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Reading through my lecture notes, and I'm stuck a bit on this concept.

Let $A$ be a set of subsets of $E$. Define $$ \sigma(A) = \{ A \subseteq E \ : \ A \in F \text{ for all } \sigma\text{-algebras } F \text{ containing }A \} .$$ Then $\sigma(A)$ is a $\sigma$-algebra, which is called the $\sigma$-algebra generated by $A$. It is the smallest $\sigma$-algebra containing $A$.

So let's say: $E = \{1,2,3\}$. All the possible subsets of $E$ will be $ = \{ \{ \emptyset \}, \{1,2,3\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\} \}$. So if we choose $A= \{1 \}$, then what would $\sigma(\{1\})$ equals to?

$$ \sigma(\{1\}) = \{ \{1 \} \subseteq E \ \colon \ \{1 \} \in F \text{ for all } \sigma\text{-algebras } F \text{ containing } \{1\} \} .$$

How would I find all the $\sigma$-algebras $F$ containing $\{1\}$? Thanks.

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    $\begingroup$ The definition you gave is not correct. What you wrote is the definition of $\sigma(F)$ not $\sigma(A)$. The set $F$ must be a subset of powerset $2^E$. E.g., you might ask what is $\sigma(F)$ for $F=\{\{1\}\}$. See also here: en.wikipedia.org/wiki/Sigma-algebra#Generated_.CF.83-algebra $\endgroup$ Nov 13, 2011 at 18:03
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    $\begingroup$ You use $A$ for subsets of $E$ but also for collections of subsets of $E$. Predictably, chaos ensues, for example in the definition of $\sigma(A)$. $\endgroup$
    – Did
    Nov 13, 2011 at 18:11
  • $\begingroup$ Ok, I might have been wrong and maybe you really want to define a $\sigma$-algebra generated by one subset of $E$. Anyway, you're using the symbol $A$ in the definition of $\sigma(A)$ in two meanings, you should clarify/correct this definition. The definition in the original version of question is: $\sigma(A)$ = {$A\subseteq E$ : $A \in$ $F$ for all $\sigma$-algebras F containing A}. $\endgroup$ Nov 13, 2011 at 18:11

3 Answers 3

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As a general comment on this kind of construction, I might suggest reading this answer.

Generally speaking, what you describe is the "top-down" approach to the construction. It is not, generally, practical to actually know the description of the elements in the generated object. For that, you want the "bottoms-up" construction. In the question linked-to above, Asaf Kargila provides the description of the bottoms-up construction.

In your particular case, of course, since $E$ is small (3 elements), so $P(E)$ is small (8 elements), the number of possible $\sigma$-algebras is somewhat manageable (though still large).

What are all $\sigma$-algebras on $E=\{1,2,3\}$ that contain $\{1\}$ as an element? They must contain $\emptyset$, $\{1\}$, $\{2,3\}$, and $E$. There are 4 other elements in $P(E)$ which may or may not be in a $\sigma$-algebra.

  1. One $\sigma$-algebra is just $\{\emptyset, \{1\}, \{2,3\}, E\}$.
  2. If the $\sigma$-algebra contains either $\{2\}$ or $\{3\}$ in addition to those, then it must also contain the other (symmetric difference with $\{2,3\}$), and hence be all of $P(E)$.
  3. If the $\sigma$-algebra contains any other $2$-element subset, then it must contain another singleton (symmetric difference with $\{2,3\}$), hence must be all of $P(E)$.

So in fact the only $\sigma$-algebras on $E$ that contain $\{1\}$ are $\{\emptyset, \{1\},\{2,3\}, E\}$ and $P(E)$. The intersection is just $\{\emptyset,\{1\},\{2,3\},E\}$, so the $\sigma$-algebra generated by $\{1\}$ is $\{\emptyset, \{1\}, \{2,3\}, E\}$.

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  • $\begingroup$ Thanks for your help, at the moment im in the process of reading through your answer and in the other thread also, its been very helpful so i must say thanks! Will post again on stackexchange if I have any more questions, but I do grasp the concept better now. $\endgroup$
    – Thomas
    Nov 15, 2011 at 0:39
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Note that the correct definition is what Martin has written. The definition of $\sigma(A)$ in general is not very useful to construction of the $\sigma$-algebra generated by $A$, because as you have mentioned it may be hard to describe all $\sigma$-algebras which contain $A$, but when $A$ is just a single set the problem is easy.

Let $P(E)$ denote the powerset of $E$, i.e. the set of all subsets of $E$ and $a\in P(E)$ - subset of $E$. Let us consider $F:=\sigma(a)$.

  1. First of all, $E\in F$ be definition;

  2. $\emptyset\in F$ and $a^c\in F$ because any complement of element of $F$ belongs to $F$.

  3. Let us assume now that $F = \{\emptyset,a,a^c,E\}$ since we prove that these elements have to be in $F$ anyway. We need to check that $E\in F$ (done), $F$ is closed under taking complements (done) and any countable union of elements of $F$ belongs to $F$ (also easy to see).

As a result, $\sigma(a) = \{\emptyset,a,a^c,E\}$. In your case $a = \{1\}$ hence $$ \sigma(\{1\}) = \{\emptyset,\{1\},\{2,3\},\{1,2,3\}\}. $$

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    $\begingroup$ Thanks for your help, yeah I think its easier to understand the way you did it compared to my original definition. Thanks! $\endgroup$
    – Thomas
    Nov 15, 2011 at 0:40
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Actually $A=\{\{1\}\}$ since it must be a subset of the Power set. Now to find $\sigma(A)$, I would do the following:

First $\emptyset, E, \{1\}$ must be in $\sigma(A)$ and then the complement of $\{1\}$ which is $\{2,3\}$. So that $$ \sigma(A)=\{\emptyset, E, \{1\}, \{2,3\} \} $$

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