Let $[a,b]$ be a compact interval and $f: [a,b] \to \mathbb R$ a Riemann integrable function with finitely many discontinuities.

Then $F(x) := \int_a^x f(t) \, dt$ is continuous on $[a,b]$ and on all except the finitely many points differentiable with $F'(x)=f(x)$.

In the basic FTC we assume continuity of $f$ everywhere. I know that for a Riemann integrable function we can change finitely many points without affecting the integral. So my idea is to change those finitely many discontinuities so that $f$ becomes continuous. But then, not every discontinuity of $f$ needs to be a removable discontinuity, right?

Can you give me a hint?

  • $\begingroup$ Yes, discontinuities generally aren't removable. What does the basic FTC tell you about $F$ between two consecutive discontinuities of $f$? $\endgroup$ Jan 23, 2018 at 15:57
  • $\begingroup$ At the points where $f$ is continuous, $F$ is differentiable and $F'(x)=f(x)$. I remember Darboux's theorem, which says $f$ is now either continuous or has an essential discontinuity (oscillation, divergence). But that doesn't really seem to help... $\endgroup$
    – philmcole
    Jan 23, 2018 at 16:08
  • 1
    $\begingroup$ How does that not really help? If you know that $F$ is continuous, that is exactly what is missing, that, except at the finitely many points where $f$ is discontinuous, $F$ is differentiable with $F' = f$. $\endgroup$ Jan 23, 2018 at 16:17
  • $\begingroup$ By the basic FTC $F$ is differentiable and continuous wherever $f$ is continuous with $F'=f$. Since $f$ is a derivative, by Darboux's theorem it has the intermediate value property (i.e. only oscillation or divergence discontinuity). But since $f$ is also Riemann integrable, it needs to be bounded (hence no divergence discontinuities). The oscillation discontinuities of $f$ can be fixed (?) and therefore we can define a $f*$ which is continuous on all $[a,b]$ and has the same antiderivative. Then the basic FCT applied to $f*$ gives the result. Is that it? I feel like this proof is confusing. $\endgroup$
    – philmcole
    Jan 23, 2018 at 16:30
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    $\begingroup$ Forget about Darboux' theorem, it plays no role here. $\endgroup$ Jan 23, 2018 at 17:04

1 Answer 1


So because of the great help from Daniel I was able to solve the excercise. Because there are no other answers I will share my solution.

The claim that $F$ is continuous is actually not dependent on the FTC at all but rather a separate theorem. Also the potential discontinuities of $f$ play no role, because the theorem relies just on the fact that $f$ is Riemann integrable and therefore bounded. For a proof of this see here.

The second claim is then just the FTC applied to the points where $f$ is continuous. There are actually different versions of the FTC which differ in their preconditions. If the precondition is that $f$ is continuous on an interval then it is actually important that we have only finitely many points of discontinuity because otherwise we might not be able to split the interval into subintervals where $f$ is continuous (e.g. if $f$ would be discontinuous at every irrational number, we couldn't make an interval between two rational numbers where $f$ is continuous, because there would be another irrational number in it). If the preconditions is just so, that $f$ is continuous at some point $x_0$ then there can even be infinitely many discontinuities and it would still work.

Most definitions of the FTC I've seen actually require an interval where $f$ is continuous. I will include a proof of the stronger version which just requires pointwise continuity below.

Let $[a,b]$ be a compact interval and $f: [a,b] \to \mathbb R$ a Riemann-integrable function. Let $F: [a,b] \to \mathbb R$ with $F(x) = \int_a^x f(t) \, dt$ and $x_0 \in [a,b]$. If $f$ is continuous at $x_0$ then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.

We now prove that $F'(x_0) = \lim_{x \to x_0} \frac{F(x)-F(x_0)}{x-x_0} = f(x_0)$ which proves that $F$ is differentiable at $x_0$ with derivative $f(x_0)$. Recall from the definition of a limit, that

$\lim_{x \to x_0} g(x)=L$ iff for every $\varepsilon \gt 0$ there is $\delta \gt 0$ such that for all $x \in [a,b] \setminus \{x_0\}$ with $|x-x_0|\lt\delta$ we have $|g(x)-L|\lt\varepsilon$.

Let $\varepsilon \gt 0$. By continuity of $f$ at $x_0$ there exists $\delta \gt 0$, such that for all $x \in [a,b]$ with $|x-x_0|\lt\delta$ we have $|f(x)-f(x_0)|\lt\varepsilon$. Then it follow also that for all $x \in (x_0,x_0+\delta)$

$$\begin{aligned} \biggl\lvert \frac{F(x)-F(x_0)}{x-x_0}-f(x_0) \biggr\rvert &= \biggl\lvert \frac{1}{x-x_0} \Big( \int_a^x f(t) \,dt - \int_a^{x_0} f(t) \,dt \Big) - f(x_0) \biggr\rvert \\&= \biggl\lvert \frac{1}{x-x_0} \Big( \int_{x_0}^x f(t) \,dt \Big) - f(x_0) \biggr\rvert \\&= \biggl\lvert \frac{1}{x-x_0} \Big( \int_{x_0}^x f(t) \,dt \Big) - \frac{1}{x-x_0} \int_{x_0}^x f(x_0) \,dt \biggr\rvert \\&= \biggl\lvert \frac{1}{x-x_0} \int_{x_0}^x f(t)-f(x_0) \,dt \biggr\rvert \\&\le \frac{1}{x-x_0} \int_{x_0}^x \biggr\rvert f(t)-f(x_0) \biggr\rvert \,dt \\&\lt \frac{1}{x-x_0} \int_{x_0}^x \varepsilon \,dt \\&= \frac{1}{x-x_0} (x-x_0) \varepsilon = \varepsilon \end{aligned}$$

where we used the additivity and the triangle inequality property of Riemann integrals. Similarly for for all $x \in (x_0-\delta,x_0)$

$$\begin{aligned} \biggl\lvert \frac{F(x)-F(x_0)}{x_0-x}-f(x_0) \biggr\rvert &= \biggl\lvert \frac{F(x_0)-F(x)}{x_0-x}-f(x_0) \biggr\rvert \\&= \biggl\lvert \frac{1}{x_0-x} \Big( \int_a^{x_0} f(t) \,dt - \int_a^x f(t) \,dt \Big) - f(x_0) \biggr\rvert \\&= \biggl\lvert \frac{1}{x_0-x} \Big( \int_x^{x_0} f(t) \,dt \Big) - f(x_0) \biggr\rvert \\&= \biggl\lvert \frac{1}{x_0-x} \Big( \int_x^{x_0} f(t) \,dt \Big) - \frac{1}{x_0-x} \int_x^{x_0} f(x_0) \,dt \biggr\rvert \\&= \biggl\lvert \frac{1}{x_0-x} \int_x^{x_0} f(t)-f(x_0) \,dt \biggr\rvert \\&\le \frac{1}{x_0-x} \int_x^{x_0} \biggr\rvert f(t)-f(x_0) \biggr\rvert \,dt \\&\lt \frac{1}{x_0-x} \int_x^{x_0} \varepsilon \,dt \\&= \frac{1}{x_0-x} (x_0-x) \varepsilon = \varepsilon \end{aligned}$$

Thereby we proved our statement.

  • 1
    $\begingroup$ Good job. Though the absolute values come out better with \biggl\lvert … \biggr\rvert than with plain |. $\endgroup$ Jan 24, 2018 at 22:53
  • $\begingroup$ Thanks Daniel for your patience! $\endgroup$
    – philmcole
    Jan 25, 2018 at 10:35

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