Haar measure of $\operatorname{SL}_n(\mathbb{R})$ in terms of the product $K\times A\times N$

Question

I'm misunderstanding a standard argument about the interpretation of the Haar measure on $K\times A\times N$. (See Lemma 2.5 on page 145 of Bekka-Mayer for example.)

Let $G=\operatorname{SL}_n(\mathbb{R})$, $K=\operatorname{SO}(n)$ and let $B\subset G$ denote the upper triangular subgroup.

We have a diffeomorphism $\phi:K\times B \to G$ given by multiplication. Write $dg$ for the (bi-invariant) Haar measure on $G$ and let $dx$ denote the measure on $K\times B$ which pushes forward to $dg$. Alternatively, one may think of them as top forms, $dg$ pulling back to $dx$. We immediately have that $dx$ is left invariant under the families of maps $(k,b) \mapsto (k_0k,b)$ and $(k,b) \mapsto (k,bb_0)$.

Question: If $dk$ and $db$ denote left and right Haar measures on $K$ and $B$ respectively. Is it then clear that $dx$ must be the product measure $dk\times db$?

$K\times B$ viewed as a product group will certainly have a unique (upto scalar) measure invariant under the maps $(k,b)\mapsto (kk_0,bb_0)$. But does $dx$ satisfy this property? Or should I be looking at it differently?

Answer 1

Here be the comment of 'reuns' (see above):

From that $dg$ is bi-invariant then $\mu(U\times V)= \int_{U^{-1}V}dg$ is a right Haar measure on $K\times B$?