Wiener Measure
Lemma 1
- $\Omega_{n} := C_{0}([0, n])$,
- continuous funcitons
- \((\omega_{k})_{k \in \mathbb{N}}\),
- $\omega_{k} \in \Omega_{n}$,
Suppose
\[\max_{t \in [0, n]} \left( \abs{ \omega_{k}(t) - \omega_{m}(t) } \right) \rightarrow 0 \quad (k, m \rightarrow \infty) .\]Then
\[\begin{equation} \exists \omega \in \Omega_{n}, \text{ s.t. } \max_{t \in [0, n]} \left( \abs{ \omega_{k}(t) - \omega(t) } \right) \rightarrow 0 \ (k \rightarrow \infty) \label{lemma_01_existence} \end{equation} .\]In other words, continuous functions is complete metric space with sup norm.
Moreover, if
\[\begin{eqnarray} \max_{t \in [0, n]} \left( \omega_{k}(t) - \omega(t) \right) & \rightarrow & 0 \nonumber \\ \max_{t \in [0, n]} \left( \omega_{k}(t) - \omega^{\prime}(t) \right) & \rightarrow & 0, \nonumber \end{eqnarray}\]then $\omega = \omega^{\prime}$.
proof
Proposition 2
- $\Omega := C_{0}([0, \infty))$,
- continuous funcitons
(1) $\rho$ is a metric on $\Omega$.
(2) $(\Omega, \rho)$ is a separable
(3) $(\Omega, \rho)$ is a complete
proof
proof of (1)
We will show
- (1-1) $\rho(\omega_{1}, \omega_{2}) \ge 0$,
- (1-2) $\rho(\omega_{1}, \omega_{2}) = \rho(\omega_{2}, \omega_{1})$,
- (1-3) $\rho(\omega_{1}, \omega_{2}) = 0 \Leftrightarrow \omega_{2} = \omega_{1}$,
- (1-4) $\rho(\omega_{1}, \omega_{2}) \le \rho(\omega_{1}, \omega_{3}) + \rho(\omega_{3}, \omega_{2})$.
(1-1) and (1-2) is obvious because of properties of modulus. Regarding (1-3),
\[\begin{eqnarray} & & \rho(\omega_{1}, \omega_{2}) = 0 \nonumber \\ & \Leftrightarrow & \forall n \in \mathbb{N}, \ \forall t \in [0, n], \ \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega_{1}(t) - \omega_{2}(t) } \wedge 1 \right) = 0 \nonumber \\ & \Leftrightarrow & \forall n \in \mathbb{N}, \ \forall t \in [0, n], \ \max_{t \in [0, n]} \left( \abs{ \omega_{1}(t) - \omega_{2}(t) } \right) = 0 \nonumber \\ & \Leftrightarrow & \forall n \in \mathbb{N}, \ \forall t \in [0, n], \ \abs{ \omega_{1}(t) - \omega_{2}(t) } = 0 \nonumber \\ & \Leftrightarrow & \forall n \in \mathbb{N}, \ \forall t \in [0, n], \ \omega_{1}(t) = \omega_{2}(t) \nonumber \\ & \Leftrightarrow & \forall t \in [0, \infty), \ \omega_{1}(t) = \omega_{2}(t) . \nonumber \end{eqnarray}\]Lastly, we will show (1-4). It is easy to show the following inequality in both cases:
- \(\abs{\omega_{1}(t) - \omega_{3}} + \abs{\omega_{2}(t) - \omega_{3}} \le 1\),
- \(\abs{\omega_{1}(t) - \omega_{3}} + \abs{\omega_{2}(t) - \omega_{3}} > 1\).
