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Given a function $f$ that is differentiable at a point $x_0$, if we define (using the Riemann integral)

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

Can we necessarily say that $F^{\prime}(x)$ is continuous at $x_0$? Going back and forth between $f$ and $F$ confuses me a bit. I think that the Fundamental Theorem of Calculus gives us some relation between $F^{\prime}(x_0)$ and $f(x_0)$, but I'm not sure.

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    $\begingroup$ Basically, if you have some function $f$, the function $F(x) = \int_a^x f$ is "nicer" than $f$. If $f$ is integrable, then $F$ is continuous. If $f$ is continuous, then $F$ is differentiable. If $f$ is differentiable, $F$ is twice differentiable. $\endgroup$ – MathematicsStudent1122 Dec 6 '16 at 18:14
  • $\begingroup$ @MathematicsStudent1122: That is true if $f$ is continuous near $x_0$. But it was only assumed that $f$ is differentiable and thus continuous at $x_0$. $F$ is almost everywhere differentiable and $F' = f$ almost everywhere. But $F'$ may not exists on a neighborhood of $x_0$, so speaking of continuity is difficult. $\endgroup$ – user251257 Dec 6 '16 at 19:50
  • $\begingroup$ @user251257 The first fundamental theorem of calculus states that if $f$ is continuous at $x_0$ (in our case, $f$ is differentiable, so this is certainly true), then $F(x) = \int_a^x f$ is differentiable at $x_0$ and $F'(x_0) = f(x_0)$. This implies $F'$ is continuous at $x_0$. Continuity in a neighbourhood of $x_0$ is not assumed. $\endgroup$ – MathematicsStudent1122 Dec 6 '16 at 20:10
  • $\begingroup$ @MathematicsStudent1122: Just because $F'$ exists in $x_0$ doesn't make it continuous at $x_0$... $\endgroup$ – user251257 Dec 6 '16 at 20:12
  • $\begingroup$ @user251257 Never mind, you're right then. Do you have a counterexample? $\endgroup$ – MathematicsStudent1122 Dec 6 '16 at 20:14
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You can't say $F'$ is continuous at $x_0$ because $F'$ may not exist in a full neighborhood of $x_0.$ Take the interval $[-1,1]$ with $x_0=0$ for example. Choose any sequences $a_n,b_n$ you like such that $1\ge b_1 > a_1 > b_2 > a_2 > \cdots \to 0^+.$ Define $f(x) = x^2$ on each $[a_n,b_n],$ $f=0$ everywhere else. Then $f$ is Riemann integrable on $[-1,1]$ and $f'(0)=0.$ But at each $a_n$ and $b_n,$ $f$ has a jump discontinuity, hence $F'$ does not exist at these points. No matter which neighborhood of $0$ you examine, there will be lots of points, namely in the tail ends of the sequences $a_n,b_n,$ where $F'$ doesn't exist.

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    $\begingroup$ When discussing the continuity of a function, it's taken for granted that we're discussing its continuity on its domain. Is $F'$ continuous on its domain? You haven't answered that. $\endgroup$ – MathematicsStudent1122 Dec 7 '16 at 3:40
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    $\begingroup$ Nice counter-example. +1 $\endgroup$ – Paramanand Singh Dec 7 '16 at 8:47
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This rather an extensive comment than an answer.

Since $f$ is Riemann integrable, its set of continuity $$ C = \left\{ x\in[a,b] \mid \lim_{t\to x} f(t) = f(x) \right\} $$ has full measure, that is $|C| = b-a$. Thus, $F'$ exists on $C$ and agrees with $f$ on $C$. Trivially, $F'$ restricted onto $C$ is continuous.

That says nothing about the domain of $F'$, which might be larger than $C$, and whether $F'$ is continuous at $x_0$.

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As stated in the problem, $F'(x)$ would be not only continuous but differentiable at $x_0$.

The FTC allows you to conclude that for $x_0\in [a,x]$, $$ F'(x_0)=f(x_0) $$ Which means that $F'(x)$ in its region of validity has the properties $f$ has.

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  • $\begingroup$ Please comment along with the downvote, so I can improve my answer $\endgroup$ – operatorerror Dec 6 '16 at 19:49
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    $\begingroup$ The problem is $F'$ need not exist in a full neighborhood of $x_0.$ $\endgroup$ – zhw. Dec 6 '16 at 22:48
  • $\begingroup$ @zhw. That's not a problem with the definition of differentiability I learned. $x_0$ is an accumulation point of the domain of $F'$, and itself in that domain. That suffices to define differentiability of $F'$ at $x_0$, and indeed $F'$ is differentiable at $x_0$ under the given hypotheses, using that definition. I'm aware that some authors gratuitously define differentiability only at interior points of the domain of the function. Just saying that definitions differ. $\endgroup$ – Daniel Fischer Dec 13 '16 at 21:52
  • $\begingroup$ @qbert It's a little more complicated, we only need to have $F'(x) = f(x)$ almost everywhere. $F'$ may exist at some point $x_1$ with $F'(x_1) \neq f(x_1)$. $\endgroup$ – Daniel Fischer Dec 13 '16 at 21:55
  • $\begingroup$ @DanielFischer I also learned that differentiability was a point property, unlike holomorphicity in complex analysis, if this is what is meant by your response to zhw. And fair enough, I realized this after. Figured I would leave my answer up anyway (and I'm glad I did as I have learned from it). $\endgroup$ – operatorerror Dec 13 '16 at 22:01

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