A group where the Weil topology induced by the Haar measure does not coincide with the original topology Let $(G,\tau)$ be a locally compact Hausdorff topological group that is $\sigma$-finite with respect to the Haar measure $\mu:\mathcal{B}(G)\to[0,\infty]$ ($\mathcal{B}(G)$ is the Borel $\sigma$-algebra for $G$). Define $\mathcal{B}\boldsymbol{a}(G)\subseteq \mathcal{B}(G)$ to be the Baire $\sigma$-ring in $G$ (the $\sigma$-ring generated by the compact $G_\delta$'s), and furthermore assume that $G\in\mathcal{B}\boldsymbol{a}(G)$ (i.e. $\mathcal{B}\boldsymbol{a}(G)$ is a $\sigma$-algebra). Let
$$\mathcal{A}=\{EE^{-1} \mid E\in \mathcal{B}\boldsymbol{a}(G), 0<\mu(E)<\infty\}.$$ Now forget about the topology $\tau$. It is well known that $\mathcal{A}$ forms a system of neighborhoods for $e$, which induces a topology $\tau_\mu$ in $G$ which makes it a Hausdorff topological group. This topology is called Weil's topology (see [1]). Under this topology $G$ is densely embeddable in a Hausdorff locally compact group $\overline{G}$, and the Haar integral in $\overline{G}$ coincides with the integral with respect to $\mu$ for all continuous functions of compact support contained in $G$.
It can be easily shown that $\tau \subseteq \tau_\mu$, and it was shown in [2] that adding the assumption that $\mathcal{B}\boldsymbol{a}({G})$ is analytic, $\tau_\mu\subseteq \tau$.
I am trying to come up with a simple example where $\tau_\mu\not\subseteq \tau$ (evidently in the case where $\mathcal{B}\boldsymbol{a}({G})$ is not analytic), but I have not been successful. Any ideas?
Refs:
[1] Halmos, Paul R., Measure theory. 2nd printing, Graduate Texts in Mathematics. 18. New York - Heidelberg- - Berlin: Springer-Verlag. XI, 304 p. DM 26.90 (1974). ZBL0283.28001.
[2] Mackey, George W., Borel structure in groups and their duals, Trans. Am. Math. Soc. 85, 134-165 (1957). ZBL0082.11201.
 A: There are no such locally compact groups, because if $G$ is a locally compact group under the topology $\tau$, then the Weil topology $\tau_\mu$ defined by the Haar measure $\mu$ is the same as the original topology $\tau$.
To show $\tau_\mu$ is finer than $\tau$, let $N$ be a $\tau$-neighbourhood of $e$. Since the mapping $g \mapsto gg^{-1}$ is continuous $G \rightarrow G$, there is a neighbourhood $M$ of $e$ such that $MM^{-1} \subseteq N$. Since $G$ is locally compact, we can find a compact $G_\delta$ neighbourhood $K$ of $e$ such that $K \subseteq M$ and therefore $KK^{-1} \subseteq N$. Since $K$ is compact, $\mu(K) < \infty$, and since it contains an open set, $\mu(K) > 0$ and therefore $KK^{-1} \in \mathcal{A}$ and so $N$ is a $\tau_\mu$-neighbourhood of $e$.
The other direction holds by Weil's extension of Steinhaus's theorem, which states that if $\mu(E) > 0$ then $EE^{-1}$ is a $\tau$-neighbourhood of $e$. Weil proved this by what is now the standard argument that convolving $\chi_E$ by $\chi_{E^{-1}}$ produces a continuous function vanishing outside $EE^{-1}$ but taking the nonzero value $\mu(E)\mu(E^{-1})$ at $e$.

For the more general question of topological groups with Haar measures, I do not know an example of a topological group $G$ with a left-invariant Radon measure $\mu$ such that the original topology $\tau$ differs from $\tau_\mu$. However, if we drop the requirement that $\mu$ be Radon there is a simple example. Take $G = \mathbb{Q}$, and let $\tau$ be its subspace topology in $\mathbb{R}$. The counting measure $\mu$ is an invariant measure on this group. However, the Weil topology $\tau_\mu$ defined by the counting measure on $\mathbb{Q}$ is easily seen to be the discrete topology, which is strictly finer than $\tau$.
