# Borel $\sigma$-algebras on paths of bounded variation

Let $$(C, \|\cdot\|)$$ be the Banach space of continuous paths $$x: [0,1]\rightarrow\mathbb{R}^d$$ starting at zero with sup-norm $$\|\cdot\|$$.

Let further $$B\subset C$$ be the subspace of $$0$$-started continuous paths of bounded variation, i.e.

$$B = \big\{x\in C \ \big| \ \|x\|_1:=\sup\nolimits_{(t_\nu)}{\textstyle\sum_\nu}|x_{t_\nu} - x_{t_{\nu-1}}|<\infty \ \text{ and } \ x(0)=0\big\}$$

where the above sup runs over all partitions $$(t_\nu)$$ of $$[0,1]$$.

The set $$B$$ can be endowed with both the (uniform) norm $$\|\cdot\|$$ and the (1-variation) norm $$\|\cdot\|_1$$.

Do you know if the Borel $$\sigma$$-algebras on $$(B, \|\cdot\|)$$ and $$(B, \|\cdot\|_1)$$ coincide?

Any references or hints are welcome.

Edit: As pointed out by Gerald Edgar, we narrowed $$B$$ down to paths started at zero to make the question more meaningful.

Edit 2: As Gerald Edgar pointed out in the comments to his answer below, the answer to this question is no, the Borel $$\sigma$$-algebras cannot coincide because there are more $$\|\cdot\|_1$$-open sets in $$B$$ than there are $$\|\cdot\|$$-Borel sets in $$C$$.

I think there is a problem with a mere semi-norm. The constant functions have variation distance $$0$$ from each other. Any variation-open set contains either all the constants, or none of them. Therefore, any variation-Borel set contains either all the constants, or none of them.

Choose $$\mathbf u \in \mathbb R^n \setminus \{0\}$$. Take the subset $$\{0,\mathbf{u}\} \subseteq B$$ of two constant functions. It is sup-norm closed, so it is a uniform-Borel set. But it is not a variation-Borel set.

Maybe a better question would restrict to the subset of $$B$$ with $$x(0) = 0$$. Then at least we have a norm.

rmcerafl pointed out a problem with the following. For now, I will leave it here for reference.

Let $$C_0 = \{x \in C : x(0)=0\}$$ with the uniform norm. Let $$B_0 = \{x \in B : x(0)=0\}$$ with the variation norm. Both of these are complete separable metric spaces. The inclusion map $$i : B_0 \to C_0$$ is continuous and injective. Then citing a result or two from descriptive set theory*, we get: $$i(B_0)$$ is a Borel subset of $$C_0$$ and $$i : B_0 \to i(B_0)$$ is a Borel-isomorphic map. This shows, in particular, the variation-Borel subsets of $$B_0$$ coincide with the uniform-Borel subsets of $$i(B_0)$$.

$${}^*$$ Proposition 8.3.5 and Theorem 8.3.7 in:

Cohn, Donald L., Measure theory, Boston, Basel, Stuttgart: Birkhäuser. IX, 373 p. DM 42.00 (1980). ZBL0436.28001.

• Thank you, I edited the question accordingly. Jun 14 at 14:36
• I don't think $(B_0, \|\cdot\|_1)$ is separable, see for instance Theorem 1.25 in: Friz and Victoir, Multidimensional Stochastic Processes as Rough Paths. Jun 14 at 15:19
• I think you are right. References I consulted only discuss non-separability of BV and of NBV (=right-continuous BV), but not of continuous BV. If it is non-separable, I am inclinded to think that there are far too many Borel sets, then. Jun 14 at 16:12
• There are $2^{\mathfrak c}$ variation-open sets in $B_0$, but only $\mathfrak c$ uniform-Borel sets in $C_0$. Jun 14 at 18:34