$f$ integrable on $[a,b]$, differentiable at $c\in (a,b)$. $F(x)=\int_a^x f$. Out of the three ways $F'$ can be discont. at $c$, which are possible? Assume a function $f$ is integrable on $[a,b]$, differentiable at a point $c\in (a,b),$ and define
$$F(x)=\int_a^x f$$
Let's investigate the theoretical conditions under which $F'$ could be discontinuous at $c$.
Since $f$ is differentiable at $c$, this implies $f$ is continuous at $c$.
By the first fundamental theorem of calculus (FTC1) applied to $f$ on $[a,b]$, we have that $F$ is differentiable at $c \in (a,b)$ and $F'(c)=f(c)$.
Let's recall the definition of continuity

Spivak, Ch. 6, Definition: A function $f$ is continuous at $c$ if
$$\lim\limits_{x\to c} f(x)=f(c)\tag{1}$$

What are the ways a function can be discontinuous at a point $c$?

*

*$f$ isn't defined at $c$, hence $f(c)$ isn't defined and so $(1)$ doesn't make sense and is false


*$\lim\limits_{x\to c} f(x)$ does not exist, so again $(1)$ doesn't make sense and is false


*$f$ is defined at $c$ and the limit above does exist but it differs from $f(c)$.
Which of the above scenarios is possible in our setup of $f$ and $F$, for $F'$ to be discontinuous at $c$?
Here is what I came up with
In the case of our function $F$, it is differentiable wherever $f$ is continuous. If $f$ were to be continuous in an interval around $c$, then $F$ would be differentiable and hence continuous on that interval, including at $c$.
Therefore, $f$ has to be discontinuous on every interval containing $c$.
We've established that $F'(c)=f(c)$ so $F'$ is defined at $c$ and hence option 1. above is not possible.
If 3. were true, then we'd have
$$\lim\limits_{x\to c} F'(x)\neq F'(c)=f(c)\tag{2}$$
Here is a line of reasoning I am unsure about

*

*Since the limit above exists, then $F'$ is defined everywhere around $c$. Is this true? Ie, when we defined a limit

$$\lim\limits_{x\to c} f(x)$$
as
$$\forall \epsilon>0\ \exists \delta>0\ \forall x, |x-c|<\delta\implies |f(x)-f(c)|<\epsilon$$
Did the $\forall x$ mean literally for every single real number within $\delta$ of $c$, or just the $x$ where $f$ is defined?

*

*if the above is indeed true, then we have the following happening

(i) $F'$ defined in an interval containing $c$
(ii) since $f$ is integrable on $[a,b]$, $F$ continuous on $[a,b]$, and in particular at $c$
(iii) $\lim\limits_{x\to c} F'$ exists
I believe the following theorem is thus applicable

Spivak, Ch. 11, Theorem 7: Suppose $F$ is continuous at $c$, $F'(x)$
exists for all $x$ in some interval containing $c$, except possibly at
$x=c$. Suppose, moreover, that $\lim\limits_{x\to c} F'(x)$ exists.
Then, $F'(c)$ exists and $F'(c)=\lim\limits_{x\to c} F'(x)$

This theorem basically says that if a function is continuous at a point, then its derivative cannot have a jump discontinuity there.
Applying to $F$, we conclude that $F'(c)=\lim\limits_{x\to c} F'(x)$, which contradicts $(2)$, and hence scenario 3. isn't possible.
Which leaves only scenario 2., which is possible by virtue of the fact that I can come up with an example for it:
$$f(x)=\begin{cases} 0, \text{ if } x \in \left (\frac{1}{n+1},\frac{1}{n} \right ] \\ x^2, \text{ if } x \in \left (\frac{1}{n},\frac{1}{n-1} \right ] \end{cases}$$
for $n\in\mathbb{N}$.
Edit For the record I thought I'd mention this is a problem from Spivak's Calculus and there is a (terse solution) that isn't very didactical. Here it is

