Lemma 1.1.3: {{\cite[Lemma 8.1.1]{stroock}}} | Brownian Motion and Stochastic Calculus

Lemma 1.1.3 ([Lemma 8.1.1, Str24]).label The space $\sps$ is a Borel set in $C([0, \infty); \real^{d})$, and $\cb(\sps)$ is the restriction of the Borel $\sigma$-algebra on $C([0, \infty); \real^{d})$ to $\sps$.

Proof. Firstly, $\lim_{t \to \infty}\theta(t)/t = 0$ if and only if for every $n \in \nat$, there exists $k \in \nat$ such that $\abs{\theta(s)}/s \le 1/n$ for all $s \ge k$. Thus

\begin{align*}\Theta_{\infty}&= \bracs{\theta \in C([0, \infty); \real^d): \limv{t}\frac{\theta(t)}{t} =0}\\&= \bigcap_{n \in \nat}\bigcup_{k \in \nat}\underbrace{\bigcap_{s \ge k}\bracs{\theta \in C([0, \infty); \real^d): \frac{\abs{\theta(s)}}{s} \le \frac{1}{n}}}_{\text{closed}}\end{align*}

and

\[\Theta_{0} = \bracs{\theta \in C([0, \infty); \real^d): \theta(0) = 0}\]

are Borel. Thus $\sps = \Theta_{0} \cap \Theta_{\infty} \cap \Theta_{b}$ is Borel in $C([0, \infty); \real^{d})$.

Now, by lower semicontinuity of $\norm{\cdot}_{\sps}$, every closed set in $\sps$ is the intersection of $\sps$ and a Borel set in $C([0, \infty), \real^{d})$. Therefore $\cb(\sps)$ contains the induced $\sigma$-algebra from $\cb(C([0, \infty); \real^{d}))$. On the other hand, the inclusion $\iota: \sps \to C([0, \infty); \real^{d})$ is continuous. Thus $\cb(\sps)$ contains the induced $\sigma$-algebra.$\square$