Given $f(x)$ and $g(x)$ like the following: \begin{align} f(x) = \begin{cases} 0 \quad x\in[-1, 0)\\ 1 \quad x\in[0, 1] \end{cases} && \phi(x) = \begin{cases} 0 \quad x\in[-1, 0]\\ 1 \quad x\in(0, 1] \end{cases} \end{align}
For any partition $\Gamma=\{x_i\}_{i=0}^m, 0=x_0<x_1<\cdots<x_m=1$, the Riemann-Stieltjes sum is \begin{align} R_\Gamma[f, \phi]=\sum_{i=1}^m f(\xi_i)\left[\phi(x_i)-\phi(x_{i-1})\right] \end{align} where $\xi_i$ is any point in $[x_{i-1}, x_i]$.
Therefore, \begin{align} R_\Gamma[f, \phi]&=\sum_{i=1}^m f(\xi_i)\left[\phi(x_i)-\phi(x_{i-1})\right]\\ &=f(\xi_1)\left[\phi(x_1)-\phi(x_0)\right] + \sum_{i=2}^m f(\xi_i)\left[\phi(x_i)-\phi(x_{i-1})\right]\\ &=1\cdot[1-0]+ 0 \quad \text{since } [\phi(x_i)-\phi(x_{i-1})] \text{ are all zero for } i=2, 3, \cdots, m\\ &=1 \end{align}
Therefore, $$ \int_0^1 fd\phi = \lim_{\max\{\Delta x_i\}\to0} R_\Gamma[f, \phi] = 1 $$
However, my professor told us it is zero. (without any explanation.)
Is it real zero? If so, where is fault in my process?
Thanks for reading my question.
For the $f$ and $\phi$ you defined, $\int_0^1 f \; d\phi$ is indeed $1$. The instructor made a mistake.