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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0How is $\int_a^b f$ a function of $x$? – 2012-03-16
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0Oops, fixed haha. – 2012-03-16
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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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0$F$ is not continuous on $[0,1]$. I think you meant this interval. – 2012-03-16
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0There's something I don't follow. We need that $F$ is a.e. an antiderivative of $f$. Thus $F^\prime = f$ a.e., but $F^\prime$ for your example "looks" nonintegrable on $[0,1]$, so $\int f$ doesn't exist. What am I missing? – 2012-03-16
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0$f(x) = \cos(\pi/x) + (\pi/x) \sin(\pi/x)$ on $(0,1)$, and for any $\epsilon > 0$, $\int_\epsilon^1 f(x)\ dx = F(1) - F(\epsilon)$. Then as an improper Riemann integral, $\int_0^1 f(x)\ dx = \lim_{\epsilon \to 0+} (F(1) - F(\epsilon)) = F(1)$ exists. – 2012-03-16
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0In order for $g(x) = \int_0^x g^\prime (t) dt + g(0)$, it is necessary and sufficient that $g$ be absolutely continuous on $[0,1]$. On the other hand, every absolutely continuous function on $[a,b]$ is of bounded variation on $[a,b]$. But it is well known that $x\sin (1/x)$ is not of bounded variation on $[0,1]$, hence not absolutely continuous there, hence is not the integral of its derivative. In other words, $F(x) \not= \int F^\prime(x) dx$. – 2012-03-17
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0That should have been $x \cos(1/x)$ is not BV to match your $F$. – 2012-03-17
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0AsI said, it's not the Lebesgue integral, but it is an improper Riemann integral. – 2012-03-18