# An absolutely continuous cumulative distribution function that fails to have a Riemann-integrable pdf.

We know that if a real-valued random variable $X$ on a probability space has an absolutely continuous cumulative distribution function (cdf) $F$, then $X$ possesses a probability density function (pdf), i.e., a measurable function $f: (\mathbb{R},\text{Borel}(\mathbb{R})) \to (\mathbb{R}_{\geq 0},\text{Borel}(\mathbb{R}_{\geq 0}))$ such that $$\forall x \in \mathbb{R}: \quad F(x) := \mathbf{Pr}(X \leq x) = \int_{(- \infty,x]} f ~ d{\mu},$$ where $\mu$ is the standard Borel measure on $(\mathbb{R},\text{Borel}(\mathbb{R}))$.

Question: Is it possible that $X$ fails to have any Riemann-integrable pdf at all?

I was thinking that we could

1. assemble two disjoint copies of the Smith-Volterra-Cantor set on the real line (with each copy having measure $1/2$),

2. Lebesgue-integrate the indicator function of this assembly on $(- \infty,x]$ for arbitrary $x \in \mathbb{R}$, and

3. consider the resulting monotonically increasing function in $x$ as a possible candidate for a cdf that fails to have any Riemann-integrable derivative.

This idea seems highly plausible, but it might be bogus. Hence, any comments from the MSE community are greatly welcome!

-
Can you write down step 2. please? – Alecos Papadopoulos Aug 5 '13 at 20:46