1
$\begingroup$

You are given a set $\Omega$ with some subsets $A_i \subset \Omega$, for some indices $i\in I$.

We define a generated $\sigma$-algebra $\sigma(A_i)$ to be the smallest $\sigma$-algebra that contains $A_i$.

Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?

My intuition says yes, and it seems clear in the case where $A_i=\{a_i\},\,a_i\in\Omega$, i.e. singleton sets. Please provide a proof or a counterexample for the general case.

ADDENDUM: As was pointed out in comments and questions, I had mixed up the subsets vs sets of subsets. The question that I meant to answer is as stated below.

Consider instead collections of subsets $\mathcal A_i$. All the elements $A\in \mathcal A_i$ are subsets of $\Omega$, i.e. $A\subseteq \Omega$. The set of such collections are indexed by $i\in I$.

Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(\mathcal A_i) \right) = \sigma\left( \bigcup_{i\in I} \mathcal A_i \right) $?

$\endgroup$
5
  • 2
    $\begingroup$ maybe what you want to ask is if $$\sigma\left( \bigcup_{i\in I} \sigma(\mathcal{A}_i) \right) = \sigma\left( \bigcup_{i\in I} \mathcal{A}_i \right)$$ where each $\mathcal{A}_i$ is a collection of subsets of $\Omega $, othewise as you have stated the question the answer is trivial as if $A_i$ is a subset then $\sigma (A_i)=\{A_i,A_i^\complement ,\emptyset ,\Omega \}$ $\endgroup$
    – Masacroso
    Apr 12, 2021 at 19:31
  • $\begingroup$ This is not true in the case you mention ($A_i = \{a_i\}$), since $\cup A_i = \Omega$ in this case. $\endgroup$ Apr 12, 2021 at 20:04
  • $\begingroup$ You are right @Masacroso . I should have asked for collection of events. And that also corrects my error, I think.... $\endgroup$
    – LudvigH
    Apr 13, 2021 at 7:39
  • $\begingroup$ @StéphaneLaurent I should have asked about collection of events. The "Singleton" should be a collection of events that only contains a single event. In that case we get $\cup \{A_i\}$ instead, which won't produce $\Omega$. What is good practise now, correct the question or ask a new question? $\endgroup$
    – LudvigH
    Apr 13, 2021 at 7:40
  • 1
    $\begingroup$ I have posted a detailed answer to your question. Please, let me know if you have any question regarding my answer. If my answer provides relevant / helpful information regarding your question, please, upvote it. If my answer actualy answers your question, accept it too, please. To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. $\endgroup$
    – Ramiro
    Apr 13, 2021 at 11:53

2 Answers 2

2
$\begingroup$

Given a set $\Omega$ with some subsets $A_i \subset \Omega$, for some indices $i\in I$.

We define a generated $\sigma$-algebra $\sigma(A_i)$ to be the smallest $\sigma$-algebra that contains $A_i$.

Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?

For the question as stated, the answer is NO.

Counter-exemple: Let $\Omega =\{1,2\}$. Let $I =\{1,2\}$ and, for each $i\in I$, let $A_i=\{i\}$. Then, $\bigcup_{i\in I} A_i = \{1\} \cup \{2\} = \{1,2\} $. The smallest $\sigma$-algebra that contains $ \{1,2\} $ is $\{\emptyset, \{1,2\}\}$. So, we have $$ \sigma\left( \bigcup_{i\in I} A_i \right) = \{\emptyset, \{1,2\}\} \tag{1} $$

On other hand, for each $i\in I$, $$ \sigma(A_i) = \{\emptyset, \{1\}, \{2\}, \{1,2\} \} $$ It is easy to see that $$ \sigma\left( \bigcup_{i\in \Bbb N} \sigma(A_i) \right) = \sigma\left( \{\emptyset, \{1\}, \{2\}, \{1,2\} \} \right) = \{\emptyset, \{1\}, \{2\}, \{1,2\} \} \tag{2} $$

From $(1)$ and $(2)$, $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \neq \sigma\left( \bigcup_{i\in I} A_i \right) $.

