Originally Posted by

**tonio** Look at each summand defining $\displaystyle \int_a^b f(x)dg(x)$ : it is of the form $\displaystyle f(c_i)(g(x_{i+1})-g(x_i))$ , where $\displaystyle [x_i,\,x_{i+1}] $ is a subinterval of $\displaystyle [a,\,b]$ and $\displaystyle c_i\in [x_i,\,x_{i+1}]$ .

Well, now apply the mean value theorem to the function $\displaystyle g\;:\;g(x_{i+1})-g(x_i)=g'(d_i)(x_{i+1}-x_i)\,,\,\,d_i\in (x_i,\,x_{i+1})$ , substitute this in every summand of

the sum defining $\displaystyle \int_a^b f(x)dg(x)$ and get the Riemann sum of exactly what you need since we're given $\displaystyle g'(x)$ is Riemann integrable...