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.

  • $\begingroup$ What would be an example of a non-measure-continuous compact subset of a (as nice as possible) topological group? $\endgroup$ – მამუკა ჯიბლაძე May 20 '18 at 8:50
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    $\begingroup$ This is a good question. I think that this best possible topological group should be $\ell_2$ or $c_0$ as products of locally compact groups have this property. Or, there is another option that each compact subset in each Polish Abelian group is measure-continuous. $\endgroup$ – Taras Banakh May 20 '18 at 8:54

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" of V. Bogachev.

  • $\begingroup$ Is "the Hilbert–Schmidt operator" supposed to be "a Hilbert–Schmidt operator"? $\endgroup$ – LSpice May 20 '18 at 18:09
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    $\begingroup$ @LSpice You are right. It should be "a". Thank you. $\endgroup$ – Taras Banakh May 20 '18 at 18:56

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