Remark: If the question was

Given a set $\Omega$ with some collections $A_i$ of subsets of $\Omega$, for some indices $i\in I$.

Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?

Then the answer would be YES.

Proof:

For all $i \in I$, $A_i \subseteq \sigma(A_i)$. So $ \bigcup_{i\in I} A_i \subseteq \bigcup_{i\in I} \sigma(A_i)$. So, we have $$ \bigcup_{i\in I} A_i \subseteq \bigcup_{i\in I} \sigma(A_i) \subseteq \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right)$$

Since $\sigma\left( \bigcup_{i\in I} \sigma(A_i) \right)$ is a $\sigma$-algebra, we have $$ \sigma\left( \bigcup_{i\in I} A_i \right)\subseteq \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \tag{3}$$

On the other hand, for all $i \in I$, $A_i \subseteq \bigcup_{i\in I} A_i \subseteq \sigma\left( \bigcup_{i\in I} A_i \right)$. Since $\sigma\left( \bigcup_{i\in I} A_i \right)$ is a $\sigma$-algebra, we have, for all $i \in I$, $$ \sigma(A_i) \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) $$ So $$ \bigcup_{i\in I} \sigma(A_i) \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) $$ Since $\sigma\left( \bigcup_{i\in I} A_i \right)$ is a $\sigma$-algebra, we have, $$ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) \tag{4}$$

From $(3)$ and $(4)$, we have that $$ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $$

$\endgroup$
3
  • $\begingroup$ Touché! I have made a notational error. I want it to be on collection of events... As indicated in one of the comments. But you answered the question as stated. What is good practise here? Accept your answer here and ask a new question? $\endgroup$
    – LudvigH
    Apr 13, 2021 at 7:36
  • 1
    $\begingroup$ @LudvigH Yes, the good pratice would be to accept the answer ans ask another question (it is not a good practice to make significant changes in the question after it is answered). However, I have added a remark to my answer, answering the question meant to ask. Take a look and let me know if you have any question regarding the answer. $\endgroup$
    – Ramiro
    Apr 13, 2021 at 11:54
  • $\begingroup$ Great. Did so. Thanks a lot! $\endgroup$
    – LudvigH
    Apr 13, 2021 at 12:20
0
$\begingroup$

By definition, we have $$\sigma\left(\bigcup_{i\in I}\mathcal A_i\right)=\bigcap X,\quad\quad\sigma\left(\bigcup_{i\in I}\sigma(\mathcal A_i)\right)=\bigcap Y,$$

where $$X=\left\{\mathcal A\mid\text{$\mathcal A$ is $\sigma$-algebra on $\Omega$ and $\bigcup_{i\in I}\mathcal A_i\subseteq \mathcal A$}\right\}$$ and $$Y=\left\{\mathcal A\mid\text{$\mathcal A$ is $\sigma$-algebra on $\Omega$ and $\bigcup_{i\in I}\mathcal \sigma(A_i)\subseteq \mathcal A$}\right\}.$$

Hence, it suffices to prove $X=Y$. Clearly, $Y\subseteq X$, since $\mathcal A_i\subseteq\sigma(\mathcal A_i)$, so the condition in $X$ is less restrictive. Secondly, we also have $Y\subseteq X$, since $\mathcal A_i\subseteq \mathcal A$ implies $\sigma(\mathcal A_i)\subseteq\mathcal A$, as $\mathcal A$ is a $\sigma$-algebra.

$\endgroup$
2
  • $\begingroup$ I have made a notational error. I want it to be on collection of events... So that your answer is correct. What is good practise in this case? Ask a new question, so that you can answer it there? $\endgroup$
    – LudvigH
    Apr 13, 2021 at 7:37
  • $\begingroup$ @LudvigH You can also remark in your question that your initial formulation is wrong, then write down what you actually meant to say. $\endgroup$
    – Zuy
    Apr 13, 2021 at 7:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .