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From Williams' Probability with Martingales:

2.3. Examples of $(\Omega, \mathcal{F})$ pairs
We leave the question of assigning probabilities until later.

(a) Experiment: Toss coin twice. We can take $$ \Omega = \{HH, HT, TH, TT\}, \quad \mathcal{F} = \mathcal{P}(\Omega) := \text{set of all subsets of $\Omega$}. $$ In this module, the intuitive event ‘At least one head is obtained’ is described by the mathematical event (element of $\mathcal{F}$) $\{HH, HT, TH\}$.

(b) Experiment: Toin coss infinitely often. We can take $$ \Omega = \{H,T\}^{\mathbb{N}} $$ so that a typical point $\omega$ of $\Omega$ is a sequence $$ \omega = (\omega_1, \omega_2, \dotsc), \quad \omega_n \in \{H,T\}. $$ We certainly wish to speak of the intuitive event ‘$\omega_n = W$’ where $W \in \{H,T\}$, and it is natural to choose $$ \color{red}{ \mathcal{F} = \sigma( \{w \in \Omega : \omega_n = W\} : n \in \mathbb{N}, W \in \{H,T\} ) }. $$ Although $\color{red}{\mathcal{F} \neq \mathcal{P}(\Omega)}$ (accept this!), it turns out that $\mathcal{F}$ is big enough; […]

Why is it that $\mathscr{F} \ne 2^{\Omega}$ ?

What are some elements of $2^{\Omega}$ that are not in $\mathscr{F}$?

Jendrik Stelzner
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BCLC
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    Give the source. – Did Oct 30 '15 at 11:40
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    @Did Consummatum est – BCLC Oct 30 '15 at 11:41
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    "Why is it that $\mathscr{F} \ne 2^{\Omega}$ ?" For the same reason that some subsets of $(0,1)$ are not Borel. "What then is $2^{\Omega}$ if it is not $\mathscr{F}$?" Huh? $2^{\Omega}$ is $2^{\Omega}$ (and vice versa). – Did Oct 30 '15 at 11:41
  • @Did Thanks! Edited last line – BCLC Oct 30 '15 at 11:48
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    What kind of math book demands that the readers "accept this!"? +1 to the post because I can't give -1 to the textbook (and because it's a legitimate question, and exactly the sort of thing one should ask in this situation). – anomaly Oct 30 '15 at 16:24
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    @anomaly The book is excellent and to omit this proof at this moment is a perfectly legitimate choice. Systematic completeness can make for unreadable books. – Did Oct 30 '15 at 16:52

2 Answers2

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The $\sigma$-algebra generated by the events $\{\omega \in \Omega: \omega_n = W \}$ is the so-called Borel $\sigma$-algebra on $\Omega = \{H,T\}^\mathbb{N}$.

One can show, by transfinite induction (so you need some set-theory background) that there are at most $|\mathbb{R}| = 2^{\aleph_0}$ many Borel sets, while the power set of $\Omega$ has $2^{|\Omega|} = 2^{|\mathbb{R}|}$ many subsets, which is more by Cantor's theorem. So $\mathcal{F}$ is much smaller than the power set of all subsets of $\Omega$. But you only need the sets in $\mathcal{F}$. One can also use the Axiom of Choice to find non-Borel sets (the characteristic function of an ultrafilter, e.g.).

Filling in all the details requires some theory the author presumably did not want to assume the reader to know about.

Henno Brandsma
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  • Your last line explains much. thanks. edited last line of mine. – BCLC Oct 30 '15 at 11:49
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    Do you know about ultrafilters? One can't give a constructive example, but we can using ultrafilters or well-orders etc. It requires some theory. – Henno Brandsma Oct 30 '15 at 12:06
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    Otherwise believe the counting argument – Henno Brandsma Oct 30 '15 at 12:06
  • Henno Brandsma, is it like we can't give a constructive example of a set that's not Lebesgue measurable? (kind of forgot real analysis class) – BCLC Oct 30 '15 at 13:21
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    @BCLC Yes, it's quite similar. Any set you can write "with a formula" is almost certainly Borel. It's harder for Lebesgue measurable, as there a counting argument won't work. The ultrafilter example will, though. – Henno Brandsma Oct 30 '15 at 13:27
  • Thanks Henno Brandsma. I reread your 2nd paragraph. Do you mean to imply that $\mathscr{F} \ne 2^{\Omega}$ is analogous to $\mathscr{B}(\mathbb{R}) \ne 2^{\mathbb{R}}$ ? – BCLC Oct 30 '15 at 13:36
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    Yes, it's essentially the same, for another separable complete metric space $\Omega$ instead of $\mathbb{R}$. The proof is the same in either case. – Henno Brandsma Oct 30 '15 at 13:37
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    Actually, in this particular case, the situation "really is" the same, not just analogous. That's because the set of sequences of $0$s and $1$s is in bijection with $[0,1]$. Also, there is a bijection which "almost" takes those sequences with a particular digit fixed to points in a certain interval (there are only countably many "bad" sequences that end in infinitely many $1$s). So it is maybe not so surprising that the minimal $\sigma$-algebra which makes the projections measurable corresponds almost exactly to the minimal $\sigma$-algebra under which intervals are measurable in $[0,1]$. – Ian Oct 30 '15 at 16:26
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Blackwell & Diaconis ["A non-measurable tail set", in Statistics, probability and game theory, pp. 1–5, IMS Lecture Notes Monograph Series, vol. 30, 1996] give an example of a subset of $2^{\Bbb N}$ that is not an element of $\mathscr F$. Their construction uses a free ultrafilter $\mathscr U$ on $\Bbb N$. Let $E\subset 2^{\Bbb N}$ consist of those $a=(a_1,a_2,\ldots)\in 2^{\Bbb N}$ (thus each $a_i$ is either $0$ or $1$) such that $N_a:=\{i\in\Bbb N:a_i=1\}\in\mathscr U$. Then $E\notin\mathscr F$.

John Dawkins
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