Let $\mu$ be a probability measure on a subset $C \subset \mathbb{R}^\infty$ of the space of sequences, and assume, for simplicity, that $C$ is closed and convex.

We say that $\mu$ admits shifts if for any $\xi \in \mathbb{R}^\infty_0$ (finitely many nonzero coordinates) the shifted measures $\mu_{\xi} := (\cdot + \xi)_\ast \mu$ have the property

- $\mu_{\varepsilon \xi} \to \mu$ in total variation, as $\varepsilon \to 0$

In other words, shifting by $\varepsilon \xi$ produces a measure that is almost $\mu$-absolutely continuous.

Call $\mu$ Lebesgue (on $C$ with shift space $S$) if:

- It admits shifts, and those shifts act ergodically
- $\frac{d\mu_\xi}{d \mu} = 1$ on $C \cap (C - \xi)$ for all $\xi \in \mathbb{R}^\infty_0$, i.e. $\mu$ is "as invariant as it could be", given that it lives on $C$.

Examples of such measures include the (unique) Lebesgue measure on $[0,1]^\infty$, or on any other product of finite-dimensional sets. For many other sets - e.g. all unit balls of *separable* Banach spaces - they don't exist.

Question:Is $\mu$ uniquely determined by $C$ (if it exists)?