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Q: If $f$ is Riemann integrable over $[0,1]$, then there exists a differentiable function $F$ on $[a, b]$ such that $F'=f$ and $\int_a^b f(x)dx = F(b)-F(a)$.

Here's what I'm thinking so far:

  • If $\alpha = x$, then you can say that $f$ is also Riemann-Stieltjes integrable on $[0,1]$.
  • Then, since $f$ must be a bounded real-valued function due to being Riemann integrable, we can use the theorem that states "For $0\le x \le 1$, put $F(x) = \int_0^x f(t) dt$. Then $f$ is continuous on $[0,1]$
  • In order for $F$ to be differentiable, $f$ must be continuous on $[0,1]$, so we add that as an assumption
  • Then, by Fundamental Theorem of Calculus, we can say $\int_0^1 f(x) \; dx = F(1) - F(0)$.

I realize there are lots of issues here, but I'm hoping I'm at least on the right track. Any direction or hints are appreciated.

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    I a$m$ trying to either prove the statement, or give a counterexample. I am thinking I need a counterexample at this point, but am having trouble writing one.2012-12-05

3 Answers 3

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Note that, A bounded function $f$ of $[a,b]$ is Riemann integrable on $[a,b]$ iff the function is continuous almost everywhere. This is known as Lebesgue's criteria.

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    @downvoter: what's the downvote for?2013-12-16
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Your question has a negative answer. For instance the function defined by $f(x)=\cases{-1,& $-1\le x<0$ \cr 1,&$ 0\le x\le 1$}$ is Riemann integrable over $[-1,1]$, but is not the derivative of any function (since derivatives have the intermediate value property as a result of Darboux's Theorem).

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    @GradStudent: Darboux's Theorem is the underlying conceptual reason behind counterexamples like this, but it's certainly not needed to check the lack of differentiability in this case. The antiderivative of a piecewise constant jump function like this is a piecewise linear function with a corner point. The left and right hand derivatives at this point are different, so the function is not differentiable.2013-12-16
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The question is clearly false. Suppose it were true, and let $f = x$ on $[0,1]$. Let $g = f$ on $[0,1)$ and $g(1) = 0$. Then we have that $\int_a^bf(x) dx = \int_a^b g(x) dx$, since the functions differ at only a finite number of points. So your requirement would necessitate a differentiable function $F$ satisfying $0 = F'(1) = 1$.

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    Why does my requirement necessitate that F satis$f$ies 0 = F'(1) = 1?2012-12-05