Defining left Haar measure on locally compact group Let $G$ is a locally compact group that is homeomorphic to an open subset (say $U$) of $\Bbb R^d$ ,and let $\varphi$ be a homeomorphism of $G$ onto $U$.
Show that if for each $a$ in $G$ the function $u \mapsto \varphi(a\varphi^{−1}(u))$ is the restriction to $U$ of an affine map $L_a : \Bbb R^d \to \Bbb R^d$, then the formula
$\mu(A)=\int_{\varphi(A)}|\det(L_{\varphi^{−1}(u)}|\lambda(du)$ defines a left Haar measure on G. ($\lambda$ is Lebesgue measure on $\Bbb R^d$)
I'm trying to prove the proposition above. I've seen some likes of this but a function as a restriction of an affine map made me confused. I know I should use change of variables for differentiable functions since affine maps are differentiable functions on finite dimensional space but I could not get the consequence with $x\in G, \ \ \ \mu(xA)=\mu(A)$. I'll be so glad if I can see an exact proof of this from any reference etc. for understanding well. 
Thanks in advance for any help. 
 A: I think we can argue like this: according to the hypothesis, we have $\varphi(g\varphi^{−1}(u))=A_gu+b_g$ where $g\in G,\ A_g\in GL_n$ and $b_g\in \mathbb R^n$. 
The proof uses a preliminary observation, which I believe is what Reuns alluded to in the comments:
if $g,g'\in G,\ z\in U,$ then
$A_gA_{g'}z+A_gb_{g'}+b_g=A_g(A_{g'}z+b_{g'})+b_g=A_g\varphi(g'\varphi^{-1}(z))+b_g=\\\varphi(g\varphi^{-1}(\varphi(g'\varphi^{-1}(z))))=\varphi(gg'\varphi^{-1}(z))=A_{gg'}z+b_{gg'}\Rightarrow \\ (A_{gg'}-A_gA_{g'})z=b_{gg'}-(b_g+A_gb_{g'}).$ 
Since this last equality is true for all $z\in U$, we conclude that 
$\tag1 A_{gg'}=A_gA_{g'}$ 
We have, for $g\in G,\ h_g:U\to \mathbb R^n:u\mapsto \varphi(g\varphi^{−1}(u))$ which satisfies 
$\tag2 h_g(U)=U\ \text{and}\ Dh_g(u)=A_g$ 
Now, $\mu$ clearly defines a measure on $G$, so to show it is a Haar measure, all we need to do is verify that 
$\int_Gf(g)d\mu=\int_Gf(yg)d\mu\ \text{for all}\ f\in C_c(G),\ y\in G.$  That is, that 
$\tag3 \int_Uf\circ h_g^{-1}(x)|\det A_g|^{-1}dx=\int_UL_y\circ f\circ h_g^{-1}(x)|\det A_g|^{-1}dx$
And $(3)$ follows readily from $(1)$ and $2)$.
