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Let us consider a stochastic process $X_{t}$ on $(\Omega, \mathcal{F}, P)$, taking values in $\mathbb{R}$, indexed by $t \in [0, \infty)$ whose paths are continuous. Let $C_{[0, \infty)}$ be the set of all continuous functions $[0, \infty) \to \mathbb{R}$.

Next, let $\mathcal{B}_{1}$ be the smallest $\sigma$-algebra on $C_{[0, \infty)}$ such that for all $t_{0} \in [0, \infty)$ the coordinate mappings $X_{t_{0}}: \omega \to X_{t_{0}}(\omega)$ are measurable. Let $\mathcal{B}_{2}$ be the Borel $\sigma$-algebra on $C_{[0, \infty)}$ generated by the topology of uniform convergence on compacts.

Prove that $\mathcal{B}_{1} = \mathcal{B}_{2}$.

J. M
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1 Answers1

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The main ingredient of the proof is separability of $C[0,\infty )$ with the topology of uniform convergence of compact sets (ucc). If $f \in C[0,\infty )$ and $n \geq 1$ there exists a polynomial $p_n$ with rational coefficients such that $\sup \{|f(x)-p_n(x)|:0\leq x \leq n\} <\frac 1 n $. It follows that $\{p_n\}$ converges to $f$ uniformly on compact sets. [ This space is a separable metric space: a metric for this topology is $d(f,g)=\sum \frac 1 {2^{n}} \frac {\sup \{|f(x)-g(x)|:0\leq x \leq n\}} {1+\sup \{|f(x)-g(x)|:0\leq x \leq n\}}$]. Now note that $f \to f(t)$ is continuous, hence measurable, w.r.t. the topology of ucc. Hence $\mathcal B_1 \subset \mathcal B_2$. To prove the other way note that any open set w.r.t. ucc is a countable union of basic open sets of the type $\{f:\sup \{|f(x)-f_0(x)|:0\leq x \leq n\}<r\}$ where $f_0 \in C[0,\infty ),n \geq 1$ and $r>0$. [This is where we use separability]. By continuity it is enough take the supremum in this set over rationals numbers $x$ in $[0,n]$ . You should now be able to see why this basic open set belongs to $\mathcal B_1$. Hence all open sets in ucc belong to $\mathcal B_1$ and $\mathcal B_2 \subset \mathcal B_1$.

Kavi Rama Murthy
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  • I am following a probability course in which it is claimed without proof that the space of continuous functions $[0,\infty) \to S$ with the topology of uniform convergence on compact sets is separable if $S$ is a separable metric space. When $S = \mathbb R$ or $\mathbb C$ this can be proven as you do, by using Stone–Weierstraß on a countable increasing covering of $[0,\infty)$ by good compact sets. This works in fact when $[0,\infty)$ is replaced by any hemicompact Hausdorff space.

    Would you know if this does indeed remain true for a general separable metric space $S$?

     – Olius Oct 15 '23 at 22:29
  • Never mind, of course right after posting the comment I finally found a reference: Engelking 3.4.16 shows that when $X$ is locally compact, $Y^X$ with the compact-open topology admits a basis of which the cardinality does not exceed that of bases for $X$ and $Y$. The source given there is Arens (1946). – Olius Oct 15 '23 at 23:00