The set $A := \{ k \in G \; | \; \mu_G(S) = \mu_G(S \cap k S) \}$ forms a closed subgroup of $G$.

Question

There is the following result:

Let $G$ be a locally compact group with Haar measure $\mu_G$. Suppose $S \subset G$ is measurable with $0 < \mu_G (S) < \infty$. Then $A := \{ k \in G \; | \; \mu_G(S) = \mu_G(S \cap k S) \}$ forms a closed subgroup of $G$.

I do not see why it should hold that $\mu_G (S) = \mu_G( S \cap xy S)$ whenever $\mu_G (S) = \mu_G (S \cap x S)$ and $\mu_G (S) = \mu_G (S \cap y S)$ with $x, y \in G$, that is, why $A$ is closed under the group operation.

Any help or comment is appreciated.

Answer 1

Clearly $S\setminus xyS\subset(S\setminus xS)\cup(xS\setminus xyS)$, where $$ A\setminus B=\{u\colon u\in A \text{ and } u\notin B\}. $$ Hence $$ \mu_G(S\setminus xyS)\le\mu_G(S\setminus xS)+\mu_G(xS\setminus xyS) =\mu_G(S\setminus xS)+\mu_G(x(S\setminus yS))=0+0=0. $$ Here $$ \mu_G(x(S\setminus yS))=\mu_G(S\setminus yS) $$ because of the invariance of the Haar measure.