I read from a book without proof the following theorem: Let $f(x)$ be a differentiable function, then $f(x)$ is strictly increasing if and only if $f'(x) \geq 0$, and $f'(x) \gt 0$ almost everywhere. Is it correct? I doubt, but I could neither prove it nor give a counter example.

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    $\begingroup$ Monotonic should be replaced by increasing. $\endgroup$ Commented Nov 1, 2011 at 1:55
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    $\begingroup$ Do you agree that $f$ is increasing iff its derivative is non negative? Is it the "almost everywhere positive" part that is posing a problem? The first part has one obvious direction : if you are increasing then the derivative must be everywhere non negative. The other direction (non negative derivative implies increasing) follows from Rolle : if $a<b$ then there exists $a<c<b$ with $f(b)-f(a)=f'(c)(b-a)\geq 0$. $\endgroup$ Commented Nov 1, 2011 at 1:58
  • $\begingroup$ yes, the problem is the "almost everywhere positive" part. $\endgroup$
    – Chao
    Commented Nov 1, 2011 at 2:07
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    $\begingroup$ One direction is not too hard : if $f'$ is $\geq 0$ almost everywhere $>0$ then $f$ must be stricly increasing. For take $a<b$. Since $f'\geq 0$, by what precedes you have $f$ increasing. On the other hand, since $f'>0$ almost everywhere, you can find $a<c<b$ with $f'(c)>0$ and by definition of $f'(c)$ as a limit you can find $c<d<b$ close to $c$ with $$\frac{f(d)-f(c)}{d-c}>\frac{1}{2}f'(c)>0.$$ Thus $$f(a)\leq f(c)<f(d)\leq f(b)$$ which shows that $f$ is stricly increasing. The two $\leq$ come from $f$ being increasing, while the $<$ comes from the difinition of $c$ and $d$. $\endgroup$ Commented Nov 1, 2011 at 2:13
  • $\begingroup$ So actually it's a sufficient additional condition that $f'(x)>0$ on a dense set. This is also necessary, due to the mean value theorem. $\endgroup$ Commented Nov 1, 2011 at 2:18

2 Answers 2


A necessary and sufficient condition is that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense subset of $\mathbb{R}$. The condition stated in the question is sufficient, but not necessary. Consider the following differentiable and strictly increasing function for which $f^\prime > 0$ does not hold almost everywhere. Let $C$ be the fat Cantor set (which is closed with positive measure, but does not contain any nontrivial open intervals). Let $g(x)$ be the distance from any point $x$ to $C$. Then, $g$ is continuous and vanishes precisely on $C$. The integral $$ f(x)=\int_0^xg(y)\,dy $$ is strictly increasing and continuously differentiable, but $f^\prime=0$ on $C$.

We can show that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense set is indeed necessary and sufficient for $f$ to be strictly increasing (see also Olivier Bégassat's comments to the question).

Sufficiency: If $f^\prime\ge0$ then the mean value theorem implies that $f$ is increasing. If it was not strictly increasing then we would have $f(a)=f(b)$ for some $a < b$ implying that $f^\prime=0$ on $[a,b]$, contradicting the condition that $f^\prime > 0$ on a dense set. This proves sufficiency of the conditions.

Necessity: In the other direction, suppose that $f$ is strictly increasing. Then, $f^\prime\ge0$ follows directly from the definition of the derivative. Also, $f(a) < f(b)$ for any $a < b$. The mean value theorem implies that $f^\prime(x) > 0$ for some $x\in[a,b]$. So, $f^\prime > 0$ on a dense set.

  • $\begingroup$ Your example shows that $f$ stricly increasing need not imply that $f'$ be positve almost everywhere. I was wondering wether $f'=0$ on a dense set implies $f$ constant. All being the same of course : $f$ is still increasing and differentiable everywhere. $\endgroup$ Commented Nov 1, 2011 at 2:59
  • $\begingroup$ @George: Thank you for the detailed answer. $\endgroup$
    – Chao
    Commented Nov 1, 2011 at 3:31

As others have said, the proposed "almost everywhere" criterion is not correct: it is sufficient, but not necessary, for $f$ to be strictly increasing.

In his very nice answer, George Lowther has given a correct necessary and sufficient condition: namely that $f'(x) > 0$ on a dense subset of the domain. As it happens, I recently covered this topic in my "Spivak calculus class" and found that even Spivak's text didn't give quite the treatment of this point that I wanted. So I wrote something up myself. Here is the statement:

[Warning: where others here say "strictly increasing", I say "increasing"; where others say "increasing", I say "weakly increasing", and similarly for "monotone" and "weakly monotone".]

Second Monotone Function Theorem: Let $f: I \rightarrow \mathbb{R}$ be a function which is continuous on $I$ and differentiable on the interior $I^{\circ}$ of $I$.
a) The following are equivalent:
(i) $f$ is weakly monotone.
(ii) Either we have $f'(x) \geq 0$ for all $x \in I^{\circ}$ or $f'(x) \leq 0$ for all $x \in I^{\circ}$.
b) Suppose $f$ is weakly monotone. The following are equivalent:
(i) $f$ is not monotone.
(ii) There exist $a,b \in I^{\circ}$ with $a < b$ such that the restriction of $f$ to $[a,b]$ is constant.
(iii) There exist $a,b \in I^{\circ}$ with $a < b$ such that $f'(x) = 0$ for all $x \in [a,b]$.

Note that my statement of the equivalence is slightly different (and, perhaps, slightly simpler?) than George's. For the proof, please see $\S 5.1$ of these notes. Finally, note that when George appeals to the Intermediate Value Theorem I think he really means to appeal (as I do) to the Mean Value Theorem, although (confusingly, for students) these names are not completely standardized even among mathematicians.

  • $\begingroup$ I did intend to use the mean value theorem. I corrected my answer, thanks! $\endgroup$ Commented Nov 1, 2011 at 11:19

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