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?


[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.

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    $\begingroup$ It seems that what needs to be done is to take a locally compact group $G$, and find a subgroup $H$ of full outer measure, and a distinct topology on $H$ that both a group topology and has the same Baire sets as the subspace topology on $H$. Unfortunately I don't know how to do this. $\endgroup$ Sep 3, 2020 at 2:01
  • $\begingroup$ This seems like a good path, although the distinct topology on $H$ needs to be also locally compact. Halmos gives an example of a thick subgroup in $G=\mathbb{R}^2$ in [1] p. 276 exercise (6), using some basic linear algebra techniques. But I would not know either how to give this subgroup a distinct locally compact topology so the Baire sets coincide. $\endgroup$ Sep 3, 2020 at 5:39
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    $\begingroup$ If $G$ is a locally compact group, then the topology on $G$ agrees with the Weil topology. One way follows from local compactness and continuity of $(g,h) \mapsto gh^{-1}$, and the other is Weil's extension of Steinhaus's theorem to locally compact groups (that if $E$ has strictly positive Haar measure, $EE^{-1}$ is a neighbourhood of the identity). I assumed that you knew this, and thought you were looking for a (necessarily non-locally compact) topological group that had a Haar measure such that the topology differed from the Weil topology. $\endgroup$ Sep 4, 2020 at 8:55
  • $\begingroup$ You assumed wrong! Thank you so much, I completely missed this. $\endgroup$ Sep 4, 2020 at 15:25
  • $\begingroup$ Do you want me to post my comment as an answer? Then you can accept it and the question will be considered "answered". Otherwise the question recirculates to the front page in search of answers for ever and ever. $\endgroup$ Sep 5, 2020 at 9:28

1 Answer 1


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 $\tau$.


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