Suppose $G$ is a locally compact group. Then $G$ has a left-invariant measure $dg$, say, which means that $$\int f (hg) dg = \int f(g) fg$$ for any test function integrable on $G$. The left-invariant measure is unique up to a positive constant multiple; therefore, $$\int f (hg) dg = \delta(h) \int f(g) fg,$$ where $\delta(h) > 0$ depends only on $h$ because $dgh^{-1}$ is another left-invariant measure. The factor $\delta(h)$ is called the modular function of $G$. Clearly $\delta : G \to \mathbb{R}^+$ is a group homomorphism, and one also shows....

I feel totally confused about the sentence "therefore, ... because $dgh^{-1}$ is another left-invariant measure." What is the reason for "therefore"? Why is $dgh$ a left-invariant measure? (It seems right multiplication...) Also confused about why $dgh^{-1}$ is a left-invariant measure and why because of this fact, $\delta(h)>0$ depends only on $h$.

Hope someone could explain it in details. Thanks a lot!

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    $\begingroup$ I have no time for a proper answer, but let me point this out: Right multiplication and left inveriance go well together because left and right multiplication commute. In other words, $h\mapsto gh$ and $h\mapsto hg'$ commute for any $g$, $g'$. $\endgroup$ – Harald Hanche-Olsen Oct 15 '12 at 7:06
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    $\begingroup$ 1) Have you actually tried to check if it is a left-invariant Haar measure? Which parts are hard to prove? 2) It is useful to know that the Haar measure is unique up to a constant. $\endgroup$ – N.U. Oct 15 '12 at 8:04
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    $\begingroup$ Have you sometimes written $fg$ in place of $dg$ ?. $\endgroup$ – GEdgar Jan 14 '14 at 14:50

A good place to start is to make sure the definitions of things are clear.

  1. A measure $\mu$ on $G$ is left invariant if for every test function $f$ and every $h\in G$ one has $\int f(hg)\,d\mu(g) = \int f(g)\,d\mu(g)$.
  2. If $dg$ is the Haar measure on $G$ and $h_0\in G$ is a fixed element, then the measure $dgh_0^{-1}$ is by definition given by $\int f(g)\,dgh_0^{-1} := \int f(gh_0)\,dg$ for all test functions $f$.

To show the measure $dgh_0^{-1}$ is left-invariant for any fixed $h_0\in G$, we must check that the condition in definition (1) holds. For any $h\in G$ and any test function $f$ that $$\int f(hg)\,dgh_0^{-1} := \int f(hgh_0)\,dg = \int f(gh_0)\,dg := \int f(g)\,dgh_0^{-1}.$$ The first and third equalities are by definition, and the second is because $dg$ is left invariant. This proves $dgh_0^{-1}$ is left invariant.

It is a fact (assumed in the problem) that any left invariant positive measure $\mu$ on $G$ is a multiple of the Haar measure, i.e., $\mu = \delta dg$ for some $\delta>0$ depending on $\mu$. We have shown that $dgh_0^{-1}$ is left invariant for any fixed $h_0\in G$, so there is a $\delta>0$ depending on $h_0$ such that $dgh_0^{-1} = \delta dg$.

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