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Let $G$ a topological group, $(X,\mathcal{B})$ a measurable space, and $G \curvearrowright X$ a measurable action, that is, the map $G \times B \rightarrow B$ is measurable, when $G \times X$ is endowed with the product $\sigma$-algebra of the Borel sets on $G$ and $\mathcal{B}$.

Let $K$ be a compact subset of $G$ and $B \in \mathcal{B}$.

  • Is it true that $KB \in \mathcal{B}$?
  • Can you provide sufficient topological conditions on $G$, or conditions on the $\sigma$-algebra $\mathcal{B}$, such that the statement is true?

Thoughts: When trying to produce counterexamples, like in the following attempt, the assumption of measurability of the action looks difficult to prove or disprove.

Attempt 1: If $G=X=\mathbb{R}$, if $G$ is endowed with the usual topology, if $G$ acts on $X$ by translations, if $\mathcal{B}$ is the $\sigma$-algebra of subsets of $X$ that are either countable or of countable complement, $B := \{0\}$ and $K := [0,1]$, then $KB \not \in \mathcal{B}$. However, I don't think the action is measurable: if $a \in X$, $\{(g,x) \ \vert \ g\cdot x = a\} = \{(x,y) \ \vert \ x+y = a\}$ and I don't believe that a line of slope $-1$ can be in the product $\sigma$-algebra.

Attempt 2: If $X$ is a topological space and $\mathcal{B}$ is the Borel $\sigma$-algebra, and if both $K$ and $B$ are compact, then $KB$ is a Borel set. We can form the set $\{A \in \mathcal{B} \ \vert \ KB \in \mathcal{B}\}$ and we know that it contains all the compacts subsets of $X$. It is closed under taking countable unions, but probably not under taking complements.

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  • $\begingroup$ There are $K,B\subseteq\Bbb R$ with $K$ compact, $B$ Borel and $K+B$ not Borel, so this fails even for the translation action $\Bbb R\curvearrowright\Bbb R$. I will dig up a reference and write a proper answer later. $\endgroup$ Commented Nov 17, 2021 at 19:49
  • $\begingroup$ If it is true, it is the best answer I could expect! $\endgroup$
    – Plop
    Commented Nov 18, 2021 at 9:54

1 Answer 1

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Erdös and Stone proved that there are $K,B\subseteq\Bbb R$ where $K$ is compact and $B$ is $G_\delta$ such that $K+B$ is not Borel in the following paper. The paper uses an explicit construction due to Von Neumann to produce a perfect $P\subseteq\Bbb R$ that is linearly independent over $\Bbb Q$, but the existence of such a set is an immediate consequence of Mycielski's theorem (see for example Theorem 19.1 and Exercise 19.2i in Kechris Classical Descriptive Set Theory).

Erdös, P., and A. H. Stone. “On the Sum of Two Borel Sets.” Proceedings of the American Mathematical Society, vol. 25, no. 2, American Mathematical Society, 1970, pp. 304–06, https://doi.org/10.2307/2037209.

This shows that even for the (continuous) translation action of $\Bbb R$ on itself the property you are interested in fails (and the argument in Erdös-Stone generalizes to many topological groups). A simpler construction is given in this answer.

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  • $\begingroup$ Thank you for this very useful answer. $\endgroup$
    – Plop
    Commented Nov 19, 2021 at 9:37
  • $\begingroup$ @plop while thinking about your question I started wondering what’s the smallest $\sigma$-algebra we can put on the reals that fixes this example. Being unable to work it out myself I asked on MO, you might be interested too mathoverflow.net/questions/408912/… $\endgroup$ Commented Nov 19, 2021 at 15:27

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