I look for an example of an Abelian locally compact topological group $G$ such that:

$G$ is connected and Haar measure on $G$ is not $\sigma$-finite and $\{0\} \times G \subset \mathbf{R} \times G$ has infinite measure


$G$ is connected and $G$ has a Haar measurable subset $S$ of non-$\sigma$-finite measure and $\{0\} \times S \subset \mathbf{R} \times G$ has infinite measure.

Here $\mathbf{R}$ is an additive group of reals with euclidean metric.


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    $\begingroup$ I don't understand the question. A connected locally compact group is automatically $\sigma$-compact ($G = \bigcup_n K^n$ for every compact neighborhood $K$ of the identity). Hence Haar measure on $G$ is $\sigma$-finite. What is $\mathbf{R}$ here? If it is the real numbers with the usual topology then $\{0\} \times S$ is always a null set by Fubini. $\endgroup$
    – t.b.
    Sep 10 '11 at 10:36
  • $\begingroup$ If you take the group $(\mathbf{R},+)$ but with discrete topology then it is not $\sigma$-compact and Haar measure on it, which is counting measure, is not $\sigma$-finite. In my question $\mathbf{R}$ is a group of reels with euclidean metric. If I don't mistake, Fubini theorem we may use for example if both measures are $\sigma$-finite, in general no. $\endgroup$
    – Richard
    Sep 10 '11 at 10:53
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    $\begingroup$ The discrete topology is not connected; rather the opposite... In your questions then, both $\mathbf{R}$ and $G$ are $\sigma$-finite and there's no problem with Fubini. $\endgroup$
    – t.b.
    Sep 10 '11 at 10:55
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    $\begingroup$ Ok, so we agree that for $G$ connected locally compact and $K$ a compact neighborhood of the identity the set $\bigcup K^n$ is an open subgroup, hence it is closed, hence all of $G$ by connectedness. Hence $G$ is $\sigma$-compact and thus every Radon measure on $G$ is $\sigma$-finite. $\endgroup$
    – t.b.
    Sep 10 '11 at 11:00
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    $\begingroup$ Yes. Every connected subset of a locally compact abelian group is contained in a translate of the connected component of the identity. The connected component is $\sigma$-compact by the argument above, hence $\sigma$-finite and thus every subset of the connected component (or one of its translates) also has $\sigma$-finite measure. $\endgroup$
    – t.b.
    Sep 10 '11 at 11:31

I should make my comments an answer.

  1. An open subgroup $H$ of a topological group is closed: Its complement is the open set $\bigcup\limits_{g \notin H} gH$.
  2. For every neighborhood $U$ of the identity element of a topological group the subgroup $H = \bigcup_{n \in \mathbb{Z}} U^{n}$ is open (hence closed by 1.): for every $h \in H$ the set $hU$ is contained in $H$ and a neighborhood of $h$.
  3. If $G$ is connected and $U$ is a neighborhood of the identity the subgroup $H$ in 2 is open and closed, hence the entire group $G$.

This implies that a connected locally compact group $G$ is $\sigma$-compact (since we may take a compact neighborhood $U$ of the identity in the above), hence every Radon measure is $\sigma$-finite and thus there are no examples as you're asking for: if $S \subset G$ is any measurable subset then $\{0\} \times S$ is a measurable rectangle, hence measurable in $\mathbf{R} \times G$ and by Fubini-Tonelli it must have measure zero.

If connectedness is dropped then this Fubini-argument fails. The standard example is $\{0\} \times \mathbf{R}_d$ in $\mathbf{R} \times \mathbf{R}_d$ where $\mathbf{R}_d$ is the additive group of the reals equipped with the discrete topology and $\mathbf{R}$ carries the standard topology. The set $\{0\} \times \mathbf{R}_d$ is locally null, but not null (hence it has infinite measure). This is an exercise all mathematicians interested in analysis should do once in their lives, so I won't spell it out. In case of emergency consult (11.33) on p.127 in Hewitt-Ross, Abstract Harmonic Analysis, I.


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