Define $s_n$ to be the following sum: $$s_n = \sum\limits_{m = 1}^{2^n - 1} 2^{ - n} f\left( {a + m(b - a)/2^n } \right)$$ Making the substitution $m' = 2m$ and using the fact that $1 + (-1)^{m'} = 0$ when $m'$ is odd to express the sum over even values of $m'$ as a sum over all values of $m'$ , this becomes $$s_n = \sum\limits_{m = 1}^{2^{n+1} - 1} (1 + (-1)^{m'}) 2^{- 1 - n} f\left( {a + m(b - a)/2^n } \right)$$ Subtracting this sum from $s_{n+1}$ and simplifying gives $$s_{n+1} - s_n = \sum\limits_{m = 1}^{2^{n+1} - 1} (-1)^{m + 1} 2^{- n} f\left( {a + m(b - a)/2^n } \right)$$ Using the telescoping sum trick, we may write $$s_k = \sum_{n=1}^k (s_n - s_{n-1}) = \sum\limits_{n = 1}^k {\sum\limits_{m = 1}^{2^n - 1} {\left( { - 1} \right)^{m + 1} } } 2^{ - n} f\left( {a + m(b - a)/2^n } \right)$$

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