Why does the Iwasawa decomposition allow us to write $dg = dp dk$?

Question

Let $G$ be the rational points of an algebraic group over a local field. Let $P$ be a parabolic subgroup of $G$. The Iwasawa decomposition allows us to write $G = KP$ for a maximal compact subgroup $K$. In fact, the product mapping gives a homeomorphism $K \times P \rightarrow G$. Let $\mu_P$ be a right Haar measure on $P$, and $\mu_K, \mu_G$ Haar measures on $G$. Identifying $G = K \times P$ as topological spaces, why can we say that $\mu_G = \mu_K \mu_P$?

Answer 1

Let $\eta: K \times P \rightarrow G$ be the product mapping. Let $h: P \rightarrow K \setminus G$ be the map $p \mapsto Kp$. Since the projections $K \times P \rightarrow P$ and $G \rightarrow K \setminus G$ are open maps, it is easy to see that $h$ is a homeomorphism and a right $P$-equivariant map. Since $K$ and $G$ are both unimodular, $K \setminus G$ has a right $G$-invariant radon measure $\overline{\mu}$. This measure is also right $P$-invariant, so it transfers over to a right $P$-invariant Radon measure $\mu_P$ on $P$, i.e. a right Haar measure. Then for $f \in C_c(K \setminus G)$, we have

$$\int\limits_{K \setminus G} f(Kg)d\overline{\mu}(Kg) = \int\limits_{K \setminus G} f d\overline{\mu} = \int\limits_P f \circ h \, d\mu_P = \int\limits_P f(p)dp$$

Now for $f \in C_c(G)$, the map $Kg \mapsto \int\limits_K f(kg) dk$ lies in $C_c(K \setminus G)$, so we can integrate it with respect to $\overline{\mu}$. If we do that, we obtain the positive linear functional on $C_(G)$ giving a Haar measure on $G$, which after scaling we may assume to be $\mu_G$. So

$$\int\limits_G f(g)dg = \int\limits_{K \setminus G} \, \int\limits_K f(kg) \, dk \, d \overline{\mu}(Kg) = \int\limits_P \int\limits_K f(kp) \, dk \, d\mu_P(p)$$