# Is the derivative of Riemann integral $\int_0^xf$, if it exists, always equal to $f(x)$?

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 $$c$$.

• for discontinuous $f$, the limit for $F'$ may not exist, but if it does, how can it be anything else? – gt6989b Aug 13 at 2:35
• 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 $c$? – RRL Aug 13 at 2:38
• This is pretty much a statement of the Fundamental Theorem of Calculus. – Jared Aug 13 at 2:41
• @gt6989b Apparently it can. See the answer below. – ashpool Aug 13 at 2:44
• 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/…) – 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 $$F'(1/2) = f(1/2) \neq \tilde{f}(1/2)$$.