[proofplan] The conclusion is Lusin's theorem applied to the response map. We verify Lusin's hypotheses: $\mu$ must be a *Radon* probability measure and the response map must be Borel measurable. Radonness of $\mu$ follows from $\sigma$-compactness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ — which gives tightness — combined with separability and Bogachev's criterion (a separable space on which every Borel measure is tight is automatically Radon). Borel measurability of the response map $\Psi_{y_0,f}: [x] \mapsto y_b$ is established in the existence-and-uniqueness proposition for controlled differential equations. Lusin's theorem for finite Radon measures then produces the required compact set. [/proofplan] [step:Set up the response map and recall its measurability] Let the controlled differential equation \begin{align*} dy_t &= f(y_t)\, dx_t, \qquad y_a = y_0, \qquad t \in [a, b], \end{align*} be as in the existence and uniqueness proposition for CDEs, with vector field $f$ and initial condition $y_0$ fixed. Define the response map \begin{align*} \Psi_{y_0, f}: \mathcal{C}_1 &\to W \\ [x] &\mapsto y_b, \end{align*} where $W$ is the codomain of the solution at the terminal time and $y_b$ is the value at $b$ of the unique solution to the CDE driven by $x$. By the [Existence and Uniqueness for CDEs](/theorems/???), $\Psi_{y_0, f}$ is well-defined (independent of representative) and Borel measurable as a map $(\mathcal{C}_1, \mathcal{B}(\chi_{\mathrm{pr}})) \to (W, \mathcal{B}(W))$. [/step] [step:Verify $\sigma$-compactness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ to obtain tightness of $\mu$] By the [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507), there exists an increasing sequence of compact sets $K_n \subseteq \mathcal{C}_1$ with $\mathcal{C}_1 = \bigcup_{n=1}^\infty K_n$ (take $K_n := \bigcup_{r=1}^n B(r)$, a finite union of compact sets, hence compact). Fix $\delta > 0$. The sequence of complements $\mathcal{C}_1 \setminus K_n$ is decreasing with empty intersection, so by continuity of the probability measure $\mu$ from above \begin{align*} \mu(\mathcal{C}_1 \setminus K_n) \downarrow \mu\left(\bigcap_{n=1}^\infty (\mathcal{C}_1 \setminus K_n)\right) = \mu(\varnothing) = 0. \end{align*} Hence there exists $n_0$ such that $\mu(\mathcal{C}_1 \setminus K_{n_0}) < \delta/2$. We have shown that $\mu$ is **tight**: for every $\delta > 0$ there is a compact set $K_{n_0}$ with $\mu(K_{n_0}) > 1 - \delta/2$. [guided] We need tightness of $\mu$ as one of the two hypotheses of Bogachev's criterion (separability is the other). Tightness says: for every $\delta > 0$ there is a compact $K \subseteq \mathcal{C}_1$ carrying mass at least $1 - \delta$. How do we manufacture such a $K$? The key structural fact is the [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507): $\mathcal{C}_1$ is a countable union of compact sets, \begin{align*} \mathcal{C}_1 = \bigcup_{n=1}^\infty K_n, \end{align*} where we can take $K_n := \bigcup_{r=1}^n B(r)$ — the union of finitely many compact balls of radii $1, \ldots, n$ — and the $K_n$ are increasing, compact, and exhaust the space. Now fix $\delta > 0$. The complements $A_n := \mathcal{C}_1 \setminus K_n$ form a *decreasing* sequence ($A_1 \supseteq A_2 \supseteq \cdots$) with $\bigcap_n A_n = \mathcal{C}_1 \setminus \bigcup_n K_n = \varnothing$. The continuity-from-above property of probability measures (continuity at decreasing sequences with finite measure on the first term — guaranteed since $\mu(A_1) \leq 1 < \infty$) gives \begin{align*} \lim_{n \to \infty} \mu(A_n) = \mu\Bigl(\bigcap_{n=1}^\infty A_n\Bigr) = \mu(\varnothing) = 0. \end{align*} Hence there is $n_0$ with $\mu(A_{n_0}) = \mu(\mathcal{C}_1 \setminus K_{n_0}) < \delta/2$, equivalently $\mu(K_{n_0}) > 1 - \delta/2$. Why $\delta/2$ rather than $\delta$? The strategy is to set up two-step approximation: this step delivers a compact set capturing $1 - \delta/2$ of the mass, leaving a "budget" of $\delta/2$ for the next step (extracting a slightly smaller compact set on which $\Psi_{y_0,f}$ is continuous, via Lusin). We will combine the two budgets in the final step. Note that the argument uses *only* $\sigma$-compactness, not the specific identity of $\mu$. Hence the same argument shows that **every** Borel probability measure on $\mathcal{C}_1$ is tight — a fact we will exploit in Step 4 when invoking Bogachev's criterion. [/guided] [/step] [step:Verify separability of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$] The topology $\chi_{\mathrm{pr}}$ is the initial topology induced by the family of signature-projection maps \begin{align*} \pi_m: \mathcal{C}_1 &\to V^{\otimes m} \\ [x] &\mapsto S(x)^{(m)}, \end{align*} for $m \geq 0$, where each $V^{\otimes m}$ is a finite-dimensional normed space, hence second-countable. By Step 2, $\mathcal{C}_1 = \bigcup_n K_n$ is a countable union of compact sets, each of which is metrisable (a compact subspace of an initial topology induced by a countable family of maps to second-countable spaces is metrisable). Each $K_n$ is therefore separable, and a countable union of separable subspaces is separable. Hence $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is separable. [guided] We need to verify separability of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ to invoke Bogachev's criterion in the next step. Why is separability not immediate? The topology $\chi_{\mathrm{pr}}$ is the *coarsest* topology making every $\pi_m$ continuous. Coarse topologies tend to have *more* convergent sequences (easier to converge), and hence many open sets are "small" in the metric sense — but the global structure can be subtle. The argument proceeds in two stages. First, observe that each level $V^{\otimes m}$ is a finite-dimensional normed space, hence second-countable (it has a countable basis: balls of rational radius around points with rational coordinates). So the family $(\pi_m)_{m \geq 0}$ maps into spaces with countable bases. Second, on any compact subset $K \subseteq \mathcal{C}_1$, the topology $\chi_{\mathrm{pr}}|_K$ is metrisable. The reason: an initial topology induced by a countable family of continuous maps to metrisable spaces is itself metrisable on any subspace where the family separates points (this is a standard fact about embeddings into countable products of metrisable spaces). Compact metrisable spaces are separable. Now apply this to the countable cover $\mathcal{C}_1 = \bigcup_n K_n$ from Step 2. Each $K_n$ is separable in the subspace topology, so it has a countable dense subset $D_n$. The countable union $D := \bigcup_n D_n$ is dense in $\bigcup_n K_n = \mathcal{C}_1$: any non-empty open $U \subseteq \mathcal{C}_1$ meets some $K_n$ in a relatively open set, which contains a point of $D_n \subseteq D$. Hence $D$ is a countable dense subset of $\mathcal{C}_1$, proving separability. [/guided] [/step] [step:Conclude that $\mu$ is a Radon measure via Bogachev's criterion] We invoke [Bogachev's Criterion for Radonness (Bogachev, Proposition 7.2.2)](/theorems/???). The criterion states: every Borel probability measure on a separable topological space in which every Borel probability measure is tight is a Radon measure. We verify the hypotheses: \begin{enumerate} \item[(i)] **Separability**: established in Step 3. \item[(ii)] **Tightness of $\mu$**: established in Step 2. \end{enumerate} Therefore $\mu$ is a Radon measure on $(\mathcal{C}_1, \chi_{\mathrm{pr}})$. [guided] We want to apply Lusin's theorem, but the form we need requires $\mu$ to be **Radon** — i.e., every Borel set's measure can be approximated from inside by compact sets: \begin{align*} \mu(A) = \sup\{\mu(K) : K \subseteq A,\ K \text{ compact}\} \quad \text{for every } A \in \mathcal{B}(\mathcal{C}_1). \end{align*} Tightness alone is weaker than Radonness — tightness gives one compact set absorbing most of the mass of the *whole space*, but Radonness demands inner regularity at *every* Borel set. We close the gap with [Bogachev's Criterion for Radonness (Bogachev, Proposition 7.2.2)](/theorems/???). **Statement of the criterion.** Let $X$ be a separable topological space such that every Borel probability measure on $X$ is tight. Then every Borel probability measure on $X$ is Radon. **Verifying the hypotheses for $X = \mathcal{C}_1$ with $\chi_{\mathrm{pr}}$.** (i) **Separability of $X$.** Established in Step 3: the space $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ admits the countable dense set $D = \bigcup_n D_n$, where $D_n$ is a countable dense subset of $K_n$ in the metrisable subspace topology. (ii) **Every Borel probability measure on $X$ is tight.** We did the argument of Step 2 specifically for $\mu$, but it used only the $\sigma$-compactness $\mathcal{C}_1 = \bigcup_n K_n$ and the abstract continuity-from-above of probability measures. Replacing $\mu$ by an arbitrary Borel probability $\nu$ on $\mathcal{C}_1$ in Step 2 yields the same conclusion: for every $\delta > 0$ there is $n_0$ with $\nu(K_{n_0}) > 1 - \delta/2$. Hence every Borel probability on $\mathcal{C}_1$ is tight. **Conclusion.