Let F:[-1,1] →𝐑 be the function defined by setting F(x):=x^2 sin((1)/(x^3)) when x is non-zero,

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Let $F:[-1,1] \rightarrow \mathbf{R}$ be the function defined by setting $F(x):=x^2 \sin \left(\frac{1}{x^3}\right)$ when $x$ is non-zero, and $F(0):=0$. Show that $F$ is everywhere differentiable, but the deriative $F^{\prime}$ is not absolutely integrable, and so the second fundamental theorem of calculus does not apply in this case (at least if we interpret $\int_{[a, b]} F^{\prime}(x) d x$ using the absolutely convergent Lebesgue integral). See however the next exercise.

   Let $F:[-1,1] \rightarrow \mathbf{R}$ be the function defined by setting $F(x):=x^2 \sin \left(\frac{1}{x^3}\right)$ when $x$ is non-zero, and $F(0):=0$. Show that $F$ is everywhere differentiable, but the deriative $F^{\prime}$ is not absolutely integrable, and so the second fundamental theorem of calculus does not apply in this case (at least if we interpret $\int_{[a, b]} F^{\prime}(x) d x$ using the absolutely convergent Lebesgue integral). See however the next exercise.

An Introduction To Measure Theory (January 2011 Draft)

Terence Tao 1st Edition

Chapter 1, Problem 52

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Let $F:[-1,1] \rightarrow \mathbf{R}$ be the function defined by setting $F(x):=x^2 \sin \left(\frac{1}{x^3}\right)$ when $x$ is non-zero, and $F(0):=0$. Show that $F$ is everywhere differentiable, but the deriative $F^{\prime}$ is not absolutely integrable, and so the second fundamental theorem of calculus does not apply in this case (at least if we interpret $\int_{[a, b]} F^{\prime}(x) d x$ using the absolutely convergent Lebesgue integral). See however the next exercise.

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