I'm trying to construct Brownian motion using the Kolmogorov extension theorem.

I am happy with the construction of a process with the required FDDs as (the canonical process associated with) a random function in $D([0, \infty), R)$ - (the set of all functions from $R_+$ to $R$, not just cadlag functions). I am also happy with the fact that the set of continuous functions is not measurable within the $\sigma$-algebra generated by 'cylindrical sets'.

So my understanding is that it does not make sense to talk about the probability that such a process is continuous?

But on the other hand it seems we can naively apply the Kolmogorov continuity theorem the constructed process to construct a (continuous) Brownian motion.

So what is going on here? When I have constructed a process with the required FDDs can I naively apply the Kolmogorov continuity theorem the complete the construction of Brownian motion? If not, why not? What goes wrong?

Edit: By naively apply Kolmogorov continuity theorem, I mean the following:

Kolmogorov continuity theorem: (from Le Gall)

Let $X = (X_t)_{t \in I}$ be a random process indexed by a bounded interval $I$ of $R$, and taking values in a complete metric space $(E, d)$. Assume that there exist three reals $q, \epsilon, C > 0$ such that, for every $s, t \in I$,

$E[d(X_s,X_t)^q]􏰒􏰂 \leq C|t - s|^{1 + \epsilon}$ :

Then, there is a modification $\tilde{X}$ of $X$ whose sample paths are Hölder continuous with exponent $\alpha \in (0, \frac{\epsilon}{q})$: This means that, for every $\omega \in \Omega$ and every $\alpha \in (0, \frac{\epsilon}{q})$ there exists a finite constant $C_\alpha(\omega)$ such that, for every $s, t \in I$,

$d(\tilde{X}_s(\omega), \tilde{X}_t(\omega) \leq C_\alpha(\omega)|t-s|^{1+ \alpha}$

In particular, $\tilde{X}$ is a modification of $X$ with continuous sample paths (by the preceding observations such a modification is unique up to indistinguishability).

[end of theorem]

So once we have the random process with the FDDs of Brownian motion taking values in R (a complete metric space, we can just apply the distribution properties of Brownian motion to satisfy the requirements of the theorem and produce a continuous modification (which has the same FDDs since it is a modification).

So where does the above argument go wrong?

  • $\begingroup$ AlexanderR: Perhaps this helps: Say that a process on $R$ is a Brownian motion if its FDD are as required. I think you set out to prove "For every Brownian motion process, (almost) all paths are continuous" but you discovered that that is false. Instead, as the two answers below show, you get the theorem "There exists a Brownian motion process whose (almost) all paths are continuous." $\endgroup$
    – user95282
    Mar 12, 2020 at 18:11

3 Answers 3


I think it helps to look more closely into the construction. I'm going to use $\Omega = \mathbb{R}^{[0,\infty)}$ instead of $D$ to denote the space of all real-valued functions on $[0,\infty)$, since $D$ is more often used for the Skorokhod space of cadlag functions.

The Kolmogorov extension theorem gives you a probability measure $\mathbb{P}$ on $\Omega$ (i.e. on its cylindrical $\sigma$-algebra $\mathcal{F}$) with the desired finite-dimensional distributions. Of course one's first inclination would be to take the random variables $X_t = \omega(t)$ as your process. This, as you know, doesn't work, as the set of continuous functions in $\Omega$ is not measurable.

But let $Q \subset [0,\infty)$ be the nonnegative rationals (or any countable dense subset you prefer), and consider the set $$E := \{ \omega \in \Omega : \omega|_Q \text{ is uniformly continuous on bounded sets}\}.$$ This set is measurable with respect to $\mathcal{F}$ (since being in $E$ only depends on the values of $\omega$ at countably many points, namely $Q$). And the important content of the Kolmogorov continuity theorem is that $\mathbb{P}(E) = 1$. (Indeed, it shows that the set of $\omega$ for which $\omega|_Q$ is Hölder continuous with the appropriate exponent, already has probability 1; and this set is contained in $E$, since Hölder continuous functions are locally uniformly continuous.)

So now you define a different set of random variables on $\Omega$: $$B_t(\omega) = \begin{cases} X_t(\omega), & t \in Q, \omega \in E \\ \lim_{s \to t, s \in Q} X_t(\omega), & t \notin Q, \omega \in E \\ \text{whatever you want} & \omega \notin E \end{cases}$$ This has the same finite-dimensional distributions as $X_t$, and by construction, it is clear that $t \mapsto B_t(\omega)$ is continuous for every $\omega \in E$, i.e. $\mathbb{P}$-almost surely.

So in short, you're right that it didn't make sense to talk about whether the "canonical" process $X_t$ was continuous, which is why we end up proving a.s. continuity of a different process instead.

