Let $A$ be a nonempty open subset of $\mathbb{R}$.

Consider a function $f : A \rightarrow \mathbb{R}$. Given that $f$ is continuous, what is the probability that it is differentiable? I suspect it is $0$.

  • 5
    $\begingroup$ You'd have to define how to compute probabilities here... $\endgroup$
    – vonbrand
    Apr 16, 2014 at 17:56
  • 2
    $\begingroup$ Are you asking about measure theory? $\endgroup$
    – user5402
    Apr 16, 2014 at 17:59
  • $\begingroup$ If it can help : the subset of nowhere differentiable functions contains an intersection of open dense subsets. Or if your function is a Brownian motion, then the probability of being differentiable is 0. $\endgroup$
    – user10676
    Apr 16, 2014 at 18:00
  • $\begingroup$ @vonbrand What if I instead asked for your credence in $f$'s differentiability? $\endgroup$ Apr 16, 2014 at 18:00
  • 1
    $\begingroup$ That question is in a finite dimensional setting, and the question plays nicely with countable operations. Probability on infinite dimensional spaces is extremely technical. For what it's worth, I would be shocked if you could find a reasonable interpretation in which the answer was not zero. $\endgroup$ Apr 16, 2014 at 18:35

1 Answer 1


There is a very precise sense in which the answer to your question is "$0$".

Let us denote by $ND$ the set of all continuous nowhere differentiable functions on, say, the interval $[0,1]$, and by $\mathcal C([0,1]$ the Banach space of all continuous functions on $[0,1]$. Then, it can be shown that $SD:=\mathcal C([0,1])\setminus ND$ (the set of all functions which are somewhere differentiable) is a "Haar-null" set, which means that there exists a Borel probability measure $\mu$ on $\mathcal C([0,1])$ such that $$\forall f\in\mathcal C([0,1])\;:\; \mu(f+SD)=0\, .$$

Here, "$\mu(E)=0$" for a possibly non-Borel set $E$ means that $E$ is contained in a Borel set $\widetilde E$ such that $\mu(\widetilde E)=0$. (This remark is needed because $SD$ is not Borel in $\mathcal C([0,1])$).

Thus, in this sense, a continuous function is nowhere differentiable "with probability $1$".

For a proof of this result, this the original paper by Hunt: http://www.ams.org/journals/proc/1994-122-03/S0002-9939-1994-1260170-X/S0002-9939-1994-1260170-X.pdf


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