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I want to make sure I understand this rigorously:

Assume we already know that Brownian motion $B_t$ on $[0,\infty)$ exists/how to construct it. Every $\sigma$-field considered is implicitly assumed to be the standard (Borel) $\sigma$-field.

To construct Wiener measure (as a measure on the space $\mathscr{C}[0,\infty)$ of continuous functions), do we (or can we) do the following?

  1. For every f $\in \mathscr{C}[0,\infty)$, we identify it with the point in $\mathbb{R}^{\mathbb{Q \cap[0,\infty)}}$ corresponding to its value f(r) at every rational r $\in \mathbb{Q}$. I.e. we identify $\mathscr{C}[0,\infty)$ with a suitable subset $D \subset \mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$.

  2. We define the finite dimensional distributions on $\mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$ using the distributions from $B_t$, i.e. for every r $\in \mathbb{Q}$, $X_r \sim \mathscr{N}(0,r)$.

  3. Then using the fact that every continuous function is uniquely defined by its values at rational coordinates, as well as the consistency of these distributions following from the continuity of Brownian sample paths, Kolmogorov's extension theorem allows to create a measure $\mu$ defined on all of $\mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$.

  4. We define Wiener measure $\nu$ on $\mathscr{C}[0,\infty) \cong D \subset \mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$ (for every measurable set $A \subset D$ in the standard Borel $\sigma$-algebra of $\mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$) as follows - $\nu(A):=\frac{\mu(A)}{\mu(D)}$.

I have no idea how one would perform any calculations with this definition, much less be able to define stochastic integration, but this is how I understand what we went over in my lecture.

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    $\begingroup$ Also how is this a measure on the space of all continuous functions, and not just those that start at the origin at t=0? (i.e. like the Brownian sample paths) $\endgroup$ Commented Apr 3, 2016 at 3:02
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    $\begingroup$ This is a measure on the space of all continuous functions (paths), because the set of functions that does not start at the origin has measure zero. $\endgroup$
    – SM2
    Commented Apr 3, 2016 at 11:08
  • $\begingroup$ Is the definition given correct then? It still seems too hand-wavy to me. $\endgroup$ Commented Apr 3, 2016 at 17:06

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Here are some comments to your definition.

  1. You say 'the consistency of these distributions following from the continuity'; I am not sure it's true. The consistency follows from the fact that you use finite dimensional distributions of the Wiener process. You can use Kolmogorov's theorem without the assumption of continuity.

  2. You say $\mathscr{C}[0,\infty) \cong D \subset \mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$. To be able to define $\mu(D)$ you need to prove that $D \in \mathfrak{S}_1$, where $\mathfrak{S}_1$ is the $\sigma$-algebra generated by the cylinder subsets of $\mathbb{R^{\mathbb{Q \cap [0,\infty)}}}$. Moreover, it would be nice to prove that actually $\mu(D)=1$. Also, you identified $\mathscr{C}[0,\infty)$ with $D$ and now you need to prove that $\mathfrak{S}_2$, the $\sigma$-algebra generated by the cylinder subsets of $\mathscr{C}[0,\infty)$, is identified with $\mathfrak{S}_1$. Only then your scenario would work completely, and you can possibly perform some calculation with this definition.


Here's another way to construct the Wiener measure.

1) Consider the measurable space $(\mathscr{C}[0,\infty),\mathfrak{S}_2)$ and introduce the standard metric of uniform convergence on every compact subset of $[0,\infty)$.

2) Prove that $\mathfrak{S}_2$ is actually a Borel algebra.

3) For a Wiener process $W(t)$ on a space $(\Omega, \mathfrak F, \mathbb P)$ define the natural map $g: \Omega \to \mathscr{C}[0,\infty)$. It sends $\omega \in \Omega$ to the corresponding path in $\mathscr{C}[0,\infty)$.

4) Using sets from the Borel algebra $\mathfrak{S}_2$, proof that $g$ is measurable. To do so you will need to use only rational $t$'s, so that you have no problem with measurability. Thankfully it makes no difference for the metric.

5) Define a measure $\nu$ on $(\mathscr{C}[0,\infty),\mathfrak{S}_2)$ as follows - $$ \nu(A) = \mathbb P(g^{-1}(A)) $$ for every $A \in \mathfrak{S}_2$. Then $\nu$ is the Wiener measure.

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  • $\begingroup$ You're definitely right; the continuous extension of the functions based on their rational coordinates does seem to only be necessary in terms of identifying 𝒞[0,∞)≅D⊂ℝ^[ℚ∩[𝟘,∞)]. And I definitely have no idea why that set would have to be measurable anyway. My professor said to use this method instead of identify with with ℝ^ℝ in order to avoid measurability issues, so hopefully it is something that can be resolved. Thank you so much for taking the time to look this over! $\endgroup$ Commented Apr 3, 2016 at 21:49
  • $\begingroup$ No problem! I wrote about another possible way in the answer. $\endgroup$
    – SM2
    Commented Apr 3, 2016 at 23:32

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