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
