$f '$ is not Lebesgue integrable on $[-1,1]$

Question

Let f be that function from R to R defined by f(x)= 0 if x=0

                                         x^2 sin(1/x)  if x not = 0

show that the function f' is bot Lebesgue integrable on [-1,1]

Answer 1

Is it not hard to see that $$ f'(x)=\left\{ \begin{array}{lll} 2x\sin(1/x)-\cos(1/x) & \text{if} & x\ne 0, \\ 0 & \text{if} & x=0. \end{array} \right. $$ Clearly $f'$ is bounded in $[-1,1]$ and continuous everywhere except at $x=0$.

Hence, not only $f'$ is Lebesgue integrable, but it is Riemann integrable as well!

Note. If $f(x)=x^2\sin (1/x^2)$, for $x\ne 0$, then indeed $f$ is not Lebesgue integrable.