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.

  • 5
    $\begingroup$ 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. $\endgroup$ Nov 15, 2015 at 18:56

2 Answers 2


(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.


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

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.

  • 1
    $\begingroup$ Nice answer. The fact that the sets $I_k$ are pairwise disjoint is also important, which can be proved easily with simple algebra. $\endgroup$
    – Sam Wong
    Dec 3, 2020 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 $k\to\infty$.


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