Complete example of haar measure on compact groups like $GL(n,R)$ I am currently reading the proof of existence of haar measure, but I learn better mostly by examples so I would like examples of explicit computation of haar measure mainly on any $Gl(n,R)$ or any lie groups and please also explain the details as I didn't take a course in harmonic analysis I want to understand more what is going on explicitly since I understand stuff more with examples. For example given some matrix in Gl(n,R) I want to see computation done on it.
 A: Here is a useful result in Folland's Modern Real Analysis.

Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^n$ in such a way that, if we identify $G$ with $U$, left translation is an affine map, i.e. $xy = A_xy+b_x$, where $A_x$ is a linear map on $\Bbb R^n$ and $b_x\in\Bbb R^n$. Then $|\det A_x|^{-1}\,dx$ is a left Haar measure on $G$. ($dx$ is the Haar measure on $\Bbb R^n$.) Particularly, if $G$ is a matrix group, then $|\det A|^{-n}\,dA$ is a Haar measure, where $dA$ is the Haar measure on $\Bbb R^{n\times n}$.

I'm going to assume that $G$ is actually a subset of $\Bbb R^n$ so that I don't have to keep track of all of the maps. You should do it in general though as a nice exercise.
The way to see the first result is by considering the Jacobian of the transformation. If you have a measurable set $S$ in $G$, then the measure of $xS$ is nothing more than the measure of $A_xS$ - note that the affine term $b_x$ does not contribute due to the Haar measure associated to $\Bbb R^n$. Moreover, since we have a Haar measure on $G$, the measure of $xS$ must be the measure of $S$.
Suppose the Haar measure on $G$ is denoted $\mu_x$. Since $A_x$ is a linear transformation, the measure of $A_xS$ is nothing more than $|\det A_x|\cdot \mu_x(S)$. Prove this yourself by considering basic Borel sets. The only way for the measure not to change (i.e. $\mu_x(xS) = \mu_x(S)$) is if we have that $\mu_x = |\det A_x|^{-1}\,dx$ (where $dx$ is the usual Haar measure on $\Bbb R^n$).
The second result follows from the first by considering the columns of the matrix independently. You can view an $n\times n$ matrix as a vector of length $n^2$, then $A$ acts naturally on the subvectors. On each piece, you pick up a factor of $|\det A|^{-1}$, giving a total factor of $|\det A|^{-n}$. Reassembling the matrix from the elongated vector gives the result.
Since manifolds do not carry integration naturally but instead we consider the images in $\Bbb R^n$ and do integration there, pushing the group into $\Bbb R^n$ is pretty natural anyway. So the above analysis works pretty well for matrix Lie groups.
Here is a related question.
