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I am lost at the Exercise 28 (page 210) in Apostol calculus. This chapter is about the fundamental theorems of calculus.

The first part of this theorem is defined in the book as: If we have an integrable $f(x)$ on $[a, b]$, and $c, x \in [a, b]$, then the function $A(x) = \int_c^x f(t) dt$ has a derivative at each point $x$ where $f(x)$ is continuous, and $A'(x) = f(x)$.

Now the exercise:

Suppose we have an integral $A(x) = \int_a^x f(t) dt$, which exists for each $x \in [a, b]$. Let $c \in (a, b)$. Now, there are a few statements that should be true (according to Apostol answers in the back):

1) $f$ discontinuous at $c$. This implies that $A(x)$ is still continuous at $c$. I believe that this is true, however, how can this be shown?

Since $f$ is discontinuous at $c$, it should say nothing about the continuity of $A(c)$. What if $A'(c) \neq f(c)$, since $f(x)$ is discontinuous at $c$, and it is a necessary condition for the 1st fundamental theorem to hold. If the 1st theorem does not hold here, then the derivative might not exist -> nothing implies continuity of $A(x)$ at $c$.

2) $f'(c)$ exists. It DOES NOT imply that $A'(x)$ is continuous at $c$.

Why is that? The derivative $f'(c)$ exists. Hence, $f(x)$ is continuous at $c$. $f(x)$ is continuous at $c$ means that $A'(c) = f(c)$. But $A'(c)$ is essentially the same as $f(c)$, how can one be continuous at $c$, while the other is not?

3) $f'$ is continuous at $c$. It somehow IMPLIES that $A'$ is continuous at $c$. It is literally the same as point (2). Why suddenly $A'$ becomes continuous at $c$?

Is there a typo in the answers? Though I found another set of answers - the same there, so highly unlikely.

John
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    Regarding (1), $A$ is continuous at $c$ even if $f$ is not. Sketch: if $x \geq c$ then $|A(x) - A(c)| = |\int_{c}^{x}f(t)dt| \leq \int_c^x |f(t)|dx \leq \int_c^x Bdx = B(x-c)$. The RHS converges to zero as $x \to c$, hence so does the LHS. Here we have used that any Riemann integrable function is by definition bounded, and $B$ is a bound for $f$. –  Jul 15 '19 at 01:56
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    for (2), it simply says that $A'(c) = f(c)$ by FTC.. Nothing about continuity of $A'(x)$ only that their values are the same because they're essentially different functions since it can differ at neighborhoods around $c$. But (3) tells us that the neighborhood around $c$ match f(x) = A'(x) which means that A' is continuous there. – K. Y Jul 15 '19 at 02:01
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    Regarding (2) and (3), the condition in (3), namely $f'$ is continuous at $c$, is stronger than the condition in (2), namely $f'$ exists at $c$. See here for a standard example of a function which is differentiable everywhere but the derivative is discontinuous at a point. –  Jul 15 '19 at 02:02

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