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!