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I have trouble with dealing with the fundamental theorem of calculus. Here's the statement of FTC from Apostol's Calculus, page 202 in the Second Edition.

FIRST FUNDAMENTAL THEOREM OF CALCULUS

Let $f$ be a function that is integrable on $[a, x]$ for each $x$ in $[a, b]$. Let $c$ be such that $a\leq c\leq b$ and define a new function $A$ as follows:

$A(x) = \int_c^xf(t)dt$ if $a\leq x\leq b$.

Then the derivative $A'(x)$ exists at each point $x$ in the open interval $(a, b)$ where $f$ is continuous and for such $x$ we have $A'(x) = f(x)$.

The theorem says $A' = f$ where $f$ is continuous. But I heard the continuity of $f$ is not a necessary condition for the FTC. How can I avoid misunderstandings to interpret this theorem? For example, I wonder the function $A$ is differentiable on $[a, b]$ or only just the values are equal. And by the Exercise 28, on Section 5.5, the continuity of $f$ at one point does not guarantee that the continuity of $A'$ at that point. Does it imply that $A' = f$ only at that point but doesn't ensure $A'$ is continuous?

J.Bo
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    That might be what you heard, but what I said is that the continuity of $f$ on all of $[a,b]$ or even on an interval around $c$ is not necessary. Only the continuity of $f$ at $c$ is what is used to prove that $A(x)$ is differentiable at $x=c$ and $A'(c)=f(c)$. – MoonLightSyzygy Jan 21 '20 at 05:57
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    Yes. The proof of the book only uses the continuity at one point to prove the theorem. Then it leads to the conclusion that the continuity of $f$ at c only ensures that $A'(c) = f(c)$, not other points. And then the differentiability of $A$ is guaranteed only at that point. I think this is right. – J.Bo Jan 21 '20 at 06:04
  • Yes, that is what it says. – MoonLightSyzygy Jan 21 '20 at 06:08
  • OK. Therefore the continuity of $A'$ at c is not guaranteed by the continuity of $f$ at c. Thanks for your correction. – J.Bo Jan 21 '20 at 06:11
  • You might replace first sentence of your theorem with "$f$ is integrable on $[a, b] $" since this implies that $f$ is integrable on any subinterval of $[a, b] $ and in particular on $[a, x] $ for all $x\in[a, b] $. – Paramanand Singh Jan 21 '20 at 07:58
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    Also it is not mentioned, but your $A(x) $ is continuous at each point $x\in[a, b] $. – Paramanand Singh Jan 21 '20 at 07:59
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    See this answer : https://math.stackexchange.com/a/1900844/72031 – Paramanand Singh Jan 21 '20 at 10:18

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