Suppose $f(x)$ is the probability density function of a random variable $X$, which means:
$$\int_{a}^{b} f(x) dx = 1$$
Also suppose $f$ is continuous and differentiable.
Provide a non-trivial condition under which $\int_{a}^{b} |f^\prime(x)| dx$ exists.
$[a,b]$ maybe a compact interval (regular integral) or $[-\infty, \infty]$ (improper integral).
EDIT: a non-trivial condition is a condition that is satisfied by many well-known distributional families.
A sufficient condition is unimodality of $f$, namely the existence of some $c \in~]a,b[$, such that $f$ non-decreasing on $[a,c]$ and $f$ non-increasing on $[a,c]$. If this property holds, then by Fatou's lemma \begin{eqnarray*} \int_a^c |f'(x)|dx &=& \int_a^c \lim_{h \to 0+} \frac{f(x+h)-f(x)}{h}dx \\ &\le& \liminf_{h \to 0+} \int_a^c \frac{f(x+h)-f(x)}{h}dx \\ &=& \liminf_{h \to 0+} \frac{1}{h}\int_c^{c+h} f(x)dx - \frac{1}{h}\int_a^{a+h} f(x)dx \\ &=& f(c)-f(a). \end{eqnarray*} In a same way, $$\int_a^c |f'(x)|dx \le f(c)-f(b).$$ The conclusion follows.
The same argument applies when $f$ has only finitely many critical points, since, by Darboux Theorem, the sign of $f'$ remains constant between two critical points.