If we assume that $f$ is continuous in an interval around $c$, then
$F'$ will be continuous at $c$, since we will have $F'(x)=f(x)$ in
this interval, and differentiability of $f$ at $c$ implies continuity
of $f$ at $c$. But without this assumption $F'$ may not even be exist
at all points near $c$. For example, $f$ could be the function shown
below


 A: Edit:
Now that you have added Spivak’s solution, I think I see what’s going on. Let me try and explain a little.
First, I think there’s some confusion about the definition of “continuity”. Suppose that $f:(a,b) \rightarrow \mathbb{R}$ is a function. Let me give you two different definitions of continuity:
Definition 1: $f$ is continuous at $c \in (a,b)$ iff for all $\epsilon >0$, there is a $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$.
Definition 2: $f$ is continuous at $c \in (a,b)$ iff for every sequence of points $\{x_n\}$ with each $x_n \in (a,b)$, if $x_n \rightarrow c$, then $f(x_n) \rightarrow c$ as well.
These two definitions can easily be shown to be equivalent.
What happens if $f:A \rightarrow \mathbb{R}$ where $A \subseteq \mathbb{R}$ is some subset of reals, but not necessarily an interval? Well these definitions can be adapted so that continuity still makes sense. Definition 2 is the definition I used below, so we’ll adapt that one.
Definition 2’: Let $f:A \rightarrow \mathbb{R}$, and let $c \in A$. $f$ is continuous at $c$ If for every sequence of points $\{x_n\}$ with each $x_n \in D$, if $x_n \rightarrow c$, then $f(x_n) \rightarrow c$ as well.
With this in mind, we see a source of confusion. $F’$ in your original question need not be defined on an interval. Indeed, as your example shows, we can have an $F’$ that is undefined on a convergent sequences of points. But when we are asking about the continuity of $F’$, we can only answer that question within the domain on which it is defined.
This is (I think) why you don’t see why your example fails. Try working it out with definition 2’ above, and see if it works.
As for Spivak’s “proof”, he is using a definition of continuity that doesn’t allow for this kind of generality. I think it holds in the full generality though, as my answer indicates.
What follows is my original answer, before answer the questions addressed in the comments.

