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From Wikipedia

an alternative way of stating Kolomogorov's extension theorem is that, provided that the above consistency conditions hold, there exists a (unique) measure $\nu$ on $(\mathbb{R}^n)^T$ with marginals $\nu_{t_{1} \dots t_{k}}$ for any finite collection of times $t_{1} \dots t_{k}$. The remarkable feature of Kolmogorov's extension theorem is that it does not require $T$ to be countable, but the price to pay for this level of generality is that the measure $\nu$ is only defined on the product σ-algebra of $(\mathbb{R}^n)^T$, which is not very rich.

I was wondering in what sense "the measure $\nu$ is only defined on the product σ-algebra of $(\mathbb{R}^n)^T$" is a price? $\nu$ not being able to be defined on some other sigma algebras?

What does "which is not very rich" mean precisely?

Thanks and regards!

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For instance, if $T = [0,1]$, you could want to define a probability measure on the $\sigma$-algebra generated by the Borel sets of $\mathcal{C}([0,1])$ endowed with the norm of uniform convergence. A classical example is the Wiener measure. Such a construction cannot be done directly with Kolmogorov's theorem. – Ahriman Jan 16 at 8:25
@Ahriman: Thank you! Can you give some references for me to learn about the construction of "the σ-algebra generated by the Borel sets of C([0,1]) endowed with the norm of uniform convergence" and construction of a probability measure such as the Wiener measure on it? – Tim Jan 16 at 8:28
You should look at "Convergence of probability measures" by Billingsley, among many others. – Ahriman Jan 16 at 8:49
Also Shiryaev's Probability is a good source. Take a look at Chapter II, §2 and especially pages 163-169 treats Kolmogorov's extension theorem on both $\mathbb{R}^\infty$ and $\mathbb{R}^T$. – Stefan Hansen Jan 16 at 9:01

1 Answer

up vote 4 down vote accepted

Here is a typical example: Let $T=[0,1]$, the unit interval and $\lambda$ Lebesgue measure, and take the coin-flipping measure on $\{0,1\}^T$. Let $f_x$ be the $x$-coordinate of $f\in\{0,1\}^T$. You might want to know the probability of the set $$B=\big\{f:\lambda\{x\in T:f_x=1\}=1/3\big\},$$ the probability that the fraction of $1$s is exactly one third. This event is not in the product $\sigma$-algebra.

This follows from the general result that if $A\in\sigma(\mathcal{F})$, then there exists a countable family $\mathcal{C}\subseteq \mathcal{F}$ such that $A\in\sigma(\mathcal{C})$. One can show this by verifying that the family of sets generated by a countable sub-family of $\mathcal{F}$ forms a $\sigma$-algebra that contains $\mathcal{F}$.

Applied to our case, an event in the product $\sigma$-algebra must be generated by countably many coordinates. So take any $f\in B$ and let $H_f$ be the set of functions $g\in\{0,1\}^T$ such that $\{x:f_x\neq g_x\}$ is countable. Then $H_f$ intersects every nonempty set generated by countably many coordinates and has therefore outer measure $1$. Clearly, $H_f\subseteq B$, so $B$ has measure $1$ too. But by a similar argument, $B^C$ has outer measure $1$ too. It follows that $B$ is not measurable in the product $\sigma$-algebra.

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Thanks! "if A∈σ(F), then there exists a countable family C⊆F such that A∈σ(C). One can show this by verifying that the family of sets generated by a countable sub-family C of F forms a σ-algebra that contains F." So F ⊆ σ(C), and σ(F) = σ(C) because C⊆F? – Tim Jan 16 at 14:43
In other words, any sigma algebra has a countable generator? – Tim Jan 16 at 14:50
@Tim You take the family $\mathscr{C}$ of all sets $A$ such that for some countable $\mathcal{C}\subseteq\mathcal{F}$ we have $A\in\sigma(\mathcal{C})$. If $A\in\sigma(\mathcal{C})$ then $A^C\in\sigma(\mathcal{C})$ and if $A_n\in \sigma(\mathcal{C}_n)$ for all $n$, then $\bigcup_n A_n\in\sigma (\bigcup_n\mathcal{C}_n)$. Moreover, for all $A\in\mathcal{F}$, we have $A\in\sigma(\{A\})$. So $\mathcal{F}\subseteq \mathscr{C}$. It is also clear that $\mathscr{C}\subseteq\sigma(\mathcal{F})$. But not every $\sigma$-algebra is countable generated, the countable family can depend on the set you gener. – Michael Greinecker♦ Jan 16 at 15:02

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