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.
Best Answer
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 $F'(1)\neq f(1)$.