proof of Ruffa’s formula for continuous functions

Define $s_n$ to be the following sum:

$$s_n=\sum _{m=1}^{2^n-1}{2}^{-n}f\left(a+m(b-a)/2^n\right)$$

Making the substitution $m^{\prime }=2m$ and using the fact that $1+{(-1)}^{m^{\prime }}=0$ when $m^{\prime }$ is odd to express the sum over even values of $m^{\prime }$ as a sum over all values of $m^{\prime }$, this becomes

$$s_n=\sum _{m=1}^{2^{n+1}-1}(1+{(-1)}^{m^{\prime }})2^{-1-n}f\left(a+m(b-a)/2^n\right)$$

Subtracting this sum from $s_{n+1}$ and simplifying gives

$$s_{n+1}-s_n=\sum _{m=1}^{2^{n+1}-1}{(-1)}^{m+1}{2}^{-n}f\left(a+m(b-a)/2^n\right)$$

Using the telescoping sum trick, we may write

$$s_k=\sum _{n=1}^k(s_n-s_{n-1})=\sum _{n=1}^k\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f\left(a+m(b-a)/2^n\right)$$

To complete the proof, we must investigate the limit as $k\to \mathrm{\infty }$. Since $f$ is assumed continuous and the interval $[a,b]$ is compact, $f$ is uniformly continuous. This means that, for every $ϵ>0$, there exists a $\delta >0$ such that implies . By the Archimedean property, there exists an integer $k>0$ such that $2^k\delta >|a-b|$. Hence, $|f(x)-f\left(a+m(b-a)/2^n\right)|\le ϵ$ when $x$ lies in the interval $[a+(m-1)(b-a)/2^n,a+(m+1)(b-a)/2^n]$. Thus, $(a-b)s_k+|a-b|ϵ$ is a Darboux upper sum for the integral

$${\int }_a^{b}f(x)𝑑x$$

and $(b-a)s_k-|a-b|ϵ$ is a Darboux lower sum. (Darboux’s definition of the integral may be thought of as a modern incarnation of the ancient method of exhaustion.) Hence

$$|{\int }_a^{b}f(x)𝑑x-s_k|\le |a-b|ϵ$$

Taking the limit $ϵ\to 0$, we see that

$$\underset{a}{\overset{b}{\int }}f(x)𝑑x=\sum _{n=1}^{\mathrm{\infty }}{A}_n=\left(b-a\right)\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f\left(a+m(b-a)/2^n\right)$$