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!

  • $\begingroup$ The answer is yes, but let me quote wikipedia : "σ-rings can be used instead of σ-fields (σ-algebras) in the development of measure and integration theory, if one does not wish to require that the universal set be measurable. Every σ-field is also a σ-ring, but a σ-ring need not be a σ-field." $\endgroup$
    – charles
    May 30, 2018 at 13:54
  • 1
    $\begingroup$ @charles The OP asked about replacing "Baire $\sigma$-ring" with "Borel $\sigma$-algebra". The $\sigma$-algebra part is not an issue, as pointed out here, but the Borel part is (see my answer). $\endgroup$ Oct 11, 2020 at 6:13

1 Answer 1


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.

  • $\begingroup$ Thank you for your detailed response! $\endgroup$ Jan 2, 2021 at 20:02
  • $\begingroup$ You are welcome! Hope it helps despite the four-year delay. $\endgroup$ Jan 4, 2021 at 6:45

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