Is every function with the intermediate value property a derivative?

Question

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 1

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. :)

Answer 2

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.