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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!

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

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