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.

**Answer**

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. 🙂

**Attribution***Source : Link , Question Author : Dominic Michaelis , Answer Author : Community*