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I am not sure if this is true, but intuitively it seems that if a function is strictly increasing and it is also is differentiable.

It may be because there are no bumps like in the absolute value.

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Not necessarily. Counterexample: $$ f(x)=\begin{cases} x & \text{if }x<0,\\ 2x & \text{if }x\ge 0.\end{cases} $$ Is continuous, strictly increasing but not differentiable at $x=0$.

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So if a a probability distribution has this property, it may not have a density? – Usman Apr 11 '12 at 14:32
^ or its density may not be its derivative – Usman Apr 11 '12 at 14:34
Or you use a different notion of derivative than that in intro calculus. One way to avoid this (for simple functions as in the example above) is to say the probability for region is defined as the integral of the density function (not the density function is the derivative), then a finite number of discontinuities would not be a problem for the Riemann inertial (perhaps use the Lebesgue integral for a countable number). – Chris K. Caldwell Apr 11 '12 at 15:10
@ChrisK.Caldwell An interesting fact about Riemann integrals is that they can also handle countably many discontinuities. In fact, a necessary and sufficient condition for being Riemann integrable is that the set of discontinuities of the function has Lebesgue measure zero. – Chris Janjigian Apr 11 '12 at 16:19
Is it possible to construct an example which is not-differentiable [almost] everywhere? – Asaf Karagila Apr 12 '12 at 1:22

Take a look at Billingsley' Probability and Measure, example 31.1

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Answers should be somewhat self-contained. References and links are fine, but descriptions and explanations in the answer help in case references are not available or links go stale. – robjohn Jan 30 '13 at 13:24

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