As it is well known every continuous function has the intermediate value property, but even some discontinuous functions like $$f(x)=\left\{ \begin{array}{cl} \sin\left(\frac{1}{x}\right) & x\neq 0 \\ 0 & x=0 \end{array} \right.$$ are having this property.
In fact we know that a derivative always have this property.

My question is, if for every function $f$ with the intermediate value property exists a function $F$ such that $F'=f$. And if so, is $F$ unique (up to a constant you may add)?

My attempts till now: (Just skip them if you feel so)

The most natural way of finding those functions would be integrating, so I guess the question can be reduced to, if functions with the intermediate value property can be integrated.
This one depends heavy on how you define when a functions is integrable, we (my analysis class) said that we call a function integrable when it is a regulated function (the limits $x\to x_0^+ f(x)$ and $x\to x_0^- f(x)$ exists ) .
As my example above shows, not every function with the intermediate value property is a regulated function. But if we say a function is integrabel when every riemann sum converges the above function is integrable, so it seems like this would be a better definition for my problem.

Edit: As Kevin Carlson points out in a commentar that being a derivative is different from being riemann integrable, he even gave an example for a function which is differentiable but it's derivative is not riemann integrable. So we can't show that those functions are riemann integrable as they are not riemann integrable in general. Now I have no clue how to find an answer.

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    $\begingroup$ Though it's a deep point, it's important to point out that being a derivative is not the same thing as being integrable. The main example is the derivative of Volterra's function, which you can read about on Wikipedia. This is why a precise statement of the fundamental theorem of calculus supposes that $f$ is integrable, or often that it's even continuous. (Here by "integrable" I mean the Riemann sums converge.) $\endgroup$ Mar 11, 2013 at 7:48
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    $\begingroup$ what about "Conway base 13 function"? $\endgroup$
    – Laura
    Mar 11, 2013 at 8:06
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    $\begingroup$ @Tai yeah that sounds good, but how to proof that conway base 13 function is not a derivative? $\endgroup$ Mar 11, 2013 at 8:12
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    $\begingroup$ FYI, derivatives are also Baire $1$ functions. Bruckner's book Differentiation of Real Functions discusses in detail many of the attractive properties that functions simultaneously Darboux (have the Intermediate Value Property) and Baire $1$ have, along with how far in many other ways such functions can be from being a derivative. $\endgroup$ Mar 11, 2013 at 19:05
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    $\begingroup$ Don't read too much into this analogy, but you can think of the collection of Baire $1-$ Darboux functions as nicely complementing each of the separate properties Baire $1$ and Darboux in much the same way that the combined topological property compact Hausdorff is much better than you'd expect when examining separately the properties compact and Hausdorff. $\endgroup$ Mar 11, 2013 at 19:09

3 Answers 3


If you compose $ \tan^{-1} $ with Conway’s Base-$ 13 $ Function, then you get a bounded real-valued function on the open interval $ (0,1) $ that satisfies the Intermediate Value Property but is discontinuous at every point in $ (0,1) $. Therefore, by Lebesgue’s theorem on the necessary and sufficient conditions for Riemann-integrability, this function is not Riemann-integrable on any non-degenerate closed sub-interval of $ (0,1) $.

Now, it cannot be the derivative of any function either, because by the Baire Category Theorem, if a function defined on an open interval has an antiderivative, then the function must be continuous on a dense set of points. This thread may be of interest to you. :)

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    $\begingroup$ Denote Conway’s Base-$ 13 $ Function by $ f $. Then $ \tan^{-1} \circ f $ is bounded between $ - \dfrac{\pi}{2} $ and $ \dfrac{\pi}{2} $. Next, $ \tan^{-1} \circ f $ cannot be continuous, otherwise $ f = \tan \circ \tan^{-1} \circ f $ would be continuous, which is a contradiction. $\endgroup$ Mar 11, 2013 at 21:54
  • $\begingroup$ Is Conway's Base-13 Function measurable? $\endgroup$
    – JSchlather
    Mar 11, 2013 at 22:15
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    $\begingroup$ @JSchlather: According to this MathOverflow thread, Conway’s Base-$ 13 $ Function is Borel-measurable. $\endgroup$ Mar 11, 2013 at 23:18
  • $\begingroup$ Thanks the link not only answered my question but led me to an interesting discussion about whether or not the absence of AC in constructing functions implies measurability. $\endgroup$
    – JSchlather
    Mar 12, 2013 at 0:11

Here is a (bounded, Baire class 1) "natural" example of a function with the intermediate value property that is not a derivative. I mentioned it also in this answer.

