Let $f:[0, 1]\to\mathbb{R}$ and $ f(x) = \begin{cases} \frac{1}{n}, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\ 0, & \text{other where} \end{cases}$

I want to find $\int_{0}^{1} f\, dx.$ I think the answer is $0$.

Any ideas or insight would be greatly appreciated

  • $\begingroup$ @Mhenni Benghorbal Ok, but I don't know how to do, would you explain your answer please $\endgroup$ – user62498 Feb 28 '16 at 9:56
  • $\begingroup$ I misread the question! $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 10:16
  • $\begingroup$ There is a problem with the website! The writing is mixed up! $\endgroup$ – Mhenni Benghorbal Feb 28 '16 at 10:42
  • $\begingroup$ Intuition: Look at the graph of $f$. At every $1/n$, you have a point at height $1/n$, and it's everywhere $0$ otherwise. So given any partition $\mathcal{P}$ (say, consisting of intervals $I_n$ around $1/n$), lower Riemann sum is $0$, and upper Riemann sum is $\sum \ell(I_n) \cdot 1/n$ (area of the rectangles around $1/n$ of decreasing height). Note that you can modify $\mathcal{P}$ so that $\ell(I_n)$ is as small as possible, which makes the sum of the areas of the rectangles as small as possible. As the integral is squashed between $0$ and something which is arbitrarily small, it is $0$. $\endgroup$ – Balarka Sen Feb 28 '16 at 14:50

For any $\epsilon > 0$, choose $n > \dfrac{2}{\epsilon}$, define a partition $P_n = \{0,\dfrac{1}{n}, 1-\dfrac{1}{n}, 1\}$, we have: $U(f,P_n)-L(f,P_n)= \dfrac{1}{n}\left(\dfrac{1}{n}-0\right)+0\left(1-\dfrac{1}{n}-\dfrac{1}{n}\right)+1\left(1-\left(1-\dfrac{1}{n}\right)\right)- 0= \dfrac{1}{n}+\dfrac{1}{n^2}< \dfrac{2}{n}< \epsilon$. Thus $f$ is Riemann integrable on $[0,1]\Rightarrow \displaystyle \int_{0}^1 f(x)dx = \displaystyle \lim_{n\to \infty} U(f,P_n) = \displaystyle \lim_{n\to \infty} \left(\dfrac{1}{n}+\dfrac{1}{n^2} \right)= 0$.

  • $\begingroup$ @Dear mathcounterexamples.net, Joanpemo and Joanpemo Thanks for your time and your answer. $\endgroup$ – user62498 Feb 28 '16 at 10:14

You have $0 \le f \le \bar{f}$ where

$$ \bar{f}(x) = \begin{cases} 1, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\ 0, & \text{else} \end{cases}$$

The Riemann integral of $\bar{f}$ is equal to $0$, hence $f$ Riemann integral is also equal to $0$.

To prove it consider the step functions $$ \bar{f_p}(x) = \begin{cases} 1, & \text{if $x \in (\frac{1}{n}-\frac{1}{2^p n(n+1)},\frac{1}{n}+\frac{1}{2^p n(n+1)}) \cap [0,1] \ \ n=1, 2, \dots ,p$ } \\ 1, & \text{if $x \in [0,\frac{1}{p+1})$ } \\ 0, & \text{else} \end{cases}$$

For all $p \ge 1$ you have $$0 \le f \le \bar{f} \le \bar{f_p}$$ and $$\int_0^1 \bar{f_p(x)} \ dx = \frac{1}{p+1} + \frac{1}{2^{p+1}}\sum_{k=1}^p \frac{1}{k(k+1)}$$

As the RHS is converging to $0$ as $p \to \infty$, you are done.


For any partition $\;P=\{x_i\}_{i=1}^n\;$ of $\;[0,1]\;$ define $\;K_P:=\{ 1\le i\le n\;:\;\;\exists\,\frac1m\in[x_{i-1},x_i]\}\;$.

Then for any points $\;c_i\in[x_{i-1},x_i]\;$ , we have$${}$$

$$\sum_{i=1}^n f(c_i)(x_i-x_{i-1})=\begin{cases}\sum\limits_{k\in K_P} (x_k-x_{k-1}),\,&\text{if}\;\;c_k=\frac1m\in[x_{i-1},x_i]\;\text{for some}\;i\\{}\\0,\,\text{otherwise}\end{cases}$$

For the integral to exist the limit of the above sum when $\;n\to\infty\;$ and $\;\text{mesh_P}:=\max_i(x_i-x_{i-1})\to 0\;$ must exists finitely without being dependent on the points $\;c_i\;$ chosen, and since

$$0\le\sum_{k\in K_P}(x_k-x_{k-1})\le|K_P|\text{mesh_P}\longrightarrow0$$

the integral exists and its value is zero.


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