Theorem 16.8 (Schilling: Brownian Motion). Let $f$ be an $\mathscr{F}_t$ adapted right-continuous process with left limits. Then $f \in \mathcal{L}_{T,\text{loc}}^2$. For every sequence $(\Pi_k)_{k\ge 1}$ of partitions of $[0,T]$ such that $\lim_{k\to \infty} |\Pi_k|=0$, the Riemann-Stieltjes sums $$Y_t^{\Pi_k}:= \sum_{s_j, s_{j+1}\in \Pi_k} f(s_j) (B_{s_{j+1}\wedge t} - B_{s_j \wedge t})$$converge uniformly (on compact $t$-sets) in probability to $f\bullet B_t, t\le T:$ $$\lim_{k\to \infty} \mathbb{P}\Bigg( \sup_{0\le t\le T} \Bigg|Y_t^{\Pi_k} - \int_0^t f(s)dB_s \Bigg|>\epsilon\Bigg)=0 \;\text{for all} \;\epsilon>0.$$
There are some parts of the proof I'm unclear about. In the line that starts with "For $0\le s \le \tau_n \wedge T$ the integrand is bounded by $4n$...", why does the integrand converge to $0$, that is, why do we have $\lim_{k\to \infty} f^{\Pi_k}(s) = f(s)$? Is this guaranteed by the right continuity? Why is this guaranteed even when $f$ may have discontinuous points at the left hand? And, why is the last sentence here relevant? That is, why does the author state that the set of discontinuity points is a Lebesgue null set? I don't see what purpose this line serves here.
Definitions. 
