has (which you need to prove one-by-one) these three properties:

  1. $\mathcal{C} \subset \sigma(\mathcal{C})$;
  2. $\sigma(\mathcal{C})$ is a $\sigma$-field;
  3. if $\mathcal{C} \subset \mathcal{G}$ and $\mathcal{G}$ is a $\sigma$-field, then $\sigma(\mathcal{C}) \subset \mathcal{G}$.
  • $\begingroup$ Thank you so much! This.is exactly what I need! Just need to clarify to things: 1) What do you mean by primal? (sorry my English is not very good... not sure if this word carries any special meaning...) 2) when you made the comments that there is no limit concept involved. When I say that there may exist an infinite amount of $\sigma$-field which contains $\mathcal{C}$, then take the intersection, since there are infinitely many intersections, I think there might be concept of limit involved... are there any misconceptions? $\endgroup$
    – JoZ
    Commented Sep 5, 2020 at 2:53
  • $\begingroup$ Also, can I prove the minimal statement using contradiction? If it is not the minimal, there must exist another $\mathcal{F}$, smaller than all the proved one and $\mathcal{C}\subset\mathcal{F}$. However, we should have exhausted all the $\mathcal{F}_\alpha$s in construction, there should not exists a smaller $\mathcal{F}$ $\endgroup$
    – JoZ
    Commented Sep 5, 2020 at 3:31
  • $\begingroup$ @JoZ Not all infinite scenario results "limit", in its narrow sense. As you may know from calculus, there are numerous infinite sequence that do not have a limit. Here, in the set sense, a collection of countably many sets may not have a "limit", but you can still talk about their union. Note the "limit" of sets exists only if the sets are increasing. Therefore, just focus on checking all three requirements needed for sigma field without introducing the limit notion here. Maybe I am a little too demanding but it is better to be very precise when doing math. $\endgroup$
    – Zhanxiong
    Commented Sep 5, 2020 at 4:49
  • $\begingroup$ I think your argument may be fine but I personally prefer mine as it is more direct and clearer. Your contrapositive assumption seems a little unclear. Also, as a matter of taste/philosophy, whenever a direct proof is available, then it should be always considered more superior than the proof by contradiction. $\endgroup$
    – Zhanxiong
    Commented Sep 5, 2020 at 4:58
  • $\begingroup$ I think I understand your point... do you mean when I consider all the fields satisfying the requirement and take their intersections. Even though I don't know how many of them exist I am not taking any limits there but simply taking intersections of them all? $\endgroup$
    – JoZ
    Commented Sep 6, 2020 at 13:25

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