If at some point on an integral the left and right hand limits do not exist, then what sort of "continuity" does the integral possess.

Question

Let us consider any bounded function $f$. The simpler $f$'s expression is, the better, but the exact properties of $f$ (including whether a closed form exists) are not relevant to this question.

Consider the integral $F(x) = \int_a^x f(t) dt$ where $a$ is some real number. Intuitively, aside from infinite area functions/distributions such as dirac delta and $\frac {1}{x}$ the graph of the integral should not jump. The left and right hand hand limits should be equal and they should equal the integral at that point. I should note that this is just my prediction but I suspect like most good things in calculus, there is a counter-example.

Query:

Is there a bounded $F$ such that at some point the left and right hand limits do not exist at one single point (I would prefer it to not be everywhere discontinuous)?

If so, what sort of "continuity" does the integral satisfy at that point? Are we merely required to have that lim sup and lim inf be equal on each side of the point of discontinuity or is there something deeper going on?

The reason I ask is because in anothet question of mine I proposed an integration method that relies on the integral being continuous and if that is true then my method needs to augmented and expanded so that it takes into consideration any other types of discontinuity that might come up.

Furthermore, I am legitimately intrigued as to what sort of functions have discontinuous integrals. It seems unintuitive yet I know it has to occur as otherwise the Riemann and Lebesgue integrals would handle almost everything. Yet, I know there has to be still unintegratable functions.

Answer 1

It is always continuous.

The fundamental theorem of calculus states (among other things) that for any Riemann integrable $f : [a, b] \to \mathbb R$,

$$F(x) = \int_a^x f(x) dx$$

is continuous on $[a, b]$.

In the Lebesgue theory, the analogous theorem is this: If $f : [a, b] \to \mathbb R$ is Lebesgue integrable then $F$ (as above) is absolutely continuous on [a, b] (absolute continuity is a stronger form of continuity.)

Answer 2

How about the example $f(x)=\sin(1/x)$?

This function is bounded between -1 and 1. Moreover, it seems clear that you can compute the area of $f$ on any interval not containing zero—formally, $f$ is bounded and has only one discontinuity, which means that it is Riemann integrable. However, if you attempt to compute

$$\lim_{x\rightarrow 0}\int_{a}^xf(x)\,dx,$$

it seems (?) as if neither the left nor the right hand limit will exist because the function and hence the area will oscillate arbitrarily frequently near 0.