Let $ (G,\cdot,e) $ be a group, and suppose that there are a $ \sigma $-ring $ \Sigma $ on $ G $ and a measure $ \mu: \Sigma \to [0,\infty] $, non-trivial, such that the following properties hold:
- $ \Sigma $ is left-invariant w.r.t. $ \cdot $, i.e., $ x \cdot S \in \Sigma $ for every $ x \in G $ and $ S \in \Sigma $.
- $ \mu $ is left-invariant w.r.t. $ \cdot $, i.e., $ \mu(x \cdot S) = \mu(S) $ for every $ x \in G $ and $ S \in \Sigma $.
- The map $ \left\{ \begin{matrix} G \times G & \to & G \times G \\ (x,y) & \mapsto & (x,x \cdot y) \end{matrix} \right\} $ is $ (\Sigma \times \Sigma,\Sigma \times \Sigma) $-measurable.
- For each $ x \in G \setminus \{ e \} $, there exists an $ S \in \Sigma $ with $$ 0 < \mu(S) < \infty \qquad \text{and} \qquad 0 < \mu((x \cdot S) \triangle S) < \infty, $$ where $ \triangle $ denotes the symmetric difference of sets.
Then Weil’s converse to Haar’s Theorem states that there exists a topological group $ ((G',\bullet,e),\tau) $ with the following properties:
- $ \tau $ is a locally compact and Hausdorff group topology on $ G' $.
- $ (G,\cdot,e) $ is a subgroup of $ (G',\bullet,e) $, so that $ G \subseteq G' $ and $ \cdot = \bullet|_{G \times G} $.
- If $ \mathscr{B} $ denotes the $ \sigma $-ring on $ G' $ generated by the $ G_{\delta} $ compact (w.r.t. $ \tau $) subsets of $ G' $, then $$ \{ B \cap G \in \mathcal{P}(G) \mid B \in \mathscr{B} \} \subseteq \Sigma. $$ Note: We call $ \mathscr{B} $ the $ \tau $-induced Baire $ \sigma $-ring on $ G' $.
- There exists a (Baire) Haar measure $ \mu': \mathscr{B} \to [0,\infty] $, associated with $ ((G',\bullet,e),\tau) $, such that $$ \forall B \in \mathscr{B}: \qquad \mu(B \cap G) = \mu'(B). $$ This implies that $ G $ is a $ \mu' $-thick subset of $ G' $, as $ B \in \mathscr{B} $ and $ B \cap G = \varnothing $ imply $ \mu'(B) = 0 $.
The version of Weil’s result presented here is taken from Halmos’s Measure Theory, which is rather antiquated but still remains a classic.
Now, I would like to determine if one can simply replace every instance of ‘$ \sigma $-ring’ by ‘$ \sigma $-algebra’, as well as replace all Baire $ \sigma $-rings by Borel $ \sigma $-algebras, i.e., $ \sigma $-algebras on a set that are generated by a given locally compact and Hausdorff topology.
Could someone kindly provide an authoritative reference to aid my query? Thank you very much!
Maybe this answer comes too late, but better late than never.
The answer is no. First note that you are missing a key hypothesis: $\mu$ is $\sigma$-finite. (This is necessary to talk about the product measure). But I digress. To answer your question, you may replace "$\sigma$-ring" by "$\sigma$-algebra". This carries no problems other than losing a bit of generality. The real deal breaker is replacing "Baire" by "Borel".
The theorem does not hold for Borel sets. To show this, consider a compact Hausdorff topological group $G$ with cardinality greater than that of $\mathbb{R}$. Now, take $\Sigma$ to be the $\sigma$-algebra of Baire sets (the $\sigma$-algebra generated by the compact $G_\delta$ sets of $G$), and $\mu:\Sigma\to[0,\infty]$ the left Haar measure restricted to the Baire sets. Then you may check that indeed:
But if $G'$ is any Hausdorff group containing $G$, and $\mathcal{B}(G')$ is the Borel $\sigma$-algebra for $G'$, then $$ \{ B \cap G \in \mathcal{P}(G) \mid B \in \mathcal{B}(G') \} \not\subseteq \Sigma. $$ To show this, take $A\subseteq G$ to be any Borel set of $G$ that is not a Baire set of $G$ (for example $A=\{e\}$). Then $A\in \{ B \cap G \in \mathcal{P}(G) \mid B \in \mathcal{B}(G') \}$, but $A\not \in \Sigma$
TL;DR: You might change $\sigma$-ring for $\sigma$-algebra, but you can't change "Baire" for "Borel", as the Borel $\sigma$-algebra is too big.