I'm reading the following theorem in Folland:
Let $f$ be a bounded real-valued function on $[a,b]$. If $f$ is Riemann integrable, then $f$ is Lebesgue measurable (and hence integrable on $[a,b]$ since it is bounded), and $\int^b_a f(x)\,dx = \int_{[a,b]}f \, dm$.
It's not clear to me how Dominated Convergence Theorem is being used to conclude $$\int G \, dm = \int g \, dm = \int_a^bf(x)\,dx$$ What's dominating? Where's the sequence? Also they conclude that $G=f$ a.e. after showing $G=g$ a.e. How did they get to $G=f$ a.e.?
EDIT: $$S_Pf = \sum_1^nM_j(t_j-t_{j-1}) \quad s_Pf = \sum_1^nm_j(t_j-t_{j-1})$$ where $M_j$ and $m_j$ are max and min on a piece of the partition