Suppose that $X_t:=\int_{0}^{t}b_sds+\int_{0}^{t}\sigma_tdB_t$ where $b,\sigma \in L^{\infty}(F)$. For $\pi:0=t_0<t_1<...<t_n=T$, denote $$S_L(\pi):=\sum_{i=0}^{n-1}X_{t_i}B_{t_i,t_{i+1}}$$.
Show that $S_L(\pi)\rightarrow\int_{0}^{T}X_TdB_t$ in $L^2$, as $|\pi|\rightarrow0$.
My idea is that since $X_t$ is continuous, we can approximate it by piecewise constant functions $X_t^n$ then $\int_{0}^{T}X_TdB_t$ by definition is the limit of $\int_{0}^{T}X_T^ndB_t$. So
\begin{align*} \int_{0}^{T}X_TdB_t&=lim \int_{0}^{T}X_T^ndB_t\\ &=lim \sum_{i=0}^{n-1} \left[\int_{0}^{t_i}b_sd_s+\int_{0}^{t_i}\sigma_tdB_t\right]B_{t_i,t_{i+1}} \end{align*}
I am a little bit puzzled because it seems that by definition, the integral is defined by the limit of partial sums.