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)?


Here is a counterexample:

Take any vector $v \in \mathbb{R}^\infty \setminus \ell^1$. Then on the set $C := [0,1]^\infty + \{tv, 0 \le t \le 1\}$ there are infinitely many "Lebesgue measures", namely for any $0 \le t \le 1$ the product measure $\mu_t$ on the shifted cube $[0,1]^\infty + t v$ satisfies my definition.

The reason is that for a $\mu_t$-typical point $x \in C$ the $n$-dimensional section $C \cap (x + (\mathbb{R}^n \times \{0\}^\infty))$ is a cube. The proof is a straightforward application of the Borel-Cantelli lemma.

Intuivitely, the product measure on $[0,1]^\infty$ doesn't "recognize" $v$ as an admissible shift direction, due to the assumption $v \notin \ell^1$, so it doesn't "feel the difference" between $C$ and the shifted cube $[0,1]^\infty + tv$ in terms of the shift invariance condition.

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