1
$\begingroup$

I'm new to this concept of a $\sigma$-algebra generated by a collection of subsets.

Let $\Omega = \{a, b, c, d\}$ and $$\begin{align} &\mathcal{F}_1 = \{\Omega, \emptyset, \{a\}\} \\ &\mathcal{F}_2 = \{\Omega, \emptyset, \{a\}, \{b, c, d\}\}\text{.} \end{align}$$ I wish to show that $$\sigma\langle \mathcal{F}_1 \rangle = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathcal{F}_1)}\mathcal{F} = \mathcal{F}_2$$ where $\mathcal{I}(\mathcal{F}_1) = \{\mathcal{F}: \mathcal{F}_1 \subset \mathcal{F} \text{ and }\mathcal{F} \text{ a }\sigma\text{-algebra on }\Omega \}$.

I know immediately from this that any $\sigma$-algebra $\mathcal{F}$ must, at the very least, contain elements of $\mathcal{F}_1$: $$\{\Omega, \emptyset, \{a\}\}$$ but other than literally listing every possible collection of subsets of $\Omega$ and checking which are $\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.

Is my way of thinking correctly or is there a quicker way?

$\endgroup$
  • $\begingroup$ Complete the collection of subsets to be closed under the conditions for an sigma algebra, i.e., closed under the operations of countable union or intersection and complementary. This is the minimal sigma algebra that contain the collection. $\endgroup$ – Masacroso Feb 22 '16 at 19:26
  • $\begingroup$ @Masacroso Ah, so that's why it's called the "smallest" $\sigma$-algebra generated by a set. Thank you! $\endgroup$ – Clarinetist Feb 22 '16 at 19:29
1
$\begingroup$

Yes. but one quick shortcut is that the complements of the sets are in the sigma algebra as well, so you can easily see $\{b,c,d\}$ must be in $\sigma\langle\mathcal{F}_1\rangle$.

$\endgroup$
  • $\begingroup$ And we know that this $\{b, c, d\}$ set with everything else in $\mathcal{F}_1$ must be the "smallest" $\sigma$-algebra generated by $\mathcal{F}_1$. Thank you! ... now, the question is, how do I prove this rigorously? I don't see how else to do this other than finding every possible $\mathcal{F}$ and then intersecting them. $\endgroup$ – Clarinetist Feb 22 '16 at 19:31
  • 1
    $\begingroup$ @Clarinetist, no, it's the smallest also because it is the minimum extension of the current set to a sigma-algebra. So suffices to find one, and show that none of its subsets are sigma algebras. $\endgroup$ – gt6989b Feb 22 '16 at 19:34
  • $\begingroup$ Very good and clear explanation. Thank you! $\endgroup$ – Clarinetist Feb 22 '16 at 19:35
1
$\begingroup$

Here are some more general statements that imply the desired equation. We assume the following without proof.

  1. The intersection of $\sigma$-algebras is again a $\sigma$-algebra.
  2. A set of the form $\lbrace \emptyset, \Omega, A, A^c \rbrace$ is a $\sigma$-algebra.

Let $\mathscr{G}$ (particular case $\mathcal{F}_1$) be any set of subsets of $\Omega$. $\mathscr{A}:=\bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{G})}\mathcal{F}$ is nonempty because the powerset $\mathscr{P}(\Omega)$ has $\mathscr{P}(\Omega) \in \mathcal{I}(\mathscr{G})$. It follows that $\mathscr{G} \subset \mathscr{A}$ and that, because of assumption 1, $\mathscr{A}$ is a $\sigma$-algebra. If $\mathscr{A}'$ is another $\sigma$-algebra with $\mathscr{G} \subset \mathscr{A}'$, then, because $\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\mathscr{A} \subset \mathscr{A}'$. So $\mathscr{A}$ satisfies the definition of the smallest $\sigma$-algebra generated by $\mathscr{G}$, i.e. $\sigma(\mathscr{G}) = \mathscr{A}$.

We conclude $\sigma(\mathcal{F}_1) = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{\mathcal{F}_1})} \mathcal{F}$.

We now show that $\sigma(\mathcal{F}_1) = \mathcal{F}_2$. Note that $A \in \mathcal{F}_1$ implies $A \in \sigma(\mathcal{F}_1)$ and $\lbrace a \rbrace \in \sigma(\mathcal{F}_1)$ implies $ \lbrace b,c,d \rbrace = \lbrace a \rbrace^c \in \sigma(\mathcal{F}_1)$, so that $\mathcal{F}_2 \subset \sigma(\mathcal{F}_1)$. By assumption 2, $\mathcal{F}_2$ is a $\sigma$-algebra. We obtain $\sigma(\mathcal{F}_1) = \mathcal{F}_2$.

Note

It holds in general that $\sigma(\lbrace A \rbrace) = \sigma(\lbrace \emptyset, \Omega, A \rbrace) = \lbrace \emptyset, \Omega, A, A^c \rbrace$, we could have proved this instead of proving this for our particular case in the last step.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.