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} &amp; \mathrm{for}\ a \le x \le b, \\[8pt]
  0 &amp; \mathrm{for}\ x&lt;a\ \mathrm{or}\ x&gt;b
  \end{cases} $

$\begin{align}\displaystyle \int_{\mathbb{R}^n} \rho(x) dx &amp;=\displaystyle \int_{-\infty}^a \rho(x) dx+\displaystyle \int_{a}^b \rho(x) dx+\displaystyle \int_{b}^\infty \rho(x) dx\\ &amp;=0+\displaystyle \int_{a}^b \frac{1}{b - a} dx+0\\ &amp;= \left.\frac{x}{b - a}\right |_a^b\\ &amp;=1\end{align}$
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} &amp; \mathrm{for}\ a \le x \le b, \\[8pt]
  0 &amp; \mathrm{for}\ x&lt;a\ \mathrm{or}\ x&gt;b
  \end{cases} $

$\begin{align}\displaystyle \int_{\mathbb{R}^n} \rho(x) dx &amp;=\displaystyle \int_{-\infty}^a \rho(x) dx+\displaystyle \int_{a}^b \rho(x) dx+\displaystyle \int_{b}^\infty \rho(x) dx\\ &amp;=0+\displaystyle \int_{a}^b \frac{1}{b - a} dx+0\\ &amp;= \left.\frac{x}{b - a}\right |_a^b\\ &amp;=1\end{align}$