2
$\begingroup$

Suppose we have a sample space of outcomes.

How do we define a suitable $\sigma$-algebra of events on this sample space, such that we can then create a good probability measure on the events? Is there a general procedure for doing so?

An example from my book (A First Look at Rigorous Probability Theory, by J Rosenthal) is to first define a probability measure $P$ on a suitable algebra $\mathcal{A}$ of events and then extend $P$ to the generated $\sigma$-algebra $\sigma(\mathcal{A}$), letting $\sigma(\mathcal{A})$ be the set of events for our probability space. But I wonder if such an algebra can always to be found while still being sufficiently broad to cover most events we'd care about. I mean, would it be possible to just take an arbitrary set $S$ of events that we'd like, extend that into an algebra if it's not already, then a $\sigma$-algebra, and then have a nice measure defined on that??? (I know about Caratheodory's extension theorem, but that only answers my question in the case where we already have a measure on an algebra of events we'd want in our space. It doesn't answer whether we can just extend an arbitrary set of events into an algebra, or whether we can necessarily find a good measure on that algebra.)

Sorry if this question is somewhat unfocused, but there's a general feeling of confusion I have about this topic so any answer that clears it up would be greatly appreciated.

$\endgroup$
1
$\begingroup$

The following result may help you.

Theorem. (Durrett, A.1 (1.3), p.439) Let $\mathcal{S}$ be a semi-algebra and let $\mu$ defined on $\mathcal{S}$ have $\mu(\emptyset) = 0$. Suppose (i) if $S \in \mathcal{S}$ is a finite disjoint union of sets $S_i \in \mathcal{S}$, then $\mu(S) = \sum_i \mu(S_i)$, and (ii) if $S_i, S \in \mathcal{S}$ with $S$ the countable disjoint union of the $S_i$, then $\mu(S) \leq \sum_i \mu(S_i)$. Then $\mu$ has a unique extension $\bar{\mu}$ that is a measure on $\bar{\mathcal{S}}$, the algebra generated by $\mathcal{S}$. Moreover, if the extension is $\sigma$-finite, then by the Caratheodory theorem there is a unique extension $\nu$ that is a measure on $\sigma(\mathcal{\bar{\mathcal{S}}})$.

$\endgroup$
0
$\begingroup$

Algebra generated by a family of subsets $\mathcal M \subseteq 2^\Omega$ is $Alg(\mathcal M)=(\mathcal M \cup \mathcal M^c)_{ds}$, which is the closure of $\mathcal M$ with respect to complement $c$, finite intersections $d$ and finite unions $s$. By convention, the empty union is $\emptyset$ and the empty intersection is $\Omega$

The measure defined on $\mathcal M$ can be extended on $Alg(\mathcal M)$ by using this constructive approach.

$\sigma$-algebra generated by a family of subsets has a much more complicated construction, which I don't understand fully myself. $\mathcal M$ is repeatedly closed under complements and countable unions $\sigma$:

$\mathcal M_1=\mathcal M$

...

$\mathcal M_{i+1}=(\mathcal M_i \cup \mathcal M_i^c)_\sigma$

Then $\sigma$-algebra generated by $\mathcal M$ is $\bigcup \mathcal M_i$

But the union is until the first ordinal with an uncounted number of predecessors (!) Otherwise this construction does not exhaust all elements of the $\sigma$-algebra

This is why it is much harder to extend a measure to the $\sigma$-algebra generated. In fact it's impossible to extend the Lebesgue measure to all subsets of $\mathbb R$

$\endgroup$
  • $\begingroup$ OP is asking about extending a measure on a arbitrary set to an algebra/sigma algebra generated by the set. $\endgroup$ – Momo Nov 6 '16 at 20:31
  • $\begingroup$ I was in the process of editing it. Besides, this is not the definition of the algebra generated by a family of sets, but a construction of it that justifies the extension. $\endgroup$ – Momo Nov 6 '16 at 20:34

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.