Jtdylktuy

How to prove that $f(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on if $xneq0$ 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 $lfloor wrfloor$ be the greatest integer less than or equal to $w$. Then the function
$$xmapsto begin{cases} lfloor 2/xrfloor & text{if } lfloor 1/xrfloorle n \[8pt]
n & text{otherwise} end{cases}$$ is simple. It is $le 2/x$ and its integral over $(0,1]$ approaches $infty$ as $ntoinfty$.

Now with the upper gaussian brackets and using that $-1le sin(frac{1}{x^2}),cos(frac{1}{x^2})le 1$. But I don't get it

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 $x mapsto 2xsinfrac{1}{x^2}$ 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,

$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$

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.

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 $j_1toinfty$, while


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