Since $\rho$ always converge absolutely, we have
\[\begin{eqnarray} & & \rho(\omega_{1}, \omega_{2}) \nonumber \\ & \le & \sum_{n \in \mathbb{N}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \left( \abs{ \omega_{1}(t) - \omega_{3}(t) } + \abs{ \omega_{3}(t) - \omega_{2}(t) } \right) \wedge 1 \right) \nonumber \\ & \le & \sum_{n \in \mathbb{N}} \left( \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega_{1}(t) - \omega_{3}(t) } \wedge 1 \right) + \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega_{3}(t) - \omega_{2}(t) } \wedge 1 \right) \right) \nonumber \\ & = & \rho(\omega_{1}, \omega_{3}) + \rho(\omega_{3}, \omega_{2}) . \nonumber \end{eqnarray}\]proof of (2)
By the Weierstrass approximation theorem, $C([0, n])$ is separable for all $n \in \mathbb{N}$.
Let $\mathcal{P}_{n}$ be the set of countable polynomial functions which is dense in $C([0, n])$. Since polynomials can be defined over $\mathbb{R}$, we extend the domain of the polynomials to $C([0, \infty))$. Let
\[\mathcal{P} := \bigcup_{n \in \mathbb{N}} \mathcal{P}_{n} .\]$\mathcal{P}$ is countable. Let $B(\omega^{\prime}, r)$ be open ball in $(\Omega, \rho)$. That is,
\[r > 0, \ \omega^{\prime} \in \Omega, \ B(\omega^{\prime}, r) := \{ \omega \mid \rho(\omega, \omega^{\prime}) < r \} .\]Taking $n_{0} \in \mathbb{N}$ and $\epsilon > 0$ such that
\[\frac{1}{2^{n_{0}-1}} < \epsilon < r .\]By the Weierstrass approximation theorem, there exists $p \in \mathcal{P}$ such that
\[\frac{1}{2^{n_{0}}} \max_{t \in [0, n_{0}]} \abs{ p(t) - \omega^{\prime}(t) } < \frac{\epsilon}{2} \frac{\epsilon}{2^{n_{0}}} .\]Note that
\[\begin{eqnarray} \forall n \in [1:n_{0}], \ \frac{1}{2^{n}} \left( \max_{t \in [0, n]} \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) & \le & \frac{1}{2^{n}} \max_{t \in [0, n_{0}]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) \nonumber \\ & = & 2^{n_{0}-n} \frac{1}{2^{n_{0}}} \max_{t \in [0, n_{0}]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) \nonumber \\ & < & 2^{n_{0}-n} \frac{\epsilon}{2} \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \frac{1}{2^{n}} . \nonumber \end{eqnarray}\]Hence
\[\begin{eqnarray} \sum_{n \in \mathbb{N}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) & = & \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) + \sum_{n = n_{0} + 1}^{\infty} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) \nonumber \\ & \le & \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ p(t) - \omega^{\prime}(t) } \wedge 1 \right) + \frac{1}{2^{n_{0}}} \nonumber \\ & \le & \sum_{n = 1}^{n_{0}} \frac{\epsilon}{2} \frac{1}{2^{n}} + \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \frac{ \frac{1}{2} \left( 1 - \frac{1}{2^{n_{0}}} \right) }{ 1 - \frac{1}{2} } + \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \left( 1 - \frac{1}{2^{n_{0}}} \right) + \frac{1}{2^{n_{0}}} \nonumber \\ & \le & \frac{\epsilon}{2} + \frac{1}{2^{n_{0}}} \nonumber \\ & < & \frac{\epsilon}{2} + \frac{\epsilon}{2} \nonumber \\ & = & \epsilon \nonumber \\ & < & r \nonumber \end{eqnarray}\]Thus, $p \in B(\omega^{\prime}, r)$. Therefore, the proof completed.
