# Riemann integration with countable discontinuities

I have a question about 'choosing partition points' for Riemann integrability.

For example,

Let $$f:[0,1]\to\mathbb{R}$$ be a function defined by $$f(x)\left\{\begin{array}{cl} 0, & x\in E, \\ 1, & \text{otherwise}. \end{array}\right.$$ where $$E=\{\frac{1}{n}\,:\,n\in\mathbb{N}\}$$.

Indeed, for any given $$\varepsilon>0$$, choose a number $$N\in\mathbb{N}$$ so that $$1/N<\varepsilon/2$$, and a partition $$P\in\mathscr{P}[0,1]$$ so that $$P=\{0=x_{0}<\cdots with $$||P||<\frac{\varepsilon}{4N}$$. Then, if $$n\ge N$$, we have \begin{align*} U(f,P)-L(f,P)&=\sum_{k=1}^{n}(M_{k}-m_{k})\Delta{x_{k}}\\ &<\sum_{k=1}^{n}1\cdot\Delta{x_{k}}\\ &<\frac{\varepsilon}{4N}\sum_{k=1}^{n}1=\frac{\varepsilon}{4N}\cdot n<\varepsilon. \end{align*} where $$M_{k}$$ and $$m_{k}$$ is the supremum and infimum on each subinterval of $$[0,1]$$, respectively.

First of all, I think my proof is not true. But, it seems to be true..

Actually, i consider only the value of $$M_{k}-m_{k}$$ on each subintervals is at most $$1$$, so i didn't consider the points of discontinuities of $$f$$ when taking a partition.

Why this proof is false? Can anyone explain it?

• $\sum \Delta x_k =1$ for any partition and you can never make this less than $\epsilon$. – Kavi Rama Murthy May 7 at 9:30
Hint: split $$\sum (M_k-m_k)\Delta x_k$$ into the sum over $$k <\epsilon$$ and the remaining sum. In the second sum there are only at most $$\frac 1 {\epsilon}$$ values of $$k$$ where $$M_k-m_k\neq 0$$. Make the norm of the partition less than $$\epsilon^{2}$$.