** Both hypotheses are satisfied, so Bogachev's criterion applies. Hence $\mu$ is a Radon measure on $(\mathcal{C}_1, \chi_{\mathrm{pr}})$: for every Borel $A \subseteq \mathcal{C}_1$, \begin{align*} \mu(A) = \sup\{\mu(K) : K \subseteq A,\ K \text{ compact}\}. \end{align*} **Why does the criterion need separability?** Tightness gives only a single global compact set; promoting it to inner regularity at *every* Borel set requires being able to approximate Borel sets themselves from inside. Separability provides a countable basis of the topology on each compact $K_n$, which lets one perform a countable open-cover argument inside the tight reference compact and refine to inner-regular compact approximations at every Borel set. Without separability, one would have only the global tightness statement and could not localise it. [/guided] [/step] [step:Apply Lusin's theorem for finite Radon measures] We invoke [Lusin's Theorem for Finite Radon Measures (Bogachev, Theorem 7.1.13)](/theorems/???). The statement we use: if $\nu$ is a finite Radon measure on a topological space $X$ and $g: X \to Y$ is a Borel measurable map into a metric space $Y$, then for every $\delta > 0$ there exists a compact set $K \subseteq X$ such that \begin{align*} \nu(X \setminus K) < \delta \quad \text{and} \quad g\big|_K \text{ is continuous}. \end{align*} We verify the hypotheses with $X = \mathcal{C}_1$, $\nu = \mu$, $g = \Psi_{y_0, f}$, and $Y = W$: \begin{enumerate} \item[(i)] **$\mu$ is a finite Radon measure**: $\mu$ is a Borel probability measure (so finite, with $\mu(\mathcal{C}_1) = 1$) and Radon by Step 4. \item[(ii)] **$\Psi_{y_0, f}$ is Borel measurable**: established in Step 1. \item[(iii)] **$Y = W$ is a metric space**: $W$ is a finite-dimensional normed vector space (the codomain of solutions of the CDE), in particular metrisable. \end{enumerate} By Lusin's theorem applied to this data, for the given $\delta > 0$ there exists a compact set $K \subseteq \mathcal{C}_1$ with \begin{align*} \mu(\mathcal{C}_1 \setminus K) < \delta \quad \text{and} \quad \Psi_{y_0, f}\big|_K = ([x] \mapsto y_b)\big|_K \text{ is continuous}. \end{align*} This is the desired conclusion, completing the proof. [guided] We have assembled all the inputs needed for Lusin: a Radon probability $\mu$ (Step 4), a Borel-measurable response map $\Psi_{y_0, f}$ (Step 1), and a metric target space $W$. Now we cash these in. We invoke [Lusin's Theorem for Finite Radon Measures (Bogachev, Theorem 7.1.13)](/theorems/???). **Statement.** If $\nu$ is a finite Radon measure on a topological space $X$ and $g: X \to Y$ is a Borel measurable map into a metric space $Y$, then for every $\delta > 0$ there exists a compact set $K \subseteq X$ with \begin{align*} \nu(X \setminus K) < \delta, \qquad g\big|_K \text{ continuous (in the subspace topology of } K\text{)}. \end{align*} **Verification of hypotheses with $X = \mathcal{C}_1$, $\nu = \mu$, $g = \Psi_{y_0,f}$, $Y = W$.** (i) **$\mu$ is a finite Radon measure on $\mathcal{C}_1$.** Finiteness: $\mu$ is a probability measure, so $\mu(\mathcal{C}_1) = 1 < \infty$. Radonness: established in Step 4 via Bogachev's criterion. (ii) **$\Psi_{y_0, f}: \mathcal{C}_1 \to W$ is Borel measurable.** Established in Step 1 by invoking the [Existence and Uniqueness for CDEs](/theorems/???); the response map is well-defined (independent of representative) and Borel measurable as a map $(\mathcal{C}_1, \mathcal{B}(\chi_{\mathrm{pr}})) \to (W, \mathcal{B}(W))$. (iii) **$Y = W$ is a metric space.** $W$ is a finite-dimensional normed vector space — the codomain of the CDE solution at the terminal time — hence metrisable in its norm topology. **Apply the theorem.** With all hypotheses verified, Lusin's theorem produces, for the given $\delta > 0$, a compact $K \subseteq \mathcal{C}_1$ with \begin{align*} \mu(\mathcal{C}_1 \setminus K) < \delta \qquad \text{and} \qquad \Psi_{y_0,f}\big|_K \text{ continuous}. \end{align*} This is exactly the statement of the theorem we set out to prove, so the proof is complete. **Why is this the right form of Lusin?** There is a more familiar version for $X = \mathbb{R}^n$ with $\mu$ Lebesgue, but it relies on inner regularity that comes for free from local compactness and second countability. Our space $\mathcal{C}_1$ has neither in any obvious form (it sits inside an infinite-dimensional ambient), so we needed to recover Radonness by hand via Bogachev's criterion before invoking the abstract Lusin. [/guided] [/step]