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I have to show that the following function $f:[0,1]\rightarrow\mathbb{R}$ is Riemann Integrable:

$$f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x = \frac{1}{n} \\ 0 & \mbox{otherwise} \end{array} \right.$$

For the upper and lower Riemann sum I am using the following definitions:

$$S_{l}(f,V)=\sum^{n}_{j=1}\inf_{I(j)}(f)(x_j-x_{j-1})$$

With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V=\{0,x_1,...,1\}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $\epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<\epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set $\{\frac{1}{n}:n\in\mathbb{N}\}$. But I can't make the proof concrete. Could anybody help me out?

Gehaktmolen
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2 Answers2

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Try the following:

The set $F=\{x\in [0,1]: f(x)>\epsilon \}$ is finite for every $\epsilon>0$. Then you can form a partition such that if an interval contains some $x\in F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $x\in F$ is $<\epsilon$. Separate the interval wich cover $F$ and those which don't.

Can you continue from this?

EQJ
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$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.

For every $\epsilon>0$, there exists an integer $N_0$ such that $1/N_0<\epsilon$. Let $N=\max\{N_0,5\}$.

Partition $[0,1]$ with $V=\{0,x_1, x_2,\ldots,x_{4N-5}\}$ where $x_1=\frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=\frac{1}{N^3}$, and $x_{2k}<\frac{1}{2N-k}<x_{2k+1}$.

Then, $$S_u(f,V)= 1\cdot \frac{1}{2N} + \frac{1}{N^3} \cdot (2N-1)<\frac{1}{2N}+\frac{2N}{N^3}=\frac{1}{2N}+\frac{2}{N^2}$$.

Since $N\ge 5$, $\frac{2}{N^2}<\frac{1}{2N}$. Therefore, $S_u(f,V)<\frac{1}{N}<\epsilon$.

Badoe
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