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

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.