# 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!

• 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." May 30, 2018 at 13:54
• @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). Oct 11, 2020 at 6:13

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.

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