Let $f:[0,1]\to\mathbb{R}$ be a Riemann integrable function. Define a function $F:[0,1]\to\mathbb{R}$ by $$F(x)=\int_0^xf.$$ Suppose that $F$ is differentiable at $c\in(0,1)$. Then is it necessarily true that $F'(c)=f(c)$ ? Note that it is true if $f$ is continuous at $c$.

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    $\begingroup$ for discontinuous $f$, the limit for $F'$ may not exist, but if it does, how can it be anything else? $\endgroup$ – gt6989b Aug 13 at 2:35
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    $\begingroup$ Suppose $f$ is continuous, then $F'(x) = f(x)$ everywhere in $[0,1]$. What happens if you change the value of $f$ at a single point $c$? $\endgroup$ – RRL Aug 13 at 2:38
  • $\begingroup$ This is pretty much a statement of the Fundamental Theorem of Calculus. $\endgroup$ – Jared Aug 13 at 2:41
  • $\begingroup$ @gt6989b Apparently it can. See the answer below. $\endgroup$ – ashpool Aug 13 at 2:44
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    $\begingroup$ You can actually relax the condition of continuity at $c$ a bit to get the same result, i.e. $F'(c)=f(c)$. $c$ needs to be a so called Lebesgue point of $f$ (see en.wikipedia.org/wiki/Lebesgue_point and en.wikipedia.org/wiki/…) $\endgroup$ – humanStampedist Aug 13 at 11:36

Just take any continuous function with its value changed at a single point. For example let $f:[0, 1] \to \mathbb R$ be any continuous function and define $\tilde{f}: [0, 1] \to \mathbb R$ by $$ \tilde{f}(x) = \begin{cases} f(x) + 1 & \text{if $x = 1/2$}, \\ f(x) & \text{if $x \neq 1/2$}. \end{cases}$$ Then since $f$ and $\tilde{f}$ differ by only a single point, they have the same indefinite integral $$F(x) = \int_0^x f(x) \, \text{d}x = \int_0^x \tilde{f}(x) \, \text{d}x.$$ However $F'(1/2) = f(1/2) \neq \tilde{f}(1/2)$.

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