Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 '13 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 '13 at 8:28
You should look at "Convergence of probability measures" by Billingsley, among many others. – Ahriman Jan 16 '13 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 '13 at 9:01
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.

share|cite|improve this answer
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 '13 at 14:43
In other words, any sigma algebra has a countable generator? – Tim Jan 16 '13 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 '13 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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