I have a real function that satisfies:

  1. $f:\mathbb R\rightarrow\mathbb R$ is differentiable at $x$ for $x\neq x_0$.
  2. There is a full-measure set $T$ such that for any sequence $t_n\in T$ with $t_n\stackrel{n\rightarrow\infty}{\longrightarrow} x_0$ we have $\lim_{n\rightarrow\infty} f^\prime (t_n) = c$.
  3. $f$ is non-decreasing

I'm trying to show that $\mathbf{f}$ must also be differentiable at $\mathbf{x_0}$.

My idea is that if there is some sequence $x_n$ with $x_n\stackrel{n\rightarrow\infty}{\longrightarrow} x_0$ and $\lim_{n\rightarrow\infty}\frac{f(x_n)-f(x_0)} {x_n-x_0} \neq c$ then I can approximate the $x_n$ and $x_0$ with points in $T$ but I have to be careful with the choice of $\varepsilon$ and $\delta$ in approximation because we have two limits here (the derivative itself and the limit of the derivatives). I'm not sure if this would work because it seems that I would need to change the order of limits. Besides, in real analysis, there are some many pathological cases that perhaps there is even a counter-example. Any suggestions on how to attack this problem?

  • $\begingroup$ As the accept answer shows we must also have $f$ continuous. The accept answer uses Theorem 7.15 (Rudin, Real & Complex Analysis, 3rd ed, p. 143). $\endgroup$ Jul 29, 2015 at 0:47

1 Answer 1


If you let $f(x)=0$ for $x\le 0$ and $f(x)=1$ for $x>0$ you get a counterexample. The result you want requires that you assume that $f$ is continuous at $x_0$.

Assuming that, there's probably a more elementary proof than the following simple but not quite so elementary proof:

Say $x_0=0$ for convenience. Subtract a constant so $f(-1)=0$.

Now since $f$ is continuous at $0$ and differentiable elsewhere it is continuous. So there exists a finite measure $\mu$ on $[-1,1]$ such that $$f(x)=\mu([-1,x])\quad(x\in[-1,1]).$$

Now $\mu=\mu_a+\mu_s$ where $\mu_a$ is absolutely continuous and $\mu_s$ is singular (wrt Lebesgue measure). We must have $\mu_s([-1,1]\setminus\{0\})=0,$ since otherwise there would exist $x\ne0$ such that $f$ was not differentiable at $x$. And $\mu_s(\{0\})=0$ since $f$ is continuous at $0$.

So $\mu_s=0$, which says that $f$ is absolutely continuous. In particular, $$\frac{f(x)-f(0)}{x}=\frac1x\int_0^xf'(t)\,dt,$$and now the assumption about $f'$ tending to $c$ on a set of full measure shows that $f'(0)=c$.

  • 1
    $\begingroup$ well, you don't need $\mu$. Since $f'$ is bounded near $x_0$, $f$ is Lipschitz continuous on a neighborhood of $x_0$, thus absolutely continuous. $\endgroup$
    – user251257
    Jul 28, 2015 at 17:50
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    $\begingroup$ @user251257 Exactly how does it follow that $f$ is Lipschitz? Can't use fundamental theorem of calculus since we don't know $f$ is absolutely continuous. Mean value theorem is at least problematic since we don't know that $f'$ is bounded near $x_0$, just on a set of full measure. Which is enough. Why is it enough, exactly? You may well be right - I don't see it offhhand... $\endgroup$ Jul 28, 2015 at 17:54
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    $\begingroup$ @user251257 While you're thinking, keep the "Cantor function" in mind. Nondecreasing, continuous, $f'=0$ almost everywhere, not constant. Not a counterexample to the theorem because it's not differentiable, but a counterexample to various wrong proofs of the theorem... $\endgroup$ Jul 28, 2015 at 18:05
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    $\begingroup$ I suppose that would be appropriate, yes. (Feel free to send money instead. heh-heh) $\endgroup$ Jul 28, 2015 at 18:12
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    $\begingroup$ @DavidC.Ullrich So you didn't use monotonicity (or it is hidden somewhere in (implicit) assumption on $\mu$)? I guess Lebesgue decomposition is valid for signed measures too. $\endgroup$
    – A.Γ.
    Jul 28, 2015 at 18:23

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