Invariance of measure on upper half plane The upper half plane has the measure $|y|^{-2}dxdy$. Show that it is invariant under the action of $SL(2, \mathbb{R})$.
I don't understand what any of this means. First, I don't understand what they mean when they say that $|y|^{-2}dxdy$ is a measure. Second, I don't even know what I am being asked to show. 
 A: Let $T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2,\mathbb{R})$, let $T(z) = x_1 + iy_1$ where $z=x_2 + iy_2$. Then, for any Borel set $A \subset\mathbb{H}^2$, consider
$$
\mu(T(A)) = \int_{T(A)} \frac{dx_1 dy_1}{y_1^2} 
= \int_A \frac{\partial(x_1,y_1)}{\partial(x_2,y_2)} \frac{dx_2 dy_2}{y_1^2}
= \int_A \left| \frac{dT}{dz} \right|^2 \frac{|cz+d|^4}{y_2^2} dx_2 dy_2,
$$
where $\frac{\partial(x_1,y_1)}{\partial(x_2,y_2)} = \left| \frac{dT}{dz} \right|^2$ follows from the Cauchy-Riemann equations and $y_1 = \frac{y_2}{|cz+d|^2}$ comes from the change of variables. Now, remark that
$$
\frac{dT}{dz} = \frac{d}{dz} \left( \frac{az+b}{cz+d} \right)
= \frac{a(cz+d)-c(az+b)}{(cz+d)^2} 
= \frac{ac-bd}{(cz+d)^2} 
= \frac{1}{(cz+d)^2},
$$
so we can conclude that $\mu(T(A)) = \int_A \frac{dx_2dy_2}{y_2^2} = \mu(A)$; that is, the action of the fractional linear transformation $T$ on $\mathbb{H}^2$ preserves the hyperbolic measure $\mu$.
A: When dealing with manifolds, volume forms are often conflated with measures. Here, the volume form on the upper half plane is $|y|^{-2}dxdy$, so the measure of a Borel set is its integral with respect to that volume form.
The group $SL(2,\mathbb{R})$ acts on the upper half plane by fractional linear transformations, 
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix}z = \frac{az + b}{cz + d}.$$
Your job is to show that this action is measure-preserving.