NOTE: I am looking at $F’$ as a function $F’:A \rightarrow \mathbb{R}$, where $A$ is the set of points on which $F$ is differentiable.
You’re proof above is not quite incorrect, sadly. The big problem is that your example at the very end is not correct. $F’$ is continuous at $0$ in that example.
Let $f:[a,b] \rightarrow \mathbb{R}$ be integrable. Let $F(x) = \int_a^x f(t) \: dt$. Let $c \in (a,b)$, and suppose that $f$ is differentiable at $c$. We are curious about the ways in which $F’$ could be discontinuous at $c$. Let’s break the problem into a series of claims:
Claim 1:
$F$ is differentiable at $c$, and $F’(c) = f(c)$
Proof: This follows from the fact that $f’(c)$ exists implies that $f$ is continuous at $c$, and the FTC. You observed this above.
From claim 1, it follows that $F’(c)$ will always exist, and we need not worry about a “missing point” discontinuity.
Claim 2:
Suppose that $f$ is continuous in a neighborhood of $c$. Then $F’$ is continuous at $c$.
Proof: This follows immediately from the FTC.
Hence as you noted, the only way that that $F’$ could be discontinuous at $c$ is if $f$ is not continuous in any neighborhood of $c$.
Here’s a question: Is it possible that there is a neighborhood of $c$ in which $F$ is only differentiable at $c$?
Claim 3:
There is a sequence of points $a_n \rightarrow c$ such that $f$ is continuous at each $a_n$.
Proof: Suppose there is no such sequence. Then there is an interval $(c- \epsilon, c+ \epsilon) \subseteq [a,b]$ on which $f$ is discontinuous. Let $D(f)$ denote the set of discontinuities of $f$. It follows that $m(D(f)) > 0$, and hence $f$ is not Riemann integrable. See here for more details: https://en.wikipedia.org/wiki/Riemann_integral#Integrability
From claims 1 and 3, it follows that there is necessarily a sequence of points $a_n \rightarrow c$ on which $F’$ exists. Hence it makes sense to talk about the continuity of $F’$ at $c$ by examining limits. If there was no such sequence, then $F’$ would be trivially continuous at $c$.
Claim 4:
$F’$ is continuous at $c$.
Proof: Suppose that $a_n \rightarrow c$ is a sequence of points such that $F’(a_n)$ exists. We may compute:
$$\begin{align*} \lim_{n \rightarrow \infty}F’(a_n) &= \lim_{n \rightarrow \infty} \lim_{h \rightarrow 0} \frac{F(a_n + h) - F(a_n)}{h} \\
& = \lim_{n \rightarrow \infty} \lim_{h \rightarrow 0} \frac{ \int_{a_n}^{a_n + h} f(t) \: dt}{h} \\
& = \lim_{h \rightarrow 0} \lim_{n \rightarrow \infty} \frac{ \int_{a_n}^{a_n + h} f(t) \: dt}{h} \quad (*) \\
& = \lim_{h \rightarrow 0} \frac{\int_c^{c+h} f(t) \: dt}{h} \\
& = F’(c)   \end{align*}$$
The line with the $(*)$ needs to be justified. Why can we switch those limits? This follows from the Moore-Osgood Theorem (https://en.wikipedia.org/wiki/Iterated_limit). I worked this out on paper, but if there’s an error in my argument it is here.
A: Let me give my own (I believe rigorous) answer, taking into account all the previous answers and comments.
Case 1 scenario: The fundamental theorem of Calculus in its most common form assumes continuity of $f$ on an interval around a point. This is not our case. Spivak gives a proof based on suprema and infima of the function $f$ on $[c,c+h]$. But this is NOT rigorous (as in many other cases in Spivak) because suprema and infima do NOT necessarily coincide in the limit, as in the continuous case. The assumption of boundedness is not sufficient. A rigorous proof should go as follows:
Let $\epsilon >0$. Then (by continuity of $f$ at $c$), there is a
$\delta>0$ such that, $|x-c|<\delta$ implies :
$f(c)-\epsilon<f(x)<f(c)+\epsilon$. That implies that,
$(f(c)-\epsilon)h<\int_{c}^{c+h}f(t)dt<(f(c)+\epsilon )h$ and dividing by $h$ we get:
$\left|\int_{c}^{c+h}f(t)dt/h\,\,-f(c) \right|\,<\,\epsilon$ for
$0<h<\delta$ and likewise from the left side, so we get that $F'(c)$
exists and is equal to $f(c)$. So scenario $1$ can be rejected.
Now we go to the case 3 scenario. Since existence of the derivative is assumed in a neighborhood of $c$, (except possibly at $c$) we can apply the mean value theorem on $[c,c+h]$, ($F$ is continuous on the interval and differentiable in $(c,c+h)$), therefore
$(F(c+h)-F(c))/h=F'(\xi_{h} )$ and since $h$ tends to zero, $\xi_{h}$ also tends to zero and it is equal to $\displaystyle \lim_{ x\to c}F'(x)$ which we have assumed it exists, hence $F'(c)$ exists too and is equal to $f(c)$. So scenario 3 can also be rejected.
In the case 2 scenario, under the given assumptions we have existence of  $ F'$ ALMOST EVERYWHERE. In other words there is no guarantee that $F'$ exists near $c$ and therefore this is the only acceptable scenario! A counterexample I found is the function $f(x)=1/n^3$ on $(1/n+1,\,1/n]$, $f(0)=0$. I am almost certain that $f$ is differentiable at $0$, $f$ is integrable, $F(x)=\int_{0}^{x}f(t)dt$ is continuous and differentiable at $0$. Yet,  $F'(x)$ does not exist at points $1/n$ because there is a jump of the integral! Thus, there are points at which  $F'$ does not exist arbitrarily close to $0$. So we can only accept scenario $2$.
