Given $F(x) = \int_a^x f$, $F$ is continuous on $[a,b]$, and $F$ is differentiable almost everwhere, is it necessarily true that F' \in L^1(a,b)?. I'm pretty sure this is false, but I am having trouble showing it. Does anyone have a quick counterexample? Thanks!
Counterexample to $F(x) = \int_{a}^{x} f(t) dt \implies F' \in L^1(a,b)$
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real-analysis
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2This is a very uninformative title. The 'real-analysis' tag already indicates that it's a real analysis question. – 2012-03-16
1 Answers
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If the integral is considered as an improper Riemann integral, then it is false. Try $F(x) = \frac{\cos(\pi/x)}{x}$.
EDIT: Oops, missed the "continuous on $[a,b]$". Make that $F(x) = x\cos(\pi/x)$ (on $[0,1]$), noting that \int_0^1 |F'(x)|\ dx \ge \sum_{n=1}^\infty \left|F\left(\frac1n\right) - F\left(\frac{1}{n+1}\right)\right| = \sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{n+1}\right) = \infty
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0AsI said, it's not the Lebesgue integral, but it is an improper Riemann integral. – 2012-03-18