When does the Borel $\sigma$-algebra of compact convergence coincide with the product $\sigma$-algebra?

Question

Let $X$, $Y$ be topological spaces, and $C(X,Y)$ the set of continuous functions $ X \to Y $, equipped with the compact-open topology. Let $\newcommand\Bco{\mathcal B_{\textrm{c-o}}} \Bco$ be the corresponding Borel $\sigma$-algebra on $C(X,Y)$, and $\newcommand\Spt{\mathcal \Sigma_{\textrm{pt}}} \Spt$ the product $\sigma$-algebra, to wit the smallest one for which evaluation at $x$ is measurable $ C(X,Y) \to Y $ for every $ x \in X $, when $Y$ carries its Borel $\sigma$-algebra.

Since points are compact, $ \Spt \subseteq \Bco $. When does $ \Spt = \Bco $ hold?

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This question is motivated by a previous one in the context of probability (hence the tag). Indeed, in the above if $Y$ is a uniform space, then the topology of compact convergence on $C(X,Y)$ is precisely the compact-open topology.

Note furthermore that the Borel $\sigma$-algebra $\newcommand\Bpt{\mathcal B_{\textrm{pt}}} \Bpt$ induced by the product topology lies between $\Spt$ and $\Bco$: $$ \Spt \subseteq \Bpt \subseteq \Bco \text. $$

Answer 1

Sufficient conditions are

• $X$, $Y$ second-countable,
• $X$ locally compact (each point has a local base of compact sets),
• $Y$ regular.

Second-countability and local compactness render $C(X,Y)$ second-countable, and regularity and first-countability allow to control elements of the countable basis of $C(X,Y)$ through the evaluation maps.

1. For $ A \subseteq X $ and $ B \subseteq Y $, let $ M(A,B) \subseteq C(X,Y) $ be the set of functions which send $A$ into $B$. The sets $M(K,V)$ with $K$ compact and $V$ open form a subbase for the compact-open topology. When $X$ is locally compact, the compact-open topology on $C(X,Y)$ admits a subbase of sets $M(K,V)$ of which the cardinality does not exceed that of infinite subbases for $X$ and $Y$ [Engelking, General Topology, Theorem 3.4.16].

   Thus if $X$, $Y$ are second-countable, so is $C(X,Y)$, and to conclude $ \newcommand\Spt{\mathcal \Sigma_{\textrm{pt}}} \newcommand\Bco{\mathcal B_{\textrm{c-o}}} \Spt = \Bco $ we can show $ M(K,V) \in \Spt $ for $K$, $V$ as above.

2. Let $ V_n \subseteq Y $ form a countable base. As $Y$ is regular, each open $ V \subseteq Y $ is a countable union of some of the closures, $$ V = \bigcup_{\overline{V_n} \subseteq V} \overline{V_n} = \bigcup_{\overline{V_n} \subseteq V} V_n \text. $$ We can suppose that the $V_n$ are closed under finite unions; then if $ L \subseteq Y $ is compact, $ L \subseteq V $ if and only if $ L \subseteq \overline{V_n} $ for some $ \overline{V_n} \subseteq V $. Hence with the above notation, $$ M(K,V) = \bigcup_{\overline{V_n} \subseteq V} M(K,\overline{V_n}) \text, $$ and to conclude $ M(K,V) \in \Spt $ we can show $ M(K,F) \in \Spt $ for closed $ F \subseteq X $.

3. Let $ Q \subseteq X $ be a countable dense subset. Then $$ M(K,F) = M(Q \cap K, F) = \bigcap_{q \in Q \cap K} M(q,F) \text, $$ and $ M(q,F) \in \Spt $. Hence $ M(K,F) \in \Spt $, and $ \Spt = \Bco $.