The first form of the Fundamental Theorem of Calculus (for the Riemann integral) is stated as follows (in my textbook):
FTC: Suppose there is a finite set $E \subset [a, b]$ and functions $f, F : [a, b] \to \mathbb{R}$ such that:
1. $F$ is continuous on $[a, b]$
2. $F'(x) = f(x)$ for all $x \in [a, b] \setminus E$
3. $f$ is Riemann Integrable.
Then
$$\int_{a}^{b} f(t) \ dt = F(b)-F(a)$$
Note that nothing is said about uniqueness of such $F$. This is the motivation of my question. Let me create a definition for this question's purposes:
Definition. Let $f : [a, b] \to \mathbb{R}$ be Riemann integrable. A function $F : [a, b] \to \mathbb{R}$ will be called a pseudo-primitive of $f$ if:
- $F$ is continuous on $[a, b]$
- There exists a finite set $E \subset [a, b]$ such that $F'(x) = f(x)$ for all $x \in [a, b] \setminus E$
Now my question is: Is there a function $f : [a, b] \to \mathbb{R}$, Riemann integrable, that has two different pseudo-primitives (disregarding constant shifts such as $F_2 = F_1 + C$)? (If not, how to prove it?)
As requested by the OP I repost my comment as an answer.
Let $F$ and $G$ be two pseudo-primitives according to the definition given in the question, clearly the fundamental theorem of calculus applies to them. We claim that the quantity $F(x) - G(x)$ is constant for any $x \in [a,b]$. To show it let $x <y$ be elements of $[a,b]$, then, by the fundamental theorem of calculus $$F(x) - F(y) = \int_y^x f(t) \, dt = G(x) - G(y).$$
Thus $F(x) - G(x) = F(y) - G(y)$ as claimed.