Let $G$ be a connected, reductive group, split over a local field $F$. Let $B = TU$ be a Borel subgroup defined over $F$ with maximal torus $T$ and unipotent radical $U$. Let $P$ be a parabolic subgroup of $G$ containing $F$, and let $N = R_u(P)$. For $\alpha \in \Phi(T,U)$, let $U_{\alpha}$ be the root subgroup corresponding to $\alpha$.

Fix an $F$-isomorphism $x_{\alpha}: \mathbb{A}^1\rightarrow U_{\alpha}$ such that $t x_{\alpha}(a)t^{-1} = x_{\alpha}(\alpha(t)a)$ for all $t \in T, a \in \mathbb{A}^1$. Each $x_{\alpha}$ allows us to transfer a Haar measure on $F$ to one on $U_{\alpha}$. Since the product map (in any order)

$$\prod\limits_{\alpha \in \Phi(T,N)} U_{\alpha}(F) \rightarrow N(F)$$

is a homeomorphism, one can transfer the Radon product measure on $\prod\limits U_{\alpha}(F)$ to a Borel measure $\mu$ on $N(F)$. Then $\mu$ is a Haar measure on $N(F)$.

One can verify this directly for parabolic subgroups of split classical groups, but how can one argue that $\mu$ is a Haar measure for arbitrary split groups? More generally in the non-split case, is it possible to define a Haar measure on $N(F)$ using root subgroups for relative roots?