• $(M,d)$ be a separable metric space
  • $E$ be a $\mathbb R$-Banach space
  • $\alpha\in(0,1]$

Moreover, let $$\left\|f\right\|_{C^{0+\alpha}(K,\:E)}:=\sup_{x\in K}\left\|f(x)\right\|_E+\sup_{\substack{x,\:y\:\in\:K\\x\:\ne\:y}}\frac{\left\|f(x)-f(y)\right\|_E}{{d(x,y)}^\alpha}\;\;\;\text{for }f:M\to E$$ for $K\subseteq M$ and $$C^{0+\alpha}(M,E):=\left\{f:M\to E\mid\left\|f\right\|_{C^{0+\alpha}(K,\:E)}<\infty\text{ for all compact }K\subseteq M\right\}$$ be equipped with the topology generated by $$\left\{\left\|\;\cdot\;\right\|_{C^{0+\alpha}(K,\:E)}:K\subseteq M\text{ is compact}\right\}.$$

I want to show that $$\mathcal B\left(C^{0+\alpha}(M,E)\right)=\left.{\mathcal B(E)}^{\otimes M}\right|_{C^{0+\alpha}(M,\:E)};$$ at least in the special case $M=\Lambda$ for an open subset $\Lambda\subseteq\mathbb R^n$, $n\in\mathbb N$.

The desired result is motivated by the following fact: If $M=[0,\infty)$ and $E=\mathbb R$, the space $\Omega:=C^0([0,\infty))$ of continuous functions from $[0,\infty)$ to $\mathbb R$, equipped with the topology of uniform convergence on compact subsets, is a Polish space. Let $$X_t:\Omega\to\mathbb R\;,\;\;\;\omega\mapsto\omega(t)$$ for $t\ge0$. Then we are able to show that $$\left.{\mathcal B(\mathbb R)}^{\otimes[0,\infty)}\right|_\Omega=\sigma(X_t,t\in[0,\infty))=\mathcal B(\Omega).$$

$^1$ Let $\mathcal B(X)$ denote the Borel $\sigma$-algebra on a topological space $X$. If $I$ is a nonempty set and $(\Omega_i\mathcal A_i)$ is a measurable space for all $i\in I$, let $\bigotimes_{i\in I}\mathcal A_i$ denote the product $\sigma$-algebra on $\prod_{i\in I}\Omega_i$. If $(\Omega_i\mathcal A_i)=(\Omega,\mathcal A)$ for all $i\in I$, write ${\mathcal A}^{\otimes I}$ instead of $\bigotimes_{i\in I}\mathcal A$. Moreover, for any measurable space $(\Omega,\mathcal A)$ and $A\subseteq\Omega$, let $\left.\mathcal A\right|_A$ denote the trace $\sigma$-algebra on $A$.

  • $\begingroup$ I don't have a counterexample, but I'm skeptical. The $\sigma$-algebra on the right side could be viewed as being generated by the evaluation maps $\pi_t(\omega) = \omega(t)$, $t \in M$. In particular, the RHS is contained in the "cylinder" $\sigma$-algebra generated by the continuous linear functionals on your space $X=C^{0+\alpha}$. This coincides with the Borel $\sigma$-algebra for, say, separable Banach spaces, but for non-separable spaces it's typically smaller. And $C^{0+\alpha}$ is non-separable. $\endgroup$ Commented Jul 4, 2018 at 17:10
  • $\begingroup$ @NateEldredge The non-separability is exactly my problem. The proof of the mentioned result $\left.{\mathcal B(\mathbb R)}^{\otimes[0,\infty)}\right|_\Omega=\mathcal B(\Omega)$ that I'm aware of relies on the separability of $\Omega$. $\endgroup$
    – 0xbadf00d
    Commented Jul 4, 2018 at 22:29

1 Answer 1


It's not true.

For brevity, let me take $M=[0,1]$ (or $(0,1)$ if you prefer), $E = \mathbb{R}$, $X = C^{0+\alpha}(M,E)$, $\mathcal{B}$ its $\sigma$-algebra, and $\mathcal{F}$ the $\sigma$-algebra appearing on the right side of your desired equation. I claim that $|\mathcal{F}| = \mathfrak{c}$ while $|\mathcal{B}| \ge 2^{\mathfrak{c}}$.

For $t \in [0,1]$, let $\pi_t : X \to \mathbb{R}$ be the evaluation map $\pi_t(\omega) = \omega(t)$ which is continuous. Let $\mathcal{U}$ be a countable base for $\mathbb{R}$. Then $\mathcal{F}$ is generated by $\{\pi_t^{-1}(U) : t \in [0,1], U \in \mathcal{U}\}$. In fact, I claim it is generated by the countable collection $\{\pi_q^{-1}(U) : q \in [0,1] \cap \mathbb{Q}, U \in \mathcal{U}\}$. For let $\mathcal{F}_0 \subseteq \mathcal{F}$ be the $\sigma$-algebra generated by the latter collection; note that $\pi_q$ is $\mathcal{F}_0$-measurable for $q \in [0,1] \cap \mathbb{Q}$. If $t \in [0,1] \setminus \mathbb{Q}$, choose a sequence $[0,1] \cap \mathbb{Q} \ni q_n \to t$; then note that $\pi_{q_n} \to \pi_t$ pointwise on $X$, because $X$ is a space of continuous functions. So $\pi_t$ is $\mathcal{F}_0$-measurable, and thus $\mathcal{F}_0$ contains the sets $\pi_t^{-1}(U)$. Hence $\mathcal{F} \subset \mathcal{F}_0$.

We have thus shown that $\mathcal{F}$ is countably generated, so its cardinality is $\mathfrak{c}$. However, as noted here, the Hölder space contains continuum many functions at pairwise distance $\ge 1$ (in your version of the norm, the distances are a little greater still). Any set of these functions is closed and thus Borel, so there are at least $2^{\mathfrak{c}}$ Borel sets in $X$.

Moral: $C^{0+\alpha}(M)$ is not really similar to $C^0(M)$ and one shouldn't expect the same properties.

  • $\begingroup$ Then I'm absolutely puzzled with the whole regularity theory of martingales with a spatial parameter as presented in Kunita's book Stochastic Flows and Stochastic Differential Equations. I've got profound problems with the book anyhow (see this question, if you're interested), but one elementary thing is that I'm not sure what he means by a $C^{0+\alpha}$-valued process $Y$ on a measurable space $(S,\mathcal S)$. Usually, this should mean that $Y$ is a collection of $(\mathcal S,\mathcal B(C^{0+\alpha}))$-measurable random variables. $\endgroup$
    – 0xbadf00d
    Commented Jul 4, 2018 at 22:30
  • $\begingroup$ But there is an ambiguity and since you've shown that the result of this question is not true, he doesn't seem to mean that. However, than I think that some things he's doing won't work. (I know that it's not really related to this question, but could you tell me, if you know any other book which considers the regularity of martingales with a spatial parameter? It seems like Kunita's book is the only one out there and everybody is citing him.) $\endgroup$
    – 0xbadf00d
    Commented Jul 4, 2018 at 22:34
  • $\begingroup$ Sorry, I don't know another book. One other interpretation of $C^{0+\alpha}$ process could just be "a process, i.e. a family of $E$-valued random variables indexed by $M$, whose sample paths are in $C^{0+\alpha}$". It could also be the case, as with Brownian motion, that the paths are a.s. in some separable closed subspace $X_0$ of $C^{0+\alpha}$, in which case the Borel $\sigma$-algebra on $X_0$ may be better behaved. $\endgroup$ Commented Jul 5, 2018 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.