If $f \colon[ a, b] \rightarrow \mathbb{R}$ is bounded and Riemann-integrable, is $F(x):=\int_a^xf(t)dt$ always differentiable?

Question

Is the following statement true or false?

Let $ f \colon [ a, b] \rightarrow \mathbb{R}$ be a bounded and Riemann-integrable function. Define $$ [a,b]\ni x\mapsto F(x) = \int_{a}^{x} f(t) dt.$$ Then the function $ F$ is differentiable on $(a,b)$.

My attempt: I thinks this statement is false. As a counterexample I am thinking of

$$ f(x) = \begin{cases} |x|, & x \in [-1,1], \\ 1,& \text{otherwise.} \end{cases}$$

Is this correct?

Answer 1

Your example doesn't work because $f$ is continuous. Take $f(x)=-1$ for $-1 \leq x \leq 0$ and $f(x)=1$ for $0 < x \leq 1$. If you compute $F$ you will see that the right hand derivative of $F$ at $0$ is $+1$ and the left hand derivative is $-1$.

Answer 2

Your counterexample won't work because the derivative of $F$ is just $f$ on $(a,b)$.

Take $f\colon[0,2]\to \Bbb R$ as the step function with $f(x)=0$ for $x<1$ and $f(x)=1$ for $1\le x\le 2$. Then $f$ is Riemann integrable, and $F$ is a piecewise linear function which is $0$ from $0$ to $1$, and then $F(x)=x-1$ for $1\le x\le 2$. In particular $F$ is not differentiable at $1$.