Conditions stronger than differentiability or weaker than integrability

Question

Let $f: [a,b] \to \mathbb R$. If $f$ is (Riemann-)integrable on $[a,b]$, then define $F: [a,b] \to \mathbb R$ by $$F(x) = \int_a^x f.$$

We have the following:

$$\begin{array}{ccccccc} f \text{ differentiable} & \implies & f \text{ continuous} & \implies & f \text{ integrable} & \implies & ?_1 \\ \Big\Downarrow & & \Big\Downarrow & & \Big\Downarrow & & \Big\Downarrow \\ ?_2 & \implies & F \text{ differentiable} & \implies & F \text{ continuous} & \implies & F \text{ integrable} \end{array}$$

Of course, we can define $$\mathcal F(x) = \int_a^x F, \qquad \mathfrak F(x) = \int_a^x \mathcal F, \qquad \text{etc.}$$ for this chain of implications to grow indefinitely.

My questions are

1. ($?_1$) Is there a property of $f$, strictly weaker than integrability, that guarantees that $F$ is integrable?
2. ($?_2$) Similar.

Edit: Ignore $?_1$. That was just plain stupidity on my part. If $f$ isn't integrable, then $F$ isn't even defined. Oops.

Answer 1

If $f$ is differentiable, then $F$ will be twice differentiable. I don't think you can get anything more than that.

Answer 2

To get an $F$ that is not continuous, you would want to integrate not a function, but a measure. That is, if $\mu$ is a (signed) Borel measure, $F(x) = \int_{[a,x]} d\mu(t)$ is a right-continuous function of bounded variation (and in particular is Riemann integrable on bounded intervals).