By using the Iwasawa decomposition, one obtains a (bi-invariant) Haar measure on $G:=\mathrm{SL}(2,\mathbb{R})$ which can be symbolically written as $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, the measures appearing on the right-hand side being the usual ones. This actually means $$\int\_G f(x)\,\mathrm{d}x=\int\_K \int\_N \int\_A f(ank)\,\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$$ for suitable $f$. (Here the order in $ank$, which is expressed by the notation $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, is important, whereas the order of the triple integral is immaterial.) Analogously, one can define another Haar measure $\mathrm{d}\_1 x=\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a$ so that $$\int\_G f(x)\,\mathrm{d}\_1 x=\int\_A \int\_N \int\_K f(kna)\,\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a.$$ We must have $\mathrm{d}\_1 x=c\cdot\mathrm{d}x$ for some $c>0$. Is $c$ equal to 1?



How do these two Haar measures on SL(2,R) compare?
Yes, since you get one from the other by applying the inversion and there is a general rule for the inversion that says $\int\_Gf(x^{-1})dx=\int\_Gf(x)dx\Delta(x)$, where $\Delta(x)$ is the modular function.


