ModularFunction 2008-04-04 21:55:55 2008-04-08 23:49:32 Definition Haar measure % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure. Although left and right Haar measures in $G$ always exist, they generally do not coincide, i.e. a left Haar measure is usually not invariant under right translations. Nevertheless, the right translations of a left Haar measure can be easily described as explained in the following theorem. $\,$ {\bf Theorem -} Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure in $G$. Then, there exists a continuous homomorphism $\Delta:G \longrightarrow \mathbb{R}^+$ such that, for every $t \in G$ and every measurable subset $A$ \begin{displaymath} \mu(At) = \Delta(t^{-1})\mu(A) \end{displaymath} Moreover, if $f:G \longrightarrow \mathbb{C}$ is an integrable function then \begin{displaymath} \Delta(t)\int_Gf(st) \mu(s) = \int_G f(s) \mu(s) \end{displaymath} $\,$ The function $\Delta$ is called the {\bf modular function} of $G$ (notice that, by uniqueness up to scalar multiple of left Haar measures, $\Delta$ only depends on $G$). Other names for $\Delta$ that can be found are: \emph{Haar modulus}, or \emph{modular character} or \emph{modular homomorphism}. We now prove the above theorem, except the continuity of $\Delta$ (which is slightly harder to obtain). $\,$ {\bf \emph{Proof (except continuity of $\Delta$):}} Let $t \in G$. The function $\nu$, defined on measurable subsets $A$ by \begin{displaymath} \nu (A):= \mu(At) \end{displaymath} is easily seen to be a measure in $G$. Moreover, $\nu$ is left invariant (since $\mu$ is left invariant) and satisfies the additional conditions to be a left Haar measure. By the uniqueness of left Haar measures, $\mu$ must be a multiple of $\nu$, i.e. $\mu=\Delta(t)\nu$ for some positive scalar $\Delta(t) \in \mathbb{R}^+$. Thus, we have proven that for every measurable subset $A$ \begin{displaymath} \mu(At)= \Delta(t)^{-1}\mu(A) \end{displaymath} Now for $s, t \in G$ we have that $\mu(Ast)=\Delta(st)^{-1}\mu(A)$, but also \begin{itemize} \item $\mu(Ast)= \Delta(t)^{-1}\mu(As)$, and \item $\mu(As) = \Delta(s)^{-1}\mu(A)$ \end{itemize} So, we can see that, for every measurable subset $A$, \begin{displaymath} \Delta(st)^{-1}\mu(A) = \Delta(t)^{-1}\Delta(s)^{-1}\mu(A) \end{displaymath} Hence, $\Delta(st) = \Delta(s)\Delta(t)$. Thus, $\Delta$ is an homomorphism. The statement about integrals of functions follows easily by approximation by simple functions. For simple functions it is easy to see it is true using the now established condition $\mu(At) = \Delta(t^{-1})\mu(A)$. $\square$