You can, if you like, go on to show that $B_t$ is a modification of $X_t$ (for any fixed $t$ you can apply the previous step to $Q \cup \{t\}$), but nobody really cares about this, since $X_t$ itself is unimportant once $B_t$ is constructed.

Note that the random variables $X_t$, $t \notin Q$, never really got used in this construction, so we needn't have bothered to construct them in the first place. Indeed, many people (including me) feel this construction is cleaner if you start instead by applying the Kolmogorov extension theorem on $\Omega = \mathbb{R}^Q$. The latter is a standard Borel space, and you need much less axiom of choice to prove KET here.

Edit. There is nothing at all wrong with the argument in your question, and it is really the same thing as what I outline above. I have just made the choice of modification explicit. My $B_t$ is your $\tilde{X}_t$.

So my understanding is that it does not make sense to talk about the probability that such a process is continuous?

It does not make sense to talk about the probability that the canonical process $X_t(\omega) = \omega(t)$ on $(\mathbb{R}^\infty, \mathcal{F})$ is continuous. But the process whose continuity we end up proving is not $X_t$; it is a modification of it.

  • $\begingroup$ Thank you for helping me with this. I now feel I understand completely a way to construct Brownian motion using these two Kolmogorov theorems. However, I don't feel I understand why the way I suggest won't work. What is the essense of what is stopping it working? I understand it's related to the lack of measurability of the set of continuous functions, but could you be a little bit more explicit? And also, the KCT doesn't have any conditions requiring this or that set to be measurable? Doesn't it need such conditions if we are not able to apply it in this case? $\endgroup$
    – user152718
    Mar 12, 2020 at 15:40
  • $\begingroup$ @AlexanderR: I think maybe I don't fully understand what you mean when you say "naively apply the Kolmogorov continuity theorem". Can you give the precise statement of the Kolmogorov continuity theorem that you're using, and explain in more detail the argument that you would suggest? $\endgroup$ Mar 12, 2020 at 15:44
  • $\begingroup$ Yes I'll edit my question to add that, thanks $\endgroup$
    – user152718
    Mar 12, 2020 at 15:45
  • $\begingroup$ @AlexanderR What I mean by "doesn't work" in the second paragraph is this: in order to have a Brownian motion, you need a process $W_t$ and an event $A$ with $\mathbb{P}(A)=1$ such that $t \mapsto W_t(\omega)$ is continuous for every $\omega \in A$. I'm saying that taking $W_t = X_t$ will not work here, because there does not exist any event $A$ with that property. Indeed, you can show that there does not exist any nonempty set $A \in \mathcal{F}$ with the property that $t \mapsto X_t(\omega)$ is continuous for all $\omega \in A$. $\endgroup$ Mar 12, 2020 at 15:51
  • $\begingroup$ Great, now I do understand a bit more. In a sense I'm now in the situation of someone who has a 'proof' of a result and who also has a counterexample, and they are trying to understand why the 'proof' is wrong even though they already understand that is must be wrong. $\endgroup$
    – user152718
    Mar 12, 2020 at 16:04

The process $X$ you mention is uniformly continuous on the rationals* in the compact interval $[0,n]$, with probability 1. So you define Brownian motion $B$ to be the unique continuous extension of: $X$ restricted to $\mathbb Q$.

*or your favorite countable dense set

  • $\begingroup$ Thank you, this is a very helpful answer and I will use this construction. Could you possibly provide any hint as to why we can't apply the Kolmogorov Continuity theorem directly to construct a continuous modifcation for the process defined on $R_+$? In the sense that the statement of the Kolmogorov Continuity theorem doesn't impose any restrictions on the space a random process is defined on? $\endgroup$
    – user152718
    Mar 11, 2020 at 15:00
  • $\begingroup$ @AlexanderR: This $B$ is the continuous modification. $\endgroup$ Mar 11, 2020 at 16:42

Another approach would be to use a version of Kolmogorov continuity theorem which gives the result in the form $\mu^* ( C ( T ) ) = 1$ and then simply restrict $\mu$ to the set of full outer measure. This approach is due to Kolmogorov and it works in many other situations.

For example, consider Theorem 7.7.4 from Bogachev's Measure Theory. Volume 2, which goes as follows.

If $( \xi_t )_{t \in T}$, $T \subset \mathbb{R}$ is a random process satisfying $$ \mathbb{E} \left| \xi_t - \xi_s \right|^\alpha \leq L | t - s |^{1 + \beta} $$ for some $\alpha, \, \beta, \, L > 0$, then $\mu^* ( C ( T ) ) = 1$.

The approach given by Nate Eldredge essentially uses the separability of the Brownian motion.


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