Surjectivity of the integral operator Let $[a,b]\subseteq\mathbb{R}$ be a closed interval and $\mathcal{R}[a,b]$ denote the linear space of Riemann integrable functions. It is well known that the integral operator 
$$R(f)(x):=(Rf)(a)+\int_a^xf(t)\,dt$$
maps elements of $\mathcal{R}[a,b]$ to the space of continuous functions on $[a,b]$ i.e. $R:\mathcal{R}[a,b]\to C[a,b]$. 
My questions are: Is this map surjective? If not, is there a counterexample?
 A: It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=\sqrt {x-a}.$ Hence the integral operator is not surjective.
A: Assume that $\mathcal{C}\big([a,b]\big)$ is taken modulo constant functions.  Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property.  A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $\text{im}(R)$. 
However, you can shift and rescale the Cantor function to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the Cantor set, which has measure $0$.  Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $\text{im}(R)$.  Indeed, $\text{im}(R)$ consists of only (but not all) absolutely continuous functions on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).
Being absolutely continuous is also insufficient.  As zhw. illustrated, the function $g(x):=\sqrt{x-a}$ is absolutely continuous, but any of its weak derivative is unbounded.  Therefore, $g'$ is not Riemann-integrable.   Every absolutely continuous function has a weak derivative.  We know that every function in $\text{im}(R)$ is absolutely continuous.  Thus, each function in $\text{im}(R)$ has a bounded weak derivative, whence Lipschitz continuous.
To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$.  Hence, $F\in\text{im}(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous).  This translates as follows: $\text{im}(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.
A: The map isn't surjective. A counterexample is the Weierstrass function which is continuous.
