Basics of Haar measure 
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!
 A: A good place to start is to make sure the definitions of things are clear. 


*

*A measure $\mu$ on $G$ is left invariant if for every test function $f$ and every $h\in G$ one has $\int f(hg)\,d\mu(g) = \int f(g)\,d\mu(g)$.

*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 $dgh_0^{-1} = \delta dg$.
