Uniformly distributed probabiity density over $\mathbb{R}^n$ I came across the following phrase in my reading:

Let $\rho(x)$ be a uniformly distributed probability density defined on $\mathbb{R}^n$.

I know that $\rho(x)$ being a probability density function means that $\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = 1$, but doesn't uniformly distributed mean that $\rho(x) = c$ for some positive constant $c$?
But, then $$\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = \displaystyle \int_{\mathbb{R}^n} c dx = \infty$$, so how is it possible for a probability density to be uniformly distributed over $\mathbb{R}^n$?
Does it implicitly assume for instance that $\rho$ vanishes outsie a set of finite measure?

Uniformly distributed means it follows a Uniform distribution.
As you correctly said, $\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = 1$.
But the density function is:
$
\rho(x)=\begin{cases}
  \frac{1}{b - a} & \mathrm{for}\ a \le x \le b, \\[8pt]
  0 & \mathrm{for}\ x<a\ \mathrm{or}\ x>b
  \end{cases} $
$\begin{align}\displaystyle \int_{\mathbb{R}^n} \rho(x) dx &=\displaystyle \int_{-\infty}^a \rho(x) dx+\displaystyle \int_{a}^b \rho(x) dx+\displaystyle \int_{b}^\infty \rho(x) dx\\ &=0+\displaystyle \int_{a}^b \frac{1}{b - a} dx+0\\ &= \left.\frac{x}{b - a}\right |_a^b\\ &=1\end{align}$
