How can I prove that the function

$f:(0, 1) \to \mathbb{R}$
$f(x) = \frac{|\sin(\frac{1}{x})|}{x}$

is not Lebesgue integrable?

Should I try the definition?

Because I need to find a function $f:A \to \mathbb{R}$ such that $A$ is open and bounded and $f$ is continuous, not bounded and not Lebesgue integrable, and I think that $f$ works, right?

  • $\begingroup$ You could just use $f(x) = 1/x^2$ on the same domain. $\endgroup$
    – user169852
    Oct 29, 2019 at 19:31

1 Answer 1


Observe that if $$x \in \left[ \frac{(6n+1)\pi}{6}, \frac{(6n+5)\pi}6 \right] = \left[ n\pi + \frac \pi 6, (n+1)\pi - \frac \pi 6 \right]$$ then $|\sin x| \ge \dfrac 12$. Define $$E_n = \left[ \frac{6}{(6n+5)\pi},\frac{6}{(6n+1)\pi} \right].$$ It follows that $$ x \in E_n \implies \frac 1x \ge \frac{(6n+1)\pi}{6} \ge n\pi \quad \text{and} \quad \left|\sin\left( \frac 1x \right) \right| \ge \frac 12$$ and consequently $$x \in E_n \implies f(x) = \frac{|\sin(\frac 1x)|}{x} \ge \frac{n\pi}2 \ge n.$$ It is simple to verify that the intervals $\{E_n\}_{n=1}^\infty$ are pairwise disjoint and that each $E_n \subset (0,1)$. This implies that for any $N$ you have $$\sum_{n=1}^N n \chi_{E_n}(x) \le f(x)$$ on $(0,1)$. Since $f \ge 0$ the definition of the Lebesgue integral gives you $$\sum_{n=1}^N n \ell(E_n) \le \int_{(0,1)} f.$$ It appears that $\ell(E_n) \ge \dfrac 1{5(n+1)^2}$ so that $$\frac 15 \sum_{n=1}^N \frac{n}{(n+1)^2} \le \int_{(0,1)} f.$$ This is true for any natural number $N$---the divergence of the harmonic series tells you that the left-hand-side is unbounded as $N \to \infty$, so $f$ cannot have finite integral.


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