Pair of nonzero continuous functions $\text{Aff}(\mathbb{R}) \to \mathbb{R}^\times$, left and right invariant measures. For $a \in \mathbb{R}^\times$ and $b \in \mathbb{R}$, let$$g_{a, b} : \mathbb{R} \to \mathbb{R}, \text{ }x \mapsto a \cdot x + b$$be an affine linear map. Let$$\text{Aff}(\mathbb{R}) = \{g_{a, b} : a \in \mathbb{R}^\times,\, b \in \mathbb{R}\}$$be the group, with respect to composition, of all such maps. Identify $\text{Aff}(\mathbb{R})$ with$$\{(a, b) \in \mathbb{R}^2 : a \neq 0\},$$an open subset of $\mathbb{R}^2$, via $g_{a, b} \mapsto (a, b)$. What is a pair of nonzero continuous functions $\phi$, $\psi : \text{Aff}(\mathbb{R}) \to \mathbb{R}^\times$, such that $\phi(a, b) \cdot da\,db$ is a left invariant, resp. $\psi(a, b) \cdot da\,db$ is a right invariant, measure on $\text{Aff}(\mathbb{R})$?
 A: As PhoemueX commented, the link How do I go about proving da db/a^(-2) is a left Haar measure on the affine group?, has the method to compute the Haar measure for any group. Just following that method gives the answer easily.
Now as $g_{c,d}\circ g_{a,b}(x) = g_{c,d}(ax+b) = c(ax+b) + d$, we get that the group multiplication is $(c,d).(a,b) = (ca,cb+d)$. Similarly $(a,b).(c,d) = (ac,ad+b)$ so the group is not commutative.
Left Haar measure: 
Let $\phi(a,b)da\,db$ be a left Haar measure, i.e. $\forall g\in G, f\in C_c(G)$ we have $\int f(g.(a,b))\phi(a,b)da\ db = \int f(a,b)\phi(a,b)da\ db $
So letting $g = (c,d)$ we get 
$\int f(ca,cb+d)\phi(a,b)da\ db = \int f(a,b)\phi(a,b)da\ db $
Now substitute, $ca= A, cb+d = B$ hence $a = A/c, b = (B-d)/c$. So the Jacobian is $1/c^2$. Renaming $A,B$ as $a,b$ we get
$ \int f(a,b)\phi(a/c,(b-d)/c) (1/c^2)da\,db = \int f(a,b)\phi(a,b)da\ db$
Hence $\phi(a/c,(b-d)/c) (1/c^2) = \phi(a,b)$. As $d$ can be varied arbitrarily, we see that $\phi(a,b)$ in fact does not depend on $b$. So $\phi(a,b) = \phi(a)$ and hence 
$\phi(a/c) = c^2\phi(a) \implies \phi(a) = (1/a^2)\phi(1)$. Hence for any constant $C>0, \phi(a,b) = C/a^2$ is a left Haar measure and any left Haar measure is of this form.
Right Haar measure: 
Let $\psi(a,b)da\,db$ be a right Haar measure, i.e. $\forall g\in G, f\in C_c(G)$ we have $\int f((a,b).g)\psi(a,b)da\ db = \int f(a,b)\psi(a,b)da\ db $
So letting $g = (c,d)$ we get 
$\int f(ac,ad+b)\psi(a,b)da\ db = \int f(a,b)\psi(a,b)da\ db $
Now substitute, $ac= A, ad+b = B$ hence $a = A/c, b = B-ad$. So the Jacobian is $1/c$. Renaming $A,B$ as $a,b$ we get
$ \int f(a,b)\psi(a/c,b-ad) (1/c)da\,db = \int f(a,b)\psi(a,b)da\ db$
Hence $\psi(a/c,b-ad) (1/c) = \psi(a,b)$. As $d$ can be varied arbitrarily, we see that $\psi(a,b)$ in fact does not depend on $b$. So $\psi(a,b) = \psi(a)$ and hence 
$\psi(a/c) = c\psi(a) \implies \psi(a) = (1/a)\psi(1)$. Hence for any constant $C>0, \psi(a,b) = C/a$ is a right Haar measure and any right Haar measure is of this form.
