The $\sigma$-algebra $ \mathscr{B}\left(C\left[0,\infty\right)\right)$ is generated by all finite-dimensional cylinders.

Question

Let $C\left[0,\infty\right)$, the space of all continuous, real-valued functions on $\left[0,\infty\right)$ with metric

$\rho\left(\omega_{1},\omega_{2}\right)=\sum_{n=1}^{\infty}\frac{1}{2^{n}}\max_{0\le t\le n}\left(\min\left\{ 1,\left|\omega_{1}\left(t\right)-\omega_{2}\left(t\right)\right|\right\} \right)$

I have proved that Under $\rho$, $C\left[0,\infty\right)$ is a complete, separable metric space

Let $\mathscr{C}$ be the collection of finite-dimensional cylinder sets i.e.,

$C=\left\{ \omega\in C\left[0,\infty\right):\left(\omega\left(t_{1}\right),\cdots,\omega\left(t_{n}\right)\right)\in A\right\} ,n\geq1,A\in\mathscr{B}\left(\mathbb{R}^{n}\right) $

where, for all $i=1,2,\cdots,n, t_{i}\in\left[0,\infty\right)$. Denote by $\mathscr{G}$ the smallest $\sigma$-field containing $\mathscr{C}$. Now we need to show that $\mathscr{B}\left(C\left[0,\infty\right)\right)=\mathscr{G}$.

I have proved that $\mathscr{G}\subset\mathscr{B}\left(C\left[0,\infty\right)\right)$. To prove the converse statement. By Karatzas&Shreve Brownian Motion and Stochastic Calculus, I just need to show that $B\left(\omega_{0},\epsilon\right)=\left\{ \omega\in C\left[0,\infty\right):\rho\left(\omega_{0},\omega\right)<\epsilon\right\} \in \mathscr{G}$. Let $Q$ be the set of rationals in $\left[0,\infty\right)$. We have

$ B\left(\omega_{0},\epsilon\right) =\left\{ \omega:\sum_{n=1}^{\infty}\frac{1}{2^{n}}\sup_{0\le t\le n,t\in Q}\left(\min\left\{ 1,\left|\omega\left(t\right)-\omega_{0}\left(t\right)\right|\right\} \right)<\epsilon\right\}$

But how can I show the set on the right is in $\mathscr{G}$? Can anyone give a hint? Many thanks in advance.

Answer 1

Statements like this are typically proved by inspecting all the "building blocks" of the object being defined, often from the inside out, and checking that they all preserve measurability.

• For each $t$, the map from $C[0,\infty)$ to $\mathbb{R}$ defined by $\omega \mapsto \omega(t)$ is measurable with respect to $\mathcal{C}$.

• The function from $\mathbb{R}$ to $\mathbb{R}$ defined by $x \mapsto \min\{1, |x-\omega_0(t)|\}$ is Borel, so its composition with measurable functions is measurable. Thus $\omega \mapsto \min\{1, |\omega(t) - \omega_0(t)|$ is measurable with respect to $\mathcal{C}$.

• The pointwise supremum of countably many measurable functions is measurable. So $\omega \mapsto \sup_{0 \le t \le n, t \in \mathbb{Q}} \min\{1, |\omega(t) - \omega_0(t)|$ is measurable with respect to $\mathcal{C}$.

• Scalar multiples and infinite sums of measurable functions are measurable. So the map $\omega \mapsto \sum_{n=1}^{\infty}\frac{1}{2^{n}}\sup_{0\le t\le n,t\in Q}\min\left\{ 1,\left|\omega\left(t\right)-\omega_{0}\left(t\right)\right|\right\} $ is measurable.

• Your set $B(\omega_0, \epsilon)$ is the preimage of the Borel set $[0,\epsilon)$ under that measurable function.

Answer 2

On the Banach space structure directly, using the fact that a continuous function is the uniform limit of a rational coefficient polynomial sequence on any segment $[a,b]$.

For $2$ cylinder sets: let $C=\left\{\left(x_t\right)_{t\in[0,T]}| \left(x_{t_1},x_{t_2}\right)\in]a_1,b_1[\times]a_2,b_2[\right\}$ a the cylinder set. We can write $C=C_1\cap C_2$ where $$ C_1=\left\{(x_t)_{t\in[0,T]}|x_{t_1}\in]a_1,b_1[\right\} $$ and $$ C_2=\left\{(x_t)_{t\in[0,T]}| x_{t_2}\in]a_2,b_2[\right\}. $$

Now, let consider $F_1$ the set of all the functions $f$ such that $f(t_1)=\frac12(a_1+b_1)$, i.e. $$ F_1=\left\{f\in\mathcal C([0,T],\mathbb R)\quad|\quad f(t_1)=\frac12(a_1+b_1)\right\}. $$

We have $C_1=\bigcap_{f\in F_1}{\rm B}(f,\frac12(b_1-a_1))$ which is not denumerable. But for $f\in F_1$, by Weierstrass' theorem, $f=\lim_{\|\cdot\|_\infty,n\to\infty}P_n$ for some sequence $(P_n)_{n\in\mathbb N}\in\mathbb R[X]^\mathbb N$.

We can also impose for all $n$ in $\mathbb N$, $P_n(t_1)=\frac12(a_1+b_1)$ by translation and impose the coefficients of $P_n$ to be rationnals by decimal approximation of the real coefficients. Thus let $$ \tilde F_1=\left\{P\in \mathbb Q[X] \quad|\quad P(t_1)=\frac12(a_1+b_1)\right\}. $$ We get that $C_1=\bigcap_{P\in \tilde F_1}{\rm B}(P,\frac12(b_1-a_1))$ which is denumerable so $C_1\in\mathcal B\left(\mathcal C([0,T],\mathbb R)\right)$. As $C_2$ too, we obtain that $C$ belongs to the Borel $\sigma$-field.

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Conversely, let consider $f\in\mathcal C([0,T],\mathbb R)$ and ${\rm B}(f,\varepsilon)$ the tube of diameter $2\varepsilon$ around $f$. Let $(t_n)_{n\in\mathbb N}$ dense in $[0,T]$, by uniform continuity we have $$ {\rm B}\left(f,\varepsilon\right) = \bigcap_{n\in\mathbb N}C_{n,\varepsilon}\quad\text{where}\quad C_{n,\varepsilon} = \left\{(x_t)_{t\in[0,T]}\;|\; |x_{t_n}-f(t_n)|<\varepsilon\right\}. $$ Thus $\mathcal B \left(\mathcal C([0,T],\mathbb R)\right)\subset\sigma(\text{cylinder sets})$ and the Borel $\sigma$-field is generated by the cylinder sets.