# Prove the derivative of $x^2 \sin (1/x^2)$ is not (Lebesgue) integrable on $[0,1]$

Prove the derivative of $x^2 \sin (1/x^2)$ is not Lebesgue integrable on $[0,1]$.

Note at $x=0$, the value of the function is defined to be $0$.

Here 'not integrable' means that the integral value approximated by simple functions from the above is not the same as that by the ones from the below.

Should I use some powerful theorem to prove this?

I don't think this question is a hard one but don't know what kind of approach I should take.

• There are two "points" to this exercise that you should notice and remember. The first is that this provides an example of an everywhere differentiable function $F$ on an interval but $F'$ is not Lebesgue integrable. The other (sometimes shocking) realization is that had you been given this problem in freshman calculus you would have said that indeed $F'$ is integrable and it is even true that $$F(1)-F(0)=\int_0^1 F'(x)\,dx.$$ The apparent mystery here is that, in freshman calculus, you used the improper Riemann integral for this problem and now you are using the Lebesgue integral. Nov 15 '15 at 18:56

(This solution is similar to the one above, but there you also need to show that the length of $$I_n$$ doesn't go to 0 fast enough -- otherwise of course the sum wouldn't diverge.)

We need to show that the following is not integrable: $$f'(0) = \cases{ 2x \sin\left( \frac{1}{x^2}\right) - \frac{2}{x}\cos\left(\frac{1}{x^2}\right), \quad x \in(0,1], \\ 0, \quad x=0. }$$

Because a continuous bounded function on an interval is measurable, and therefore integrable by Lebesgue's theorem, we know that $$2x \sin\left(\frac{1}{x^2}\right)$$ is integrable. The sum of integrable functions will be integrable, so $$f'$$ not integrable is equivalent to $$- \frac{2}{x}\cos\left(\frac{1}{x^2}\right)$$ not being integrable; which implies by definition that its absolute value is also not integrable.

Define

$$g(x):= \left\lvert \frac{2}{x}\cos\left(\frac{1}{x^2}\right)\right\rvert.$$

Observe that for $$y \in [2k\pi - \pi/4, 2k\pi +\pi/4]$$, $$\cos(y) \geq \frac{1}{\sqrt{2}}$$. For integers $$k \geq 1$$, define $$\overline x(k) := \frac{1}{\sqrt{2k\pi -\pi/4}}$$ $$\underline x(k) := \frac{1}{\sqrt{2k\pi +\pi/4}}$$

Therefore, $$g(x) \geq \frac{\sqrt{2}}{\overline x(k)}$$ for $$x \in [\underline x(k), \overline x(k)]$$. If $$I_k:= [\underline x(k), \overline x(k)]$$, then $$g \geq \sum_{k=1}^K \frac{\sqrt{2}}{\overline x(k)} \chi_{I_k}$$ for arbitrary $$K$$. Therefore $$\int g \geq \sum_{k=1}^K \sqrt{2}\left(1-\frac{\underline x(k)}{\overline x(x)}\right) = \sum_{k=1}^K \sqrt{2}\left(1-\sqrt{\frac{2k\pi -\pi/4}{2k\pi + \pi/4} }\right).$$

By applying the inequality $$\sqrt{1-a} \leq 1 - \frac{a}{2}$$, for $$a\in[0,1]$$, we get: $$\int g \geq \sum_{k=1}^K \sqrt{2}\left(\frac{\pi/4}{2k\pi + \pi/4}\right)\geq \sum_{k=1}^K \frac{\sqrt{2}}{4}\left(\frac{1}{2(k+1)}\right).$$ This sum diverges for $$K \rightarrow \infty$$, so we are done.

• Nice answer. The fact that the sets $I_k$ are pairwise disjoint is also important, which can be proved easily with simple algebra. Dec 3 '20 at 7:07

We need only look at the term $\frac{1}{x}\cos(\frac{1}{x^2})$ which occurs in the expression for the derivative of the given function. The derivative isn't integrable because of the $\frac{1}{x}$ factor (the other $x\sin(\frac{1}{x^2})$ term in the derivative is obviously integrable being continuous on $[0,1]$). You can easily construct a minorising sequence $s_k$ of simple functions whose integral blows up as $k\to\infty$. Consider this construction taking for simplicity only the positive part of $f$. Since $\cos(1/x^2)\ge 1/2$ on $I_n:= [1/(2\sqrt{(n+2/3)\pi}),1/(2\sqrt{(n+1/3)\pi})]$ (pls check and correct if required but the principle is valid), so $\frac{1}{x}\cos(\frac{1}{x^2})\ge \sqrt{(n+1/3)\pi}=:a_n$ on $I_n$. Then if $s_k = \sum_{n=1}^k a_n\chi_{I_n}$, then the integral of $s_k$ blows up as $k\to\infty$.