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).
It's hard to decide what nontrivial means. For a proper interval $[a,b]$ the continous differentiability implies boundedness of $f'(x),$ therefore the integral in question is finite.
Consider $[0,\infty).$ Assume $f(x)\searrow 0$ when $x\to \infty.$ Then $f'(x)\le 0.$ Hence $$\int\limits_0^n|f'(x)|\,dx = -\int\limits_0^nf'(x)\,dx =f(0)-f(n)\underset{n\to \infty}{\longrightarrow}f(0)$$ Therefore $$\int\limits_0^\infty|f'(x)|\,dx=f(0)$$