Probability Function and Bounded Variation Function

Question

Let $ X $ be a continuous random variable with probability function $ f_X(x) $ and distribution function $ F_X(x) $. Assume $ f_X(x) $ is a piecewise continuous function. Is it possible to prove for an arbitrary random variable $ X $, $ f_X(x) $ is a function of bounded variation?

My attempt

Let $ [a,b] \subset supp(X) $, then we can divide $ f_X(x) $ into several subinterval where it only contains strictly increasing function, strictly decreasing function, or constant function. Let say we can divide $ (a,b) $ into $ n+1 $ subinterval with $ a = t_0 < t_2 < ... < t_n = b $, where in every subinterval, $ f_X(x) $ is strictly increasing function, strictly decreasing function, or constant function. For arbitrary partition of $ [a,b] $, then we can calculate variation of $ f_X(x) $ for $ x \in [a,b] $ as follows $$ V_a^b = \sup \sum_{i=0}^{n-1} | f(x_{i+1}) - f(x_i) | $$ $$ V_a^b = | f(x_{1}) - f(x_0) | + ... + | f(x_{n}) - f(x_{n-1}) | $$ Because $ 0 \leq f(x) \leq 1 $ then for every $ x_i $ and $ x_{i+1} $, $ | f(x_{i+1}) - f(x_i) | \leq 1 $, then the upper bound of variation of $ f_X(x) $ is $$ V_a^b \leq 1 + ... + 1 = n $$ The question still remains. How to prove $ n $ is a bounded constant? But my intuition is that, if $ f_X(x) $ is a probability function, then it is unlikely for a bounded interval, it contains infinitely number of monotone function, so $ n $ must be a bounded constant. But I cannot find any source that support my claim.

I need help to support my claim (by explanation or reliable source) or maybe another way to solve it. Thank you!

Answer 1

All we know about a probability density function is that it is non-negative and integrates to 1. You add "piecewise continuous". This is certainly not enough for bounded variation.

Take any (piecewise) continuous function with unbounded variation, e.g., $\sin\frac1x$, and tweak it to be a pdf.

E.g., let $$f_0(x)=1+\sin\frac1x$$ when $x\in(0,1)$ and $0$ otherwise. Then $f_0\ge0$ and let $w:=\int_0^1f_0$.

Now, $f=\frac{f_0}{w}$ is a piecewise continuous PDF with unbounded variation.

How to prove $n$ is a bounded constant?

It is not.