Consider first the function you mentioned, $$f(x)=\left\{\begin{array}{cl}\sin\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 0&\mbox{ if }x=0.\end{array}\right.$$ This function is a derivative, because, letting $g$ be the function $$ g(x)=\left\{\begin{array}{cl}x^2\cos\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 0&\mbox{ if }x=0,\end{array}\right. $$ and setting $$h(x)=\left\{\begin{array}{cl}2x\cos\left(\frac1x\right)&\mbox{ if }x\ne 0,\\ 0&\mbox{ if }x=0,\end{array}\right.$$ then $h$ is continuous, and $f(x)=g'(x)-h(x)$ for all $x$. But continuous functions are derivatives, so $h$ is also a derivative. Now take $$j(x)=\left\{\begin{array}{cl}\sin\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 1&\mbox{ if }x=0.\end{array}\right.$$ This function still has the intermediate value property, but $j$ is not a derivative. Otherwise, $j-f$ would also be a derivative, but $j-f$ does not have the intermediate value property (it has a jump discontinuity at $0$).

The reference

A.C.M. van Rooij, and W.H. Schikhof, A second course on real functions, Cambridge University Press, 1982,

discusses this example in great detail, showing that instead of sine, one can use any periodic derivative:

If $j:[0,\infty)\to\mathbb R$ is a derivative, and $j(x+1)=j(x)$ for all $x\ge 0$, one can set $J$ be an antiderivative of $j$, let $A=J(1)-J(0)$, and define $h:[0,1]\to\mathbb R$ by $$ h(x)=\left\{\begin{array}{cl} j\left(\frac1x\right)&\mbox{ if } 0<x\le 1,\\ A&\mbox{ if }x=0.\end{array}\right. $$ One can then argue that $h$ is a derivative and, letting $i$ be any function that coincides with $h$ except at $0$, where it takes a value different from $A$ but close, we have an example of an $i$ with the intermediate value property, bounded, and of Baire class 1, that is not a derivative.

To see that $h$ is indeed a derivative, notice first that $A=0$ without loss of generality (replacing $j$ by $j-A$, $h$ by $h-A$, and $J$ by $\hat J(x)=J(x)-Ax$). Now set $$ H(x)=\left\{\begin{array}{cl}-x^2J\left(\frac1x\right)+2\int_{\frac1x}^\infty \frac{J(s)}{s^3}\,ds&\mbox{ if }0<x\le 1,\\ 0&\mbox{ if }x=0.\end{array}\right. $$ One can then verify that $H'=h$ (using that our choice of $A=0$ makes $J$ periodic and therefore bounded, to ensure that $H'(0)=0$).

Van Rooij and W.H. Schikhof then proceed to consider specific examples of functions $j$ that they use to verify the following:

  • There is a derivative $f$ such that $|f|$ is not a derivative.
  • There is a derivative $f$ such that $f^2$ is not a derivative.
  • There is derivative $f$ with, say, $1\le f\le 2$, such that $1/f$ is not a derivative.

Another conterexample is this: let $(a_n, b_n), n = 1, 2, \ldots$ be the sequence of all open intervals in $\mathbb{R}$ with rational endpoints. Let $C_1$ be some Cantor set inside $(a_1, b_1)$. Because $C_1$ is closed and has no interior, $(a_2, b_2) - C_1$ contains some open subinterval. Construct Cantor set $C_2$ inside this subinterval. We can continue this procedure (constructing new Cantor set $C_n$ in $(a_n, b_n)$ that does not intersect previously created $C_1, C_2, \ldots$) -- this is essentialy the same argument as in proof of Baire's theorem.

Now, we have a sequence of Cantor sets $C_1, C_2, ...$ such that 1) $C_i \cap C_j = \emptyset$ for $i \ne j$ and 2) $\bigcup_{i \geq 1} C_i$ is dense in $\mathbb{R}$. For each $n$, pick a bijection $f_n: C_n \to \mathbb{R}$ (there exists one, because $C_n$ has cardinality of continuum). Then construct a function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = f_n(x)$ if $x \in C_n$ and $f(x) = 0$ if $x \not \in C_n$ for any $n$. It's easy to see that $f$ has intermediate value property: if we have some $x < y$ and $f(p) < f(q)$, there's some interval with rational endpoints $(a_n, b_n) \subset (x, y)$, there's $C_n \subset (a_n, b_n)$, so for any $z \in (f(p), f(q))$ there's $w \in C_n$ such that $f(w) = z$: $w$ is just $f_n^{-1}(z)$ (remember that $f_n$ was a bijection).

More interestingly, the set $\bigcup_{n \geq 2} C_n$ is also dense in $\mathbb{R}$, so we can repeat above construction with any function $f_1: C_1 \to \mathbb{R}$, not necessarily a bijection, and we still will obtain a function with the intermediate value property. Now, there are $2^{\mathfrak{c}}$ different functions $C_1 \to \mathbb{R}$, so there are $2^{\mathfrak{c}}$ functions $\mathbb{R} \to \mathbb{R}$ with the intermediate value property. Since there are only $\mathfrak{c}$ continuous functions $\mathbb{R} \to \mathbb{R}$, there are also $\mathfrak{c}$ derivatives, and since $\mathfrak{c} < 2^{\mathfrak{c}}$, there are lots of functions with intermediate value property that are not derivatives.

  • $\begingroup$ +1 I like this one, looks nice to me will read it tomorrow more extrensive it is late now $\endgroup$ Mar 30, 2013 at 21:18

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