Is the derivative of Riemann integral $\int_0^xf$, if it exists, always equal to $f(x)$?

Question

Let $f:[0,1]\to\mathbb{R}$ be a Riemann integrable function. Define a function $F:[0,1]\to\mathbb{R}$ by $$F(x)=\int_0^xf.$$ Suppose that $F$ is differentiable at $c\in(0,1)$. Then is it necessarily true that $F'(c)=f(c)$ ? Note that it is true if $f$ is continuous at $c$.

Answer 1

Just take any continuous function with its value changed at a single point. For example let $f:[0, 1] \to \mathbb R$ be any continuous function and define $\tilde{f}: [0, 1] \to \mathbb R$ by $$ \tilde{f}(x) = \begin{cases} f(x) + 1 & \text{if $x = 1/2$}, \\ f(x) & \text{if $x \neq 1/2$}. \end{cases}$$ Then since $f$ and $\tilde{f}$ differ by only a single point, they have the same indefinite integral $$F(x) = \int_0^x f(x) \, \text{d}x = \int_0^x \tilde{f}(x) \, \text{d}x.$$ However $F'(1/2) = f(1/2) \neq \tilde{f}(1/2)$.