A question about André Weil’s converse to Haar’s Theorem on the existence of Haar measures 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!
 A: 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:

*

*$\mu$ is $\sigma$-finite (it is actually finite because $G$ is compact).

*$\Sigma$ and $\mu$ are left invariant.

*The maps $(x,y)\mapsto (x,xy)$ and $(x,y)\mapsto (x,x^{-1}y)$ are $\Sigma \otimes \Sigma$-measurable. (Note that you need both to be measurable)

*For each $ x \in G \setminus \{ e \} $, there exists an $ S \in \Sigma $ with
$$
  0 < \mu(S) < \infty \qquad \text{and} \qquad
  0 < \mu((x \cdot S) \triangle S) < \infty.
  $$
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.
