Is the function $\sin \frac{1}{x}$ Riemann integrable on an interval containing $0$?

  • $\begingroup$ Consider a change of variables. $\endgroup$
    – Pedro
    Mar 27, 2014 at 4:17
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    $\begingroup$ A function is Riemann integrable iff it is bounded and continuous almost everywhere. $\endgroup$ Mar 27, 2014 at 4:41

2 Answers 2


There are two proofs: a short one and a longer one.

First Proof: $f$ is bounded on $[0, 1]$ since $\left| \sin \frac{1}{x} \right| \leq 1$. $f$ is continuous except at $x = 0$ (since limit as $x \to 0$ doesn’t exist), so the set of discontinuities is the single point $x = 0$ and this set has measure zero. There is a well known theorem stating that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.

The second proof is based on the upper $U$ and lower $L$ sums. $f$ is integrable if given any $\varepsilon > 0$ there is a partition such that $U – L < \varepsilon$. So given $\varepsilon > 0$, consider the intervals $[0, \varepsilon]$ and $[\varepsilon, 1]$. Since $f$ is bounded and continuous on $[\varepsilon, 1]$ it is integrable, so there is a partition $P_1$ such that $U - L < \varepsilon$. On $[0, \varepsilon]$, the most $U - L$ could be (for any partition, $P_\varepsilon$) is $2 \varepsilon$ (since $\max_{x \in \mathbb{R}} \sin x = 1$, $\min_{x \in \mathbb{R}} \sin x = - 1$, and length of the interval is $\varepsilon$). Now consider the partition of $[0, 1]$ given by $P_1$ union $P_e$ and look at $U - L$. We have $|U - L|$ for $[0, 1]$ is less than or equal to the sum of $U - L$ for the two partitions $P_1$ and $P_e$, which is less than $\varepsilon + 2 \varepsilon = 3 \varepsilon$. So $U - L < 3\varepsilon$, which means it can be made arbitrarily small. Thus $f$ is integrable on $[0, 1]$.


This is late but will be useful for future readers.


Let $\epsilon>0$ be given but arbitrarily small. Choose $x_1\in(0,1]$ such that $1-x_1<\epsilon/2.$ Then, the set $P=\{0,x_1,1\} $ is a patition. The upper and lower Darboux sums are given by

\begin{align}U(f,P)=\sum^{2}_{k=1}M_k (x_k-x_{k+1}),\;\;\text{where}\;\;M_{k}= \sup\{f(x):\;x_{k-1}\leq x<x_{k}\},\end{align} \begin{align}L(f,P)=\sum^{2}_{k=1}m_k (x_k-x_{k+1}),\;\;\text{where}\;\;m_{k}= \inf\{f(x):\;x_{k-1}\leq x<x_{k}\}.\end{align} By computation \begin{align}U(f,P)&=\sum^{2}_{k=1}M_k (x_k-x_{k+1})\\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\\&= 1-x_{1},\;\;\text{where}\;\;M_{1}=0,\,M_{2}=1,\end{align} \begin{align}L(f,P)&=\sum^{2}_{k=1}m_k (x_k-x_{k+1})\\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\\&=-(1-x_{1}),\;\;\text{where}\;\;m_{1}=0,\,m_{2}=-1.\end{align} Thus, \begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<\epsilon.\end{align} So, given any $\epsilon$, choose $P=\{0,x_1,1\} $ such that $1-x_1<\epsilon/2.$ Then, \begin{align}U(f,P)-L(f,P)<\epsilon.\end{align} Hence, $f$ is integrable.

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    $\begingroup$ Im sorry im having some trouble figuring out how u got those values for Mk and mk $\endgroup$
    – Someone
    Feb 24, 2019 at 12:25
  • $\begingroup$ @Pedro Santos: It is easy to see that $M_1=0$ and $m_1=0$ since $f(0)=0$. Now, on $(0,1],$ $-1\leq \sin(1/x)\leq 1.$ So, the upper bound is $1$ while the lower bound is $0.$ $\endgroup$ Feb 24, 2019 at 22:05
  • $\begingroup$ @Pedro Santos: Let me see if you need further explanations. $\endgroup$ Feb 24, 2019 at 22:06
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    $\begingroup$ Yes i am aware of those facts my question is in the interval from $0$ to $x_1$ doesnt the function loop around from -1 to 1? How do i know that in that interval the function doenst take a value lower than zero? $\endgroup$
    – Someone
    Feb 24, 2019 at 22:08
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    $\begingroup$ This solution is incorrect. Like @whatever said, $M_1 = 1$ since on the the interval $[0,x_1]$, we know that $f(x) = 1$ infinitely many times. In addition, this solution shows that $-\epsilon/2 < -(1 - x_1) = L(f, P) \le U(f, P) = 1- x_1 < \epsilon/2$. This implies that the integral $\int_0^1 f(x)\,dx = 0$, which is false. The integral is about $0.504$. $\endgroup$
    – HiMatt
    Jul 25, 2022 at 22:17

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