Basic Confusion on Push-Forward of a Measure Let $\rho:\mathbb{R}^d\to \mathbb{R}$ be a probability density on $\mathbb{R}^d$. Let $f:\mathbb{R}^d\to \mathbb{R}^d $ be invertible. Consider the push-forward of $\rho$ by $f$, denoted $f_{\#}\rho$, see Wikipedia.
My question is the integral of $f_{\#}\rho$ always 1 ? And is $f_{\#}\rho$ guaranteed to be a density (i.e absolutely continuous with respect to Lebesgue measure ?
 A: First of all it should be pointed out that push-forward is an operation applied to measures, rather than
densities.  If $\rho $ is a density (measurable, nonnegative function) on $\mathbb{R}^n$, then the associated measure
$\mu _\rho $ is defined for every measurable set $E\subseteq \mathbb{R}^n$ by
$$
  \mu _\rho (E) = \int_E\rho (x) dx.
  $$
Calling the push-forward measure $\nu $, we have by definition
$$
  \nu (E) =   \mu _\rho (f^{-1}(E)) = \int_{f^{-1}(E)}\rho (x) dx = \int_E\rho (f^{-1}(x))|J(f^{-1})(x)| dx,
  $$
by the change of variable formula,
where $J$ refers to the Jacobian.   In other words $\nu $ is the measure given by the density function
$$
  \tau (x)=\rho (f^{-1}(x))|J(f^{-1})(x)|,
  $$
so it is absolutely continuous with respect to Lebesgue measure.
Needless to say, all of this requires $f$ to be smooth!  Otherwise strange things can happen:  there are homeomorphisms
of $\mathbb{R}$ which send a set of positive measure onto the Cantor set.  The push-forward of Lebesgue
measure (density 1) through such function assigns positive measure to the Cantor set, hence is clearly not absolutely continuous.
A: When you take a probability measure with a density w.r.t. Lebesgue measure, and push it forwards, you get a new probability measure, but this push-forward measure need not have  a density.  For the first claim here, look at the theorem recited in the wikipedia page you cite, in the special case of the function $g$ being the constant $1$. For the second claim here, let $H:[0,1]\to[0,1]$ be any strictly increasing no-where differentiable function, for which $H(0)=0$ and $H(1)=1$, and let $f$ be its inverse function.  If $X$ has uniform distribution on $[0,1]$ then $f(X)$ has $H$ as its cumulative distribution function, but (by hypothesis on $H$) has no density function.
Here is one class of examples of such $H$;  they are discussed in Billingsley's Ergodic Theory book.  Let $B_i$ be i.i.d. random bits, with $P(B_i=0)=1-p, P(B_i=1)=p$, where $0<p<1$ and $p\ne 1/2$.  Then $H(x)=P(\sum_{n>0} B_n 2^{-n}\le x)$ has the required properties.  (It is easy to plot the graph of $H$: you know $H(1/2)=1-p$, and you can fractally interpolate. Billingsley gives such a plot on p.37.)
