modular function
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^{}.
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Theorem  Let $G$ be a locally compact Hausdorff topological group and $\mu $ a left Haar measure in $G$. Then, there exists a continuous^{} homomorphism^{} $\mathrm{\Delta}:G\u27f6{\mathbb{R}}^{+}$ such that, for every $t\in G$ and every measurable subset $A$
$$\mu (At)=\mathrm{\Delta}({t}^{1})\mu (A)$$ 
Moreover, if $f:G\u27f6\u2102$ is an integrable function then
$$\mathrm{\Delta}(t){\int}_{G}f(st)\mu (s)={\int}_{G}f(s)\mu (s)$$ 
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The function $\mathrm{\Delta}$ is called the modular function of $G$ (notice that, by uniqueness up to scalar multiple of left Haar measures, $\mathrm{\Delta}$ only depends on $G$). Other names for $\mathrm{\Delta}$ that can be found are: Haar modulus, or modular character or modular homomorphism.
We now prove the above theorem, except the continuity of $\mathrm{\Delta}$ (which is slightly harder to obtain).
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Proof (except continuity of $\mathrm{\Delta}$):
Let $t\in G$. The function $\nu $, defined on measurable subsets $A$ by
$$\nu (A):=\mu (At)$$ 
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 =\mathrm{\Delta}(t)\nu $ for some positive scalar $\mathrm{\Delta}(t)\in {\mathbb{R}}^{+}$. Thus, we have proven that for every measurable subset $A$
$$\mu (At)=\mathrm{\Delta}{(t)}^{1}\mu (A)$$ 
Now for $s,t\in G$ we have that $\mu (Ast)=\mathrm{\Delta}{(st)}^{1}\mu (A)$, but also

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$\mu (Ast)=\mathrm{\Delta}{(t)}^{1}\mu (As)$, and

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$\mu (As)=\mathrm{\Delta}{(s)}^{1}\mu (A)$
So, we can see that, for every measurable subset $A$,
$$\mathrm{\Delta}{(st)}^{1}\mu (A)=\mathrm{\Delta}{(t)}^{1}\mathrm{\Delta}{(s)}^{1}\mu (A)$$ 
Hence, $\mathrm{\Delta}(st)=\mathrm{\Delta}(s)\mathrm{\Delta}(t)$. Thus, $\mathrm{\Delta}$ 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 $\mu (At)=\mathrm{\Delta}({t}^{1})\mu (A)$. $\mathrm{\square}$
Title  modular function 

Canonical name  ModularFunction 
Date of creation  20130322 17:58:18 
Last modified on  20130322 17:58:18 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  8 
Author  asteroid (17536) 
Entry type  Definition 
Classification  msc 22D05 
Classification  msc 28C10 
Synonym  Haar modulus 
Synonym  modular character 
Synonym  modular homomorphism 