# Is the derivative of an integral always continuous?

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.

• 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. Commented Dec 6, 2016 at 18:14
• @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. Commented Dec 6, 2016 at 19:50
• @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. Commented Dec 6, 2016 at 20:10
• @MathematicsStudent1122: Just because $F'$ exists in $x_0$ doesn't make it continuous at $x_0$... Commented Dec 6, 2016 at 20:12
• @user251257 Never mind, you're right then. Do you have a counterexample? Commented Dec 6, 2016 at 20:14

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.

• 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. Commented Dec 7, 2016 at 3:40
• Nice counter-example. +1 Commented Dec 7, 2016 at 8:47
• I don't understand where the discontinuity occurs, could you help me? Commented Dec 14, 2021 at 22:14

The statement is true in the sense, that $$F'$$ is continuous at $$x_0$$ on its domain, which need not to be the entire interval $$(a, b)$$. We may drop that $$f$$ is differentiable at $$x_0$$, we may also replace Riemann integrability with Lebesgue.

Claim:

Let $$D = \{ x\in (a, b) \mid F \text{ is differentiable at } x \}$$. For $$x_n\in D$$ with $$x_n\to x_0$$ it follows $$F'(x_n) \to F'(x_0) = f(x_0)$$.

Proof:

Notice that as $$f$$ is continuous at $$x_0$$ it follows $$F'(x_0) = \lim_{y\to x_0} \frac{1}{y - x_0} \int_{x_0}^y f(t) dt = f(x_0).$$

Fix $$\varepsilon > 0$$.

1. As $$f$$ is continuous at $$x_0$$, a $$\delta > 0$$ exists such that for every $$x\in [a, b] \cap (x_0 - \delta, x_0 + \delta)$$ it follows $$|f(x) - f(x_0)| < \frac\varepsilon2.$$
2. As $$x_n\to x_0$$, a $$N> 0$$ exists such that for every $$n \ge N$$ it follows $$|x_n - x_0| < \frac{\delta}{2}.$$
3. As $$F$$ is differentiable at $$x_n$$, a $$\eta_n > 0$$ exists such that for every $$y\in [a, b] \cap (x_n - \eta_n, x_n + \eta_n)$$ with $$y\ne x_n$$ it follows $$\left|\frac{F(y) - F(x_n)}{y - x_n} - F'(x_n) \right| < \frac\varepsilon2.$$

Putting these together, for every $$y$$ with $$|y - x_n| < \min(\eta_n, \frac{\delta_n}{2})$$ it follows \begin{align*} \left| F'(x_n) - f(x_0) \right| &\le \left| F'(x_n) - \frac{F(y) - F(x_n)}{y - x_n} \right| + \left| \frac{F(y) - F(x_n)}{y - x_n} - f(x_0) \right| \\ & \le \frac\varepsilon 2 + \left| \frac{1}{y - x_n}\int_{x_n}^y f(t) - f(x_0) dt \right| \\ & \le\frac\varepsilon 2 + \frac{1}{y - x_n}\int_{x_n}^y \underbrace{|f(t) - f(x_0)|}_{\le \varepsilon / 2} dt \\ &\le \epsilon, \end{align*} as $$|t - x_0| \le |t - x_n| + |x_n - x_0| \le |y - x_n| + |x_n - x_0| < \delta$$.

That is, $$F'(x_n)$$ converges to $$f(x_0) = F'(x_0)$$.

Notes:

If $$f$$ is differentiable at $$x_0$$, a similar estimation shows that $$\lim_{n\to\infty} \frac{F'(x_n) - F'(x_0)}{x_n - x_0} = f'(x_0)$$

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 $$x_0$$.

• Does this mean if we take the definition of continuity as being only applicable where $F'$ is defined, that $F'$ is continuous? Commented Jan 12, 2021 at 13:00
• I understand that, but just because there exist points in the neighbourhood of $c$ where $F'$ is not continuous, doesn't mean it's not continuous at $c$? Commented Jan 12, 2021 at 18:14
• @user251256 Maybe you can construct counterexample? Commented Jan 12, 2021 at 18:16
• @SenZen the statement made precise is in fact correct. I have updated the answer. Commented Jan 13, 2021 at 21:35

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.

• Please comment along with the downvote, so I can improve my answer Commented Dec 6, 2016 at 19:49
• The problem is $F'$ need not exist in a full neighborhood of $x_0.$
– zhw.
Commented Dec 6, 2016 at 22:48
• @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. Commented Dec 13, 2016 at 21:52
• @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 $F'(x_1) \neq f(x_1)$. Commented Dec 13, 2016 at 21:55
• @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). Commented Dec 13, 2016 at 22:01