**Definition.** A subset $K$ of a topological group $X$ is called *measure-continuous* if there exists a $\sigma$-additive Borel probability measure $\mu$ on $X$ such that for every compact subset $C\subset X$ the map $f:K\to [0,1]$, $f:x\mapsto \mu(C+x)$ is continuous.


**Remark 1.** Each measure-continuous set $K$ in a Polish group $X$ is contained in a $\sigma$-compact subgroup; so $K$ is small in a sense. What about the converse? 


**Problem.** Is each compact subset of a Banach space $X$ measure-continuous? What is the answer for classical Banach spaces $c\_0$ or $\ell\_p$? 


**Remark 2.** It can be shown that each compact subset in the countable product of locally compact topological groups is measure-continuous.



Are all compact subsets of Banach spaces small in a measure-theoretic sense?
I was informed by Vladimir Bogachev that the answer to Problem is negative at least for the Hilbert space $\ell\_2$ as every measure-continuous compact subset of $\ell\_2$ is contained in the image of a Hilbert-Schmidt operator $T:\ell\_2\to\ell\_2$ (for which there exists an orthonormal basis $(e\_n)\_{n\in\omega}$ in $\ell\_2$ such that $\sum\_{n=1}^\infty \|T(e\_n)\|^2<+\infty$). 


According to Bogachev the answer to the Problem also is negative for the Banach space $c\_0$ in which the compact subset $K=\{(x\_n)\_{n=1}^\infty\in c\_0:\sup\_{n\in\mathbb N}(|x\_n|\cdot\ln n)<\infty\}$ is not measure-continuous.


On the other hand, in nuclear Frechet spaces all compact subsets are contained in Hilbert-Schmidt ellipsoids and hence are measure-continuous.


More information on this topic can be found in Chapter 3 of the book ["Differentiable measures and Malliavin Calculus"](https://www.ams.org/books/surv/164/surv164-endmatter.pdf) of V. Bogachev.


