Proof of the measure decomposition $dg = dk \, d_lp$ using the Iwasawa decomposition for reductive groups Let $G$ be (the rational points of) a connected, reductive group over a $p$-adic field $F$.  Let $A$ be a maximal $F$-split torus of $G$, and $P$ a minimal parabolic subgroup containing $A$.  Let $M = Z(A)$ be the centralizer of $A$ in $G$, and $N(A)$ the normalizer.  Let $W = N(A)/M$.
The apartments of the building associated to $G$ correspond to maximal $F$-split tori of $G$.  Through some buildings shenanigans I don't understand, one obtains a special, good, maximal compact open subgroup $K$ of $G$, and a certain subgroup $B$ of $K$ called an Iwahori subgroup.
Each element $w$ of $W$ has a representative $\omega(w)$ in $K \cap N(A)$.  The Iwasawa decomposition says that $G$ is the disjoint union of the double cosets
$$ Pw(\omega)B$$
for $w \in W$.  In particular, $G = PK$.  Since $K$ is open in $G$, its Haar measure $dk$ is the restriction of the one on $G$, which we normalize to make $K$ have measure one.  We also normalize a left Haar measure $d_lp$ on $P$ so that $P \cap K$ has measure one.  
Apparently, one can use the Iwasawa decomposition to relate the Haar measures $dg, d_lp$, and $dk$.  This is explained in P. Cartier's article Representations of $\mathfrak p$-adic Groups in the Corvallis proceedings, volume one.  I am confused on a couple of details in the proof.

First, I understand why the function $u_h: G \rightarrow \mathbb{C}$ is well defined.  What I don't understand is how we can conclude that it is continuous and of compact support.  
Next, I don't understand how we can conclude that $h \mapsto \int\limits_G u_h(g)dg$ is left $K \times P$-invariant.  It seems difficult to compute change of variables when integrating the function $u_h(g)$, since for each $g$ one can only compute $u_h(g)$ by a choice of $p$ and $k$ such that $g = pk^{-1}$.  
 A: Not quite directly addressing Cartier's argument, but abstracting the issues slightly, which I do think clarifies what is genuinely involved:
Claim: let $G$ be a unimodular group expressible as $G=AB$ with closed subgroups $A,B$, such that $A\cap B$ is compact. Then (with the usual abuse of notation), up to normalization constants, $dg = da_L\cdot db_R=da_R\cdot db_L$, with left/right Haar measures having the obvious subscripts.
Proof: $G$ is an $A \times B$ space in at least two ways, by $(a,b)(g)=a^{-1}gb$, and by $(a,b)(g)=b^{-1}ga$. The isotropy subgroup of the point $1\in G$ is $A\cap B$, which we are assuming is compact. There is an $A\times B$-invariant measure on $G\cong (A\times B)/(A\cap B)$ if and only if the Haar measure on $A\times B$ restricted to $A\cap B$ is the Haar measure of $A\cap B$, and this is so, since $A\cap B$ is compact. Then this measure is unique, with the property that $\int_G \varphi = \int_A\int_B \varphi(a^{-1}b)\,da\,db$, etc.
In particular, it's not about p-adic groups, or totally disconnected groups, or Lie groups, or... just about general topological groups, and the property that (with different notation than above) a quotient $H\backslash G$ has a right $G$-invariant measure if and only if the modular function of $G$ restricted to $H$ is equal to the modular function of $H$, and then that invariant measure is unique.
(An extended discussion of this can be found in many standard sources. On-line, for example, in my essay http://www.math.umn.edu/~garrett/m/v/smooth_of_td.pdf as background for a discussion of smooth representations of totally disconnected groups.)
EDIT: It may seem that this procedure only shows that the resulting measure is left $A$-invariant and right $B$-invariant. However, using the unimodularity of $G$ and the uniqueness, there is (up to scalars) just one $A\times B$-invariant measure on $A\times B/A\cap B\isom G$. The two-sided Haar measure on $G$ is $A\times B$-invariant, also. So the two coincided.
