function is not Lebesgue integrable How to prove that $f(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$ on if $x\neq0$ and $f(x)=0$ if $x=0$ is not Lebesgue-integrable on $[-1,1]$? I think the reason is, that $\frac{2}{x}$ is part of the function Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$. 
I tried to construct step functions using  the step function for $1/x$ on $(0,1]$: Let $\lfloor w\rfloor$ be the greatest integer less than or equal to $w$.  Then the function
$$x\mapsto \begin{cases} \lfloor 2/x\rfloor & \text{if } \lfloor 1/x\rfloor\le n \\[8pt]
n & \text{otherwise} \end{cases}$$ is simple.  It is $\le 2/x$ and its integral over $(0,1]$ approaches $\infty$ as $n\to\infty$.
Now with the upper gaussian brackets and using that $-1\le \sin(\frac{1}{x^2}),\cos(\frac{1}{x^2})\le 1$. But I don't get it
 A: The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = \max(f(x),0)$ and $f^-(x) = \max(-f(x),0)$, must have finite-valued integrals 
$$\int_E f^+ < +\infty, \,\, \int_E f^- < +\infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $\int_E|f| < +\infty$.
In this case there is no problem with $x \mapsto 2x\sin\frac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$.  Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x \mapsto \frac{2}{x} \cos \frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x \to \frac{1}{\sqrt{t}}$ and have
$$\int_1^\infty \frac{|\cos t|}{t} \, dt = \int_0^1 \frac{2}{x} \left|\cos \frac{1}{x^2}\right| \, dx < +\infty$$
However,
$$\int_1^\infty \frac{|\cos t|}{t} \, dt > \sum_{k=1}^\infty\int_{k\pi}^{k\pi + \pi} \frac{|\cos t|}{t} \, dt  > \sum_{k=1}^\infty \frac{1}{k\pi +\pi}\int_{k\pi}^{k\pi + \pi} |\cos t| \, dt  = \frac{2}{\pi}\sum_{k=1}^{\infty} \frac{1}{k+1} \\ = +\infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$\int_0^1 \frac{2}{x} \cos \frac{1}{x^2} \, dx $$
is convergent, but this does not enforce Lebesgue integrability.
A: The derivative of $F(x)= x^2 \sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined. 
Now if $f$ were integrable, then $F$ would be absolutely continuous. 
We show that this is not the case. 
For $j=1,2,\dots$ let  $a_j= \frac{1}{\sqrt{\pi j}}$ and $b_j = \frac{1}{\sqrt{\pi( j + 1/2)}}$. 
Then  $F(a_j)=0$ and $F(b_j)=\pi (j+1/2)$. 
To conclude, for any   $j_1$: 


*

*$\sum_{j>j_1}| a_j - b_j| <  a_{j_1}\to 0$ as $j_1\to\infty$, while

*$\sum_{j>j_1} |F(a_j)-F(b_j)| = \sum_{j>j_1} \pi (j+1/2)= \infty$. 

