# Disintegration of Haar measures

Suppose I have a locally compact group $G$ and a quotient map $f:G\to G/N$. Is it true that for every Borel-measurable function $f : G → [0, +∞]$, $$\int_{G} f(g) \, \mathrm{d} \mu_G (g) = \int_{G/N} \int_{N} f(gn) \, \mathrm{d} \mu_{N} (n) \mathrm{d} \mu_{G/N} (gN),$$ where $\mathrm{d} \mu_{G}$, $\mathrm{d} \mu_{N}$ and $\mathrm{d} \mu_{G/N}$ are the Haar measures on $G$, $N$ and $G/N$ respectively?

In particular, is it true that if a subset $A\subseteq G$ is such that its intersection with $\mathrm{d}\mu_{G/N}$-almost every coset $gN$ is of full measure in $gN$ (w.r.t. to the measure on $gN$ obtained by shifting the measure of $N$, which is independent of the choice of $g$), then $A$ is of full measure in $G$? (This is implied from the equation above by taking $f$ to be the characteristic function of $A$).

I wanted to use the disintegration theorem in Wikipedia, but I'm not sure if it applies here. I'm not sure I understand the definition of a Radon space and I don't know which locally compact groups satisfy it.

I know of a more specific disintegration result, which appears in many/most introductions to Haar measures (e.g. Raghunthan's book), but it is only stated for continuous $f$ with compact support. I suppose it's not hard to get rid of the continuous part by using some Luzin argument (although I am not sure how to do it myself), but the compact support bothers me.

In any case, if this is not true for general locally compact groups, for which groups it is true? I have a reference for second-countable compact groups (Halmos's book Measure Theory; his definitions are a bit out-dated, but coincide with the modern ones for second-countable compact groups).

I don't mind assuming separability. Compactness is not awful either, but I prefer not to assume metrisability.

Thanks.

• The disintegration theorem that I know (Dellacherie & Meyer, "Probabilities and Potential", page 79-III) makes essential use of metrizabilty. The version for compact spaces, though, does not require it (see "First case" at page 78-III). (Since the authors work with probabilities, compactness seems required anyway if you want to use their result.) – Alex M. Aug 19 '16 at 9:24
• This similar discussion on MathOverflow might provide you with a bunch of results to explore, and maybe even answer your question. – Alex M. Aug 19 '16 at 9:32
• I will look at your reference. I actually saw this discussion on MathOverflow, but I already checked quite a few of their sources to no avail: they always assume $f$ is continuous and with compact support. But there are still many books I haven't checked, so not all hope is lost. – Cronus Aug 19 '16 at 10:33
• It seems to me they do requite metrizability in the First Case, don't they? They write "$\Omega$ is a compact metric space..." – Cronus Aug 19 '16 at 10:40
• Ha, yes, I totally missed that on my superficial reading. – Alex M. Aug 19 '16 at 10:49

The identity

$$\int_{G} f(g) \, \mathrm{d} \mu_G (g) = \int_{G/N} \int_{N} f(gn) \, \mathrm{d} \mu_{N} (n) \mathrm{d} \mu_{G/N} (gN),$$

does not hold under the conditions you mention. You need some extra conditions. For example, only assuming that $f : G \to [0,\infty]$ is Borel measurable does not imply that $f$ is $\mu_G$-integrable, i.e., it might be that $\int_{G} f(g) \, \mathrm{d} \mu_G (g) = \infty$. Also, the quotient $G / N$ should be locally compact in order to define a Haar measure on it. This leads to the following result:

Let $G$ be a locally compact group and let $N \subset G$ be a closed subgroup. Suppose two of the Haar measures on $G, N$ and $G / N$ be given. Then there exists a unique third measure such that $$\int_{G} f(g) \, \mathrm{d} \mu_G (g) = \int_{G/N} \int_{N} f(gn) \, \mathrm{d} \mu_{N} (n) \mathrm{d} \mu_{G/N} (gN),$$ for all $f \in L^1 (G)$.

This result is well-known, and is often called Weil's formula or the quotient integral formula. It can be looked up in any book on abstract harmonic analysis.

There are also some generalisations of this result and related results. In case you want to know more about this, I strongly recommend you to look into Reiter's Classical Harmonic Analysis and Locally Compact Groups. In chapter 8 there is an extensive discussion on quasi-invariant quotient measures.

• Thank you very much! So this in particular holds if $f$ is a characteristic function of a null set, right? So the corollary I wrote ("if a subset $A\subseteq G$ is such that its intersection with $\mathrm{d}\mu_{G/N}$-almost every coset $gN$ is of full measure in $gN$ (w.r.t. to the measure on $gN$ obtained by shifting the measure of $N$) then $A$ is of full measure in $G$") is correct? – Cronus Aug 19 '16 at 10:36
• The characteristic function is certainly in $L^1 (G)$, so the identity holds in particular for such a function. However, I don't see immediately the link with your corollary, but if you proved that it is implied by the Weil's formula with $f$ being the characteristic function, then it should be correct. – user342207 Aug 19 '16 at 10:45
• I could be mistaken, but it seems to me that if $A$ is a set satisfying the properties I mentioned, then the characteristic function $f$ of its complement $A^c$ satisfies the following: – Cronus Aug 19 '16 at 10:50
• Thanks for providing these steps. This seems totally reasonable to me! – user342207 Aug 19 '16 at 10:53
• The generalisation to all of $L^1 (G)$ is not trivial and it needs some work, but I think it is pretty straightforward. If you want to see the details: It is done at the top of page 20 of Deitmar's Principles of Harmonic Analysis (second edition). The generalisation here takes as much space as the proof for $C_c (G)$. – user342207 Aug 19 '16 at 11:01