proof of (3)
Supopse that \((\omega_{k})_{k \in \mathbb{N}} \subseteq \Omega\) and
\[\rho(\omega_{k}, \omega_{m}) \rightarrow 0 \quad (k, m \rightarrow \infty) .\]we will show
\[\exists \omega \in \Omega, \ \text{ s.t. } \rho(\omega_{k}, \omega) \rightarrow 0 \quad (k \rightarrow \infty) .\]From our assumption, we claim
\[\begin{equation} \forall n \in \mathbb{N}, \ \max_{t \in [0, n]} \left( \abs{ \omega_{m}(t) - \omega_{k}(t) } \right) \rightarrow 0 \quad (k, m \rightarrow \infty) \label{proposition_02_cauchy_sequence_01} . \end{equation}\]Indeed, Let $\epsilon \in (0, 1)$ and $n \in \mathbb{N}$ be fixed. there exsits $n_{0} \in \mathbb{N}$ such that
\[\forall m, k \ge n_{0}, \ \rho(\omega_{m}, \omega_{k}) = \sum_{l \in \mathbb{N}} \frac{1}{2^{l}} \max_{t \in [0, l]} \left( \abs{ \omega_{k}(t) - \omega_{m}(t) } \wedge 1 \right) < \frac{\epsilon}{2^{n}} .\]Thus,
\[\begin{eqnarray} & & \forall m, k \ge n_{0}, \ \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega_{k}(t) - \omega_{m}(t) } \wedge 1 \right) < \frac{\epsilon}{2^{n}} \nonumber \\ & \Leftrightarrow & \forall m, k \ge n_{0}, \ \max_{t \in [0, n]} \left( \abs{ \omega_{k}(t) - \omega_{m}(t) } \wedge 1 \right) < \epsilon \nonumber \\ & \Leftrightarrow & \forall m, k \ge n_{0}, \ \max_{t \in [0, n]} \abs{ \omega_{k}(t) - \omega_{m}(t) } < \epsilon \quad (\because \epsilon < 1) \end{eqnarray}\]Since $\epsilon$ is arbitrary, our claim proved.
Let $n^{\prime} \in \mathbb{N}$ be fixed. Define \(\omega^{n^{\prime}} := \omega_{\restriction_{[0, n^{\prime}]}}\) and \(\omega_{k}^{n^{\prime}} := \omega_{k \restriction_{[0, n^{\prime}]}}\). By \(\eqref{proposition_02_cauchy_sequence_01}\),
\[\begin{eqnarray} \max_{t \in [0, n^{\prime}]} \left( \abs{ \omega_{m}^{n^{\prime}}(t) - \omega_{k}^{n^{\prime}}(t) } \right) \rightarrow 0 \quad (k, m \rightarrow \infty) . \end{eqnarray}\]By \(\eqref{lemma_01_existence}\), there exists contunous function $\omega^{n^{\prime}}$ on $[0, n^{\prime}]$ such that
\[\begin{eqnarray} \max_{t \in [0, n^{\prime}]} \left( \abs{ \omega_{k}^{n^{\prime}}(t) - \omega^{n^{\prime}}(t) } \right) \rightarrow 0 \quad (k \rightarrow \infty) \nonumber . \end{eqnarray}\]Since $n^{\prime}$ is arbitrary, we obtain the sequence \((\omega^{n})_{n \in \mathbb{N}}\) of continuous functions. It is easy to confirm that for all $n > m$,
\[\begin{eqnarray} \max_{t \in [0, m]} \left( \abs{ \omega_{k}^{m}(t) - \omega^{n}(t) } \right) & = & \max_{t \in [0, m]} \left( \abs{ \omega_{k}^{n}(t) - \omega^{n}(t) } \right) \nonumber \\ & \le & \max_{t \in [0, n]} \left( \abs{ \omega_{k}^{n}(t) - \omega^{n}(t) } \right) \rightarrow 0 \quad (k \rightarrow \infty) \nonumber \end{eqnarray}\]By lemma 1,
\[\begin{equation} \forall t \in [0, m], \ \omega^{m}(t) = \omega^{n}(t) \label{proposition_02_uniqueness} \end{equation} .\]Now we can define
\[\omega(t) := \sum_{n \in \mathbb{N}} 1_{[n-1, n)}(t) \omega^{n}(t) .\]From \(\eqref{proposition_02_uniqueness}\), $\omega$ is well defined. Moreover, $\omega \in \Omega$. To see this, if $t \in \mathbb{N}$, $\omega$ is left-continuous and right continuous. Continuity at point $t \notin \mathbb{N}$ is obvious from continuity of $\omega^{\lceil t \rceil}$, Thus, $\omega \in \Omega$.
