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 $\mathrm{\Delta }:G⟶{ℝ}^{+}$ such that, for every $t\in G$ and every measurable subset $A$

$$\mu (At)=\mathrm{\Delta }(t^{-1})\mu (A)$$

Moreover, if $f:G⟶ℂ$ 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 $\mathrm{\Delta }(t)\in {ℝ}^{+}$. 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

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

• $\mu (As)=\mathrm{\Delta }{(s)}^{-1}\mu (A)$

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 $\mu (At)=\mathrm{\Delta }(t^{-1})\mu (A)$. $\mathrm{\square }$