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In this answer, Gerald Edgar mentions that Haar measure is naturally defined on the $\sigma$-algebra of Baire sets (the smallest $\sigma$-algebra that contains all the compact $G_\delta$ sets) of a locally compact group and that the uniqueness of Haar measure can fail for larger $\sigma$-algebras. I wonder if there is a nice example of this.

Curious, I skimmed through Halmos's classic Measure Theory, and I found that he proves the existence and uniqueness of Haar measure for the slightly larger $\sigma$-algebra generated by all compact sets. (Confusingly, Halmos defines Borel sets to be those in this $\sigma$-algebra; I will stick with the usual definition of Borel sets.)

Is there a nice example of a locally compact group where the uniqueness of Haar measure fails for the $\sigma$-algebra of Borel sets — the $\sigma$-algebra generated by open sets?

To dispell some potential confusion (see comments by Keenan Kidwell and Gerald Edgar) Haar measures are not required to be regular (for the purpose of this question).

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  • $\begingroup$ I thought the usual construction of a (left) Haar measure on a locally compact (Hausdorff) group $G$ was done by constructing a left invariant positive linear functional on $C_c(G)$ and then invoking the Riesz representation theorem, which always gives a measure whose domain includes all open sets (a Borel measure according to Rudin or Folland). $\endgroup$
    – Keenan Kidwell
    Jul 19, 2011 at 18:36
  • $\begingroup$ Thanks Keenan. This seems to prove existence but what about uniqueness? $\endgroup$
    – François G. Dorais
    Jul 19, 2011 at 18:43
  • $\begingroup$ Uniqueness holds too, and is proved e.g. as Theorem 2.20 in Folland's book on abstract harmonic analysis (books.google.com/…). Although I guess maybe this depends on people's definition of a Radon measure. For Folland, I believe a Radon measure is an outer regular Borel measure that's finite on compact sets and I think also inner regular on open sets. Anyway, given two such measures $\mu$ and $\nu$ on $G$ that are non-zero and left translation invariant (on Borel sets), they differ by a constant. $\endgroup$
    – Keenan Kidwell
    Jul 19, 2011 at 18:53
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    $\begingroup$ Uniqueness among (outer) regular measures. Regularity is automatic for the Baire sets (and thus for Borel sets in case of metrizable groups). If you do not add the hypothesis of regularity, uniqueness can fail. Two measures that agree on the Baire sets also agree on $C_c(G)$, so if they differ in Borel sets we don't care.. $\endgroup$
    – Gerald Edgar
    Jul 19, 2011 at 19:06
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    $\begingroup$ Whenever you need a thorough reference for generalities on measure theory, consult Fremlin. In this case, Chapter 44 in Volume 4 Part I. Available online on his homepage: essex.ac.uk/maths/people/fremlin/mt.htm $\endgroup$
    – Theo Buehler
    Jul 19, 2011 at 19:54

2 Answers 2

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Let's try this example. Let $\mathbb R$ be the additive group of reals with the usual topology, and let $\mathbb R_\mathrm{d}$ be the additive group of reals with the discrete topology. Our group is $G = \mathbb R \times \mathbb R_\mathrm{d}$. A big group. Any compact set in $G$ has finite projection on the $y$-axis. The Haar integral $\Lambda$, defined on $C_c(G)$, is $$ \Lambda(f) = \sum_y \int f(x,y)\,dx $$ where of course the sum is finite.

Now consider the (Borel but not Baire) set $E = \{0\}\times \mathbb R_\mathrm{d}$. It is closed, hence Borel. But any Baire set has projection on the $y$-axis either countable or co-countable, so $E$ is not a Baire set. The set $E$ has outer measure $\infty$ but inner measure $0$. Changing our "extension" of Haar measure on $E$ to any value between $0$ and $\infty$ will make sense, and still be a Borel measure that produces $\Lambda$ on the functions in $C_c(G)$.

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    $\begingroup$ This is also an example for: failure of Fubini's theorem when a measure is not sigma-finite. Also for: dual of $L^1$ not equal to $L^\infty$ when the measure is required to be regular. If, on the other hand, we requre locally null sets (like $E$ defined here) to have measure zero, then we recover the dual of $L^1$ as $L^\infty$. But lose regularity. $\endgroup$
    – Gerald Edgar
    Jul 19, 2011 at 19:38
  • $\begingroup$ I think you can also tweak the definition of $L^\infty$ to identify functions which are locally almost everywhere equal-- then I believe we get that $(L^1)^*=L^\infty$ (at least in this example, if we follow Rudin, and insist upon outer regularity, but only inner regularity of sets of finite measure or open sets). $\endgroup$
    – Matthew Daws
    Jul 19, 2011 at 19:49
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    $\begingroup$ This is a very nice group! Is there an easy way that I could convince myself that I can set the measure of $E$ to $\pi$ and maintain translation invariance? $\endgroup$
    – François G. Dorais
    Jul 19, 2011 at 19:50
  • $\begingroup$ This example is mentioned in Folland's book (already mentioned by Keenan). The basic problem is the lack of $\sigma$-compactness, which implies (and is equivalent to) non-$\sigma$-finiteness of Haar measure. He does show that even in this case Haar measure is "decomposable" (a generalization of $\sigma$-finiteness) which allows him to obtain generalizations of the theorems mentioned in the comments above, as well as the Radon-Nikodym theorem. $\endgroup$
    – Mark
    Jul 19, 2011 at 21:02
  • $\begingroup$ @François: I just claim it is an extension of the measure, and integrals on $C_c(G)$ agree with $\Lambda$. I guess if you want it also to be translation invariant, the only choice besides $\infty$ is $0$. (The restriction of the measure to the $y$-axis has to be a scalar multiple of the counting measure...) So to get something other than the outer regular measure, let sets supported in countably many columns have measure zero. $\endgroup$
    – Gerald Edgar
    Jul 19, 2011 at 22:48
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I may be missing something here but what François was looking for may be the following (it is of course already present in the previous answers, I only put it slightly differently):

Let $B$ be a Borel subset of the group $\mathbb{R} \times \mathbb{R}_d$. Define $B_y = \{x : (x,y) \in B\}$. Denote the Lebesge measure on $\mathbb{R}$ by $\lambda$.

First, let $$\mu(B) = \sum_{y \in \mathbb{R}} \lambda(B_y).$$ Second, let $$ \nu(B) = \begin{cases} \sum_{y \in \mathbb{R}} \lambda(B_y) \textrm{ if } B \textrm{ only intersects countably many sets of the form } \mathbb{R} \times \{y\},\\ \infty, \textrm{ otherwise. } \end{cases} $$

It is easy to check that both $\mu$ and $\nu$ are translation invariant $\sigma$-additive measures on the Borel sets, taking finite values on the compact sets and positive values on the nonempty open sets. (And the open sets are inner regular w.r.t. compact sets.) Do you require anything else?

Moreover, $\mu(\{0\} \times \mathbb{R}) = 0$, but $\nu(\{0\} \times \mathbb{R}) = \infty$.

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  • $\begingroup$ Hi Marton, it's been a long while since the Fields Institute in 2002! Welcome to MO! Yes, this is a great example, thank you! $\endgroup$
    – François G. Dorais
    Feb 27, 2016 at 3:02

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