Finally, we will show
\[\rho(\omega, \omega_{k}) \rightarrow 0 .\]Let $\epsilon \in (0, 1)$ be fixed. There exists $n_{0} \in \mathbb{N}$ such that
\[\frac{1}{2^{n_{0} - 1}} < \epsilon .\]Since the definition of $\omega$ and \(\eqref{proposition_02_uniqueness}\),
\[\begin{eqnarray} \max_{t \in [0, n_{1}]} \abs{ \omega(t) - \omega_{n}(t) } = \max_{t \in [0, n_{1}]} \abs{ \omega^{n_{1}}(t) - \omega_{n}(t) } \nonumber \end{eqnarray}\]Thus, for all $n_{1} \in [1:n_{0}]$, there exists $n_{2, n_{1}} \in \mathbb{N}$ such that
\[\begin{eqnarray} \forall n \ge n_{2, n_{1}}, \ \frac{1}{2^{n_{1}}} \max_{t \in [0, n_{1}]} \left( \abs{ \omega(t) - \omega_{n}(t) } \wedge 1 \right) < \frac{\epsilon}{2} \frac{1}{2^{n_{1}}} . \nonumber \end{eqnarray}\]Now, we define $n_{3} := \max_{n_{1} \in [1:n_{0}]} n_{2, n_{1}}$. We have
\[\begin{eqnarray} \forall n \ge n_{3}, \ \forall n_{1} \in [1:n_{0}], \ \frac{1}{2^{n_{1}}} \max_{t \in [0, n_{1}]} \left( \abs{ \omega(t) - \omega_{n}(t) } \wedge 1 \right) < \frac{\epsilon}{2} \frac{1}{2^{n_{1}}} . \nonumber \end{eqnarray}\]Thus, for all $k \ge n_{3}$,
\[\begin{eqnarray} \sum_{n \in \mathbb{N}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega(t) - \omega_{k}(t) } \wedge 1 \right) & = & \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega(t) - \omega_{k}(t) } \wedge 1 \right) + \sum_{n = n_{0} + 1}^{\infty} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega(t) - \omega_{k}(t) } \wedge 1 \right) \nonumber \\ & \le & \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega(t) - \omega_{k}(t) } \wedge 1 \right) + \sum_{n = n_{0} + 1}^{\infty} \frac{1}{2^{n}} \nonumber \\ & \le & \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} \max_{t \in [0, n]} \left( \abs{ \omega(t) - \omega_{k}(t) } \wedge 1 \right) + \frac{1}{2^{n_{0}}} \nonumber \\ & < & \frac{\epsilon}{2} \sum_{n = 1}^{n_{0}} \frac{\epsilon}{2^{n}} + \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \sum_{n = 1}^{n_{0}} \frac{1}{2^{n}} + \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \frac{ \frac{1}{2} \left( 1 - \frac{1}{2^{n_{0}}} \right) }{ 1 - \frac{1}{2} } + \frac{1}{2^{n_{0}}} \nonumber \\ & = & \frac{\epsilon}{2} \left( 1 - \frac{1}{2^{n_{0}}} \right) + \frac{1}{2^{n_{0}}} \nonumber \\ & \le & \frac{\epsilon}{2} + \frac{1}{2^{n_{0}}} \nonumber \\ & \le & \frac{\epsilon}{2} + \frac{\epsilon}{2} \nonumber \\ & = & \epsilon \nonumber . \end{eqnarray}\]Proof of (3) completes.
