# Is $f(x)=\sin(1/x),\;x\ne0$, $f(0)=0$, Riemann integrable on $[0,1]$? [closed]

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

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

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.

PROOF

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 \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 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.

• Im sorry im having some trouble figuring out how u got those values for Mk and mk Feb 24, 2019 at 12:25
• @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 $0.$ Feb 24, 2019 at 22:05
• @Pedro Santos: Let me see if you need further explanations. Feb 24, 2019 at 22:06
• 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? Feb 24, 2019 at 22:08
• 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 $0.504$. Jul 25, 2022 at 22:17