Cartan integral formula for a p-adic group? Let $G$ denote a reductive group over a local field $F$. Suppose that $G$ is split over $F$ and fix a maximal (split) torus $A$. Let $A^+$ denote a Weyl chamber in $A$ and let $K$ be a suitable maximal compact subgroup of $G$.
Then it is known that a Cartan decomposition holds:
$$
G=KA^+K.
$$
Hence there is a weight function $w\ge 0$ such that for the Haar-integrals one has
$$
\int_G f(x)\ dx=\int_K\int_{A^+}\int_K f(kal)\ w(a)\ dk\ da\ dl.
$$
If the field is archimedean, the function $w$ can be explicitly computed as a product over positive roots, see Knapp's book on page 141.
My question is this: Is there a similar explicit formula for $w$ in the non-archimedean case?
Normalising the measures so that the compact open subgroups get measure 1 we see that
$$
w(a)=|K\backslash KaK|,
$$
but I don't know whether that's of any help.
 A: In response to Anton for a uniform proof of your formula: note that $K\backslash KaK\cong(a^{-1}Ka\cap K)\backslash K$. Let $\pi:K\rightarrow G(\mathbb{F}_q)$  be the natural quotient map, $K_1:=\ker(\pi)$ and $P:=\pi(a^{-1}Ka\cap K)$. Then $P$ is the $\mathbb{F}_q$-points of the parabolic subgroup of $G$ determined by the walls that $a$ lies on. We have
$$|(a^{-1}Ka\cap K)\backslash K|=|P\backslash G(\mathbb{F}_q)|\cdot|(a^{-1}Ka\cap K_1)\backslash K_1|.$$
Let $\mathfrak{K}$ be the Lie algebra of $K$ and likewise for $\mathfrak{K}_1$. Choose Haar measures for the group and Lie algebra so that $\mu(K_1)=\mu(\mathfrak{K}_1)$. Since $a^{-1}Ka\cap K_1$ and $K_1$ are open compact, we have $|(a^{-1}Ka\cap K_1)\backslash K_1|=\mu(a^{-1}Ka\cap K_1)\backslash\mu(K_1)=\mu(a^{-1}\mathfrak{K}a\cap\mathfrak{K}_1)\backslash\mu(\mathfrak{K}_1)=|(a^{-1}\mathfrak{K}a\cap\mathfrak{K}_1)\backslash \mathfrak{K}_1|$.
Identifying $a$ as an element in the positive Weyl chamber in the cocharacter lattice, the last term is $q$ to the power of $\sum_{\alpha\in\Phi^+}\max(\langle a,\alpha\rangle-1,0)$, while the roots of $P$ are those $\alpha\in\Phi$ with $\langle a,\alpha\rangle\le 0$.
