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I'm taking an intermediate course in probability theory (that is without measure theory) and when defining singular random variables (after showing the devil's function), the book defines:

$X$ is a singular random variable iff $F(x)$ (cumulative distributive function) is continuous and $F'(x)=0$ a.e.

Then it says $X$ is a singular random variable if and only if there exists a lebesgue measure zero set $B$ such that $P(X\in B)=1$ and $F(x)$ is continuous.

It has no proof and I can't find one myself. Could you give me some hints?

Thanks and regards!

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    $\begingroup$ This relies on the fact that every probability measure on the Borel sigma-algebra of the real line can be decomposed uniquely as the sum of three measures, the first one discrete, the second one absolutely continuous, and the third one singular. Most books on measure theory prove this, which one(s) are you using? $\endgroup$
    – Did
    Feb 12, 2016 at 20:28
  • $\begingroup$ @Did I'm using a book written by Barry James, "Probabilidad: un curso de nivel intermedio". As far as I'm concerned it is only available in spanish and portuguese. It is intermediate in the sense that tries to prove everything rigorously without using measure theory... Thanks for your help! $\endgroup$ Feb 12, 2016 at 22:49

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