1
$\begingroup$

I have a very concrete question on the proof of the following (see below): Given a 'nice' top. space $X$ and a 'nice' group operation of $G$, say, from the right, on $X$ and a certain measure on $X$, there is a measure on $X/G$ such that

$$ \int_{X/G} \int_G f(\mathcal{C}g) dg d\mathcal{C} = \int_X f(x) dx$$

they key thing here is that I want this to be true for all $f \in L^1(X)$ and not merely for all $f \in C_C(X)$ (= continuous functions with compact support from $X$ to $\mathbb{C}$). As there is a whole zoo of definitions of what a "Haar measure" is, let me sketch the terrain on which I work on (please don't be scared off by the length, Im just trying to put everything together that we will need): Let us assume throughout that

  • $G$ is locally compact and hausdorff
  • $X$ is locally compact and hausdorff
  • $G$ operates strongly continously and properly, meaning that the map $$ \phi : X \times G \to X \times X, ~~~~~ (x, g) \mapsto (x, xg)$$ is continuous and proper (preimages of compact sets are compact again)
  • There are measures on $X$ and $G$ such that
    • $\mu(K) < \infty$ for all compact sets $K$.
    • $\mu(A) = \inf\{ \mu(U) : A \subset U ~\text{and}~ U ~\text{is open}\}$ for all sets $A \in \mathcal{B}(\{G, X\})$ where $\mathcal{B}(\{G, X\})$ is the Borel-$\sigma$-alg of $G$, respectively $X$.
    • $\mu(A) = \sup\{ \mu(K) : K \subset A ~\text{and}~ K ~\text{is compact}\}$ holds for all sets $A \in \mathcal{B}(\{G, X\})$ that are either open or of finite measure.
    • the measure $\mu_G$ on $G$ is left invariant, i.e. $\mu_G(gA) = \mu_G(A)$ for all $A \in \mathcal{B}(G)$ and $g \in G$.
    • the measure $\mu_X$ satisfies $\mu_X(Ag) = \Delta_G(g) \mu_X(A)$ for all $A \in \mathcal{B}(X)$ and $g \in G$, where $\Delta_G$ is the modularity function of $G$. This last property makes it possible to define a measure on $X/G$ such that the quotient integral formula works for all $f \in C_C(X)$.

Further, we are going to need

  • For the measure $\mu$ on $X$, $\mu(U) > 0$ for all sets $U$ that are open and non empty
  • if $V \subset G$ is open and $x \in X$ is fixed, then $xV$ is open as well

These last two properties are needed in order to make the following theorem work:

For every function $f : X \to \mathbb{C}$ such that $\int_X |f(x)| dx < \infty$, we have that $S_f := \{x \in X : f(x) \neq 0\}$ (note that I talk about functions, not classes in $L^1(X)$!!) is contained in a sigma-compact set $V = \bigcup_{n \in \mathbb{N}} K_n$ where we may assume $K_1 \subset K_2 \subset ...$.

The proof of this theorem can be found in Deitmar, Principles of Harmonic Analysis, Cor 1.3.5 (d). He does it for $X$=Group, $G=H=$subgroup but the proof works equally well if one just assumes these two properties.

What Deitmar wants to do now in order to prove the quotient integral formula is the following: He wants to approximize an $L^1(X)$-function $f$ by $f \cdot \mathbf{1}_{K_n}$. They converge monotonously against $f$ everywhere (not merely almost everywhere). So he sais that one can assume $f$ itself to be compactly supported (Im still fine with that). Now the next step is the critical one: Take a sequence of step functions $f_n$ converging monotonously against $f$. As $f$ is in $L^1(X)$, so are the $f_n$. Now Deitmar sais that we can assume that $f$ is bounded right away. Of course, he wants to show the quotient integral formula for the $f_n$ and then take the monotonous limit. The problem is the following: In order to view $f$ and the $f_n$ as bounded functions, one has to change them on a $X$-null set $N$. So here comes my simple question:


Let us assume that $f, h$ are two $L^1(X)$-functions that are equal $X$-almost everywhere and that the quotient formula holds for $f$, does it then also hold for $h$?

I believe that this is problematic: For example,

$$\int_G f(\mathcal{C}g) dg \neq \int_G h(\mathcal{C}g) dg$$

in general, take SL$_2(\mathbb{Z})$ acting on the upper half plane and $\tau_0 \in \mathbb{H}$ fixed, then $N = \bigcup_{\gamma \in \text{SL}_2(\mathbb{Z})} \{\gamma \tau_0\}$ is a zero set but $$ \int_G \mathbf{1}_{N}(\gamma\mathcal{C}) d\gamma = \begin{cases} \infty & \text{if $\mathcal{C} \in \pi(N)$} \\ 0 & \text{otherwise} \end{cases}$$

Does somebody know how to solve this?

Cheers,

FW

$\endgroup$
  • $\begingroup$ Why does one have to change $f$ in this argument? The $f_n$ are surely bounded, so once proven the quotient formula for them, it follows for $f$ by monotonous convergence, or does it? $\endgroup$ – Zero Oct 13 at 2:28
  • $\begingroup$ Depending on your teacher, $L^1$ functions might or might not take the value $\pm \infty$ itself. So let's assume that they are allowed to do that. Since they are in $L^1$ we know that the set of $x$ such that $f(x)=\infty$ is a zero set, however, without changing the function on that set you will never achieve that they are bounded. Not 100% sure anymore (It's been a few years :-)) but I think I did not feel comfortable with the relationships of the measures on $X$ and the one on $X/G$: What does the formula say? Well first of all we need to make sense of the expression $\endgroup$ – Fabian Werner Oct 13 at 21:21
  • $\begingroup$ $f(\mathcal{C}g)$, right? The function $\tilde{f}: X/G \mapsto \mathbb{R}, \tilde{f}(xG) := f(x)$ is not well defined. Hence we actually define $I_f(xG) := \int_G f(xg) dg$ which then is well defined. However, this expression needs to be independent of the chosen representative of the $L^1$ class, i.e. given $f=g$ $X$-almost everywhere, is $I_f = I_g$? And I cannot remember why right now but it must be a relationship between the measure on $X$ and the one on $X/G$ that makes this true... that was what I did not understand at the time... $\endgroup$ – Fabian Werner Oct 13 at 21:23
  • $\begingroup$ I think I meant $I_f = I_g$ at least $X/G$-almost everywhere... $\endgroup$ – Fabian Werner Oct 13 at 21:35
  • $\begingroup$ Why should the function f be bounded? This is not necessary. Even unbounded functions are monotonous limits of bounded Lebesgue step functions. So, when you know the quotient formula for Lebesgue step functions, you know it for all functions in L1. $\endgroup$ – Zero Oct 15 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.