Definition 3
\[\begin{eqnarray} \mathcal{T}_{s} & := & \{ \tilde{t} := (t_{1}, \ldots, t_{n}) \mid t_{i} \in [0, s], \ t_{i} \neq t_{j}, \ n \in \mathbb{N} \} \nonumber \\ \mathcal{T} & := & \{ \tilde{t} := (t_{1}, \ldots, t_{n}) \mid t_{i} \in [0, \infty), \ t_{i} \neq t_{j}, \ n \in \mathbb{N} \} \nonumber \end{eqnarray}\] \[\begin{eqnarray} \tilde{t} \in \mathcal{T}, \ C(A, \tilde{t}) & := & \{ \omega \in \Omega \mid \omega(\tilde{t}) \in A \}, \nonumber \\ \mathcal{C} & := & \{ C(A, \tilde{t}) \mid n \in \mathbb{N}, \ \tilde{t} \in \mathcal{T}^{n}, \ A \in \mathcal{B}(\mathbb{R}^{n}) \} \nonumber \\ \mathcal{G} & := & \sigma(\mathcal{C}) \nonumber \\ \tilde{t} \in \mathcal{T}_{s}, \ C_{s}(A, \tilde{t}) & := & \{ \omega \in \Omega \mid \omega(\tilde{t}) \in A \}, \nonumber \\ \mathcal{C}_{s} & := & \{ C_{s}(A, \tilde{t}) \mid n \in \mathbb{N}, \ \tilde{t} \in \mathcal{T}^{n}, \ A \in \mathcal{B}(\mathbb{R}^{n}) \} \nonumber \\ \mathcal{G}_{s} & := & \sigma(\mathcal{C}_{s}) \nonumber \\ B(\omega^{\prime}, r) & := & \{ \omega \mid \rho(\omega, \omega^{\prime}) < r \} . \nonumber \\ \mathcal{O} & := & \{ B(\omega, r) \mid \omega \in \Omega, \ r > 0 \} \nonumber \\ \mathcal{B}(\Omega) & := & \sigma(\mathcal{O}) \nonumber . \end{eqnarray}\]$\phi_{t}: C[0, \infty) \rightarrow C[0, \infty)$ is the mamping
\[(\phi_{t}\omega)(s) := \omega(t \wedge s) .\]Remark
\[\begin{eqnarray} \mathcal{B}(\mathbb{R}^{n}) & = & \sigma \left( \{ A_{1} \times \cdots \times A_{n} \mid A_{i} \mathcal{B}(\mathbb{R}) \} \right) \nonumber \\ & = & \sigma \left( \{ (a_{1}, b_{1}) \times \cdots \times (a_{n}, b_{n}) \mid a_{i} < b_{i} \} \right) \end{eqnarray}\]Proposition 4
(1)
\[\mathcal{G} = \mathcal{B}(\Omega) .\](2)
\[\mathcal{G}_{t} = \phi_{t}^{-1}(\mathcal{B}(\Omega)) .\]proof
proof of (1)
We claim that $\mathcal{C} \subseteq \sigma(\mathcal{O})$.
Let $C(\tilde{t}, A) \in \mathcal{C}$ where $\tilde{t} \in \mathcal{T}^{n}$ and $A \in \mathcal{B}(\mathbb{R}^{n})$ be fixed.
\[\begin{eqnarray} \bigcup_{i \in \mathbb{N}} C(\tilde{t}, A_{i}) & = & C \left( \tilde{t}, \bigcup_{i \in \mathbb{N}} A_{i} \right) \nonumber \\ \bigcap_{i \in \mathbb{N}} C(\tilde{t}, A_{i}) & = & C \left( \tilde{t}, \bigcap_{i \in \mathbb{N}} A_{i} \right) \nonumber \end{eqnarray} .\]Let \((\pi_{\tilde{t}})^{-1}(A) \in \mathcal{C}_{T}\) be fixed.
\[\]