$f$ is integrable on $[a, b]$ and $F(x) = \int_a^x f(t) \, dt$. If $F$ is differentiable at $x_0$ is it always true that $F'(x_0) = f(x_0)$?

Question

Problem. Always true or sometimes false: If $f$ is Riemann integrable on $[a, b]$ (not necessarily continuous) and $F(x) \int_a^x f(t) \, dt$ is differentiable at $x_0 ∈ [a, b]$ then $F'(x_0) = f(x_0)$?

Full disclosure: this question appeared on an open book exam for my analysis class. The exam is now over – I can no longer submit answers – so this question is purely for my interest. Also please note that this function $f$ is not necessarily continuous everywhere on $[a, b]$, so it does not satisfy all the conditions of the fundamental theorem of calculus. Please find my work on the problem below:

Obviously if $f$ is everywhere continuous on $[a, b]$ then the statement holds, so we can suppose $f$ is not continuous everywhere on $[a, b]$. I know that a function is Riemann integrable on $[a, b]$ if and only if it is continuous almost everywhere on $[a, b]$. That is, the set of points where it is not continuous is a set of measure zero. So the set $U$ of points where $f$ is not continuous is a set of measure zero. Also I have the following result from class, which is stronger than the fundamental theorem of calculus.

Lemma. Let $f$ be integrable on $[a, b]$ and let $c ∈ [a, b]$. Suppose $f$ is continuous at $x_0 ∈ [a, b]$. Let $F(x) = \int_c^x f(t) \, dt$. Then $$F'(x_0) = f(x_0).$$

So the statement given in the title certainly holds at every point where $f$ is continuous. That is, $F'(x_0) = f(x_0)$ at every point in $x_0 \in U$.

Now, the question that remains as far as I can see, since the statement we are considering includes the assumption that $F$ is differentiable at $x_0$, is whether $F$ can be differentiable at $x_0$ while $f$ is not continuous at $x_0$.

So we really just need to consider the case where $f$ is not continuous at $x_0$. This is where I am stuck. I tried to proceed by classifying the possible discontinuities at $x_0$. The fact that $f$ is integrable means $f$ is bounded, so it definitely does not have an essential discontinuity at $x_0$. But a priori it may have a jump discontinuity or a removable discontinuity at $x_0$. I think that if $f$ has a jump discontinuity at $x_0$ then $F$ will not be differentiable at $x_0$, although I can’t prove it. As for a removable discontinuity, I think the effect of this would be that $F'(x_0) \neq f(x_0)$, although I also cannot prove it.

I also tried the following to prove the statement to be true: The fact that the set $U$ of points where $f$ is discontinuous is of measure zero also means that $U$ is dense in $[a, b]$. So every subinterval of $[a, b]$ contains points in $U$. This means we can choose a sequence $x_n \to x_0$ with $x_n \neq x_0$ and $x_n ∈ U$ for all $n$. So then since $x_n ∈ U$ it follows by the lemma that $F'(x_n) = f(x_n)$ for all $n$. Thus, $$\lim_{n \to \infty} F'(x_n) = \lim_{n \to \infty} f(x_n).$$

But this gets us nowhere since we don't know if $f$ or $F'$ is continuous at $x_0$

That's all the information I have on the problem. Thank you for any assistance.

Answer 1

No, it is not. A counterexample is the function $f:[0,2]\to \mathbb{R}$ defined by $$ f(x):=\begin{cases} 1,& x=1\\ 0,& \text{ otherwise } \end{cases} $$

Then $F(x):=\int_0^x f(t) \mathop{}\!dt=0$ but $F'(1)\neq f(1)$.

Answer 2

Hint: Consider a function $f$ which is the zero function except at a single point in the interval.

Answer 3

The best approach to the problem is to understand the proof of Fundamental Theorem of Calculus. If you understand the proof well you should note that the proof is actually about this more general version:

Theorem: Let $f$ be Riemann integrable on $[a, b] $ and $F(x) =\int_{a} ^{x} f(t) \, dt$. Let $c\in[a, b] $ be such that one sided limit $f(c+) =\lim_{x\to c^{+}} f(x) $ exists. Then the right derivative of $F$ at $c$ exists and equals $f(c+) $. A similar statement holds for $f(c-) $ and left derivative of $F$ at $c$.

From the above it follows that if $L=\lim_{x\to c} f(x) $ exists then $F'(c) $ exists and equals $L$. But then this limit $L$ does not necessarily equal $f(c) $ (in other words $f$ may have a removable discontinuity at $c$) and then $F'(c) \neq f(c) $.

The above theorem also shows that if $f$ has jump discontinuity then $F$ is not differentiable at $c$ (left and right limits of $f$ are different and hence left and right derivative of $F$ are different).

Another more curious example is when $f$ has essential (oscillatory) discontinuity at $c$ and $F$ is differentiable at $c$. This is possible as exhibited by the function $F(x) =\int_{0}^{x}\sin(1/t)\,dt$. It can be proved with some effort that $F'(0)=0$.

One should also observe that the definition of $F$ as Riemann integral of $f$ over $[a, x] $ involves the behavior of $f$ in an interval. Changing the values of $f$ at a finite number of points does not affect the integral and hence does not affect $F$ and hence one should not feel surprised that properties of $F$ are not really dependent on values of $f$ at specific points. Thus one should not expect $F'(c) =f(c) $ in general. This happens in a very specific case when $f$ is continuous at $c$ otherwise this is not guaranteed.