Random time change from Oksendal's SDE textbook 

I have two questions related to the random time change introduced in Oksendal's textbook on SDEs (page 155-156). Specifically, for Lemma 8.5.6., I have no clue as to why we should define $t_j$ in terms of $\alpha_t$ in such a strange way. Perhaps the biggest confusion comes from the chain of identities in the proof of Lemma 8.5.6., it seems that the author used $$\sum_j \int_{\alpha_j}^{\alpha_{j+1}} f(s,\omega)\,\mathrm{d}B_s = \int_0^{\alpha_t} f(s,\omega)\,\mathrm{d}B_s,$$ but this relation is not that obvious in my mind (perhaps it is linked to the weird way that $t_j$ is defined...) My second question concerns a step in the proof of Theorem 8.5.7, in which the author used the relation $\Delta \tilde{B}_j = \sqrt{c(\alpha_j,\omega)}\,\Delta{B}_{\alpha_j}$. However, this relation is not clear to me, as from (8.5.13), we should have $$\Delta \tilde{B}_j = \tilde{B}_{j+1} - \tilde{B}_j = \int_{\alpha_j}^{\alpha_{j+1}} \sqrt{c(s,\omega)}\,\mathrm{d}B_s.$$ So for the claimed identity to hold, we need at least that $\alpha_{j+1} - \alpha_j$ to be sufficiently small (say of order $\mathrm{d}t$). My deepest thanks for any help on these questions.
 A: For the first question, indeed as mentioned in https://math.stackexchange.com/questions/4213957/random-time-change-from-oksendals-sde-textbook
the goal of the proof is to split the interval $[0,a_{t}]$ into dyadic intervals. So I think there is a small typo in the definition: it should be $a_{t}$ instead of $t$
$$t_{j}:=\left\{\begin{matrix}\frac{j}{2^{k}} &,\frac{j}{2^{k}}\leq a_{t} \\ a_{t} &, \frac{j}{2^{k}}>a_{t}\end{matrix}\right..$$
For the second question, there might be a shorter proof, but here is one idea.
We have that $\Delta \tilde{B}_{j}=\int_{a_{j-1}}^{a_{j}}\sqrt{c_{s}}dB_{s}$ and so we add and substract that term. This means we have to show that the following goes to zero
$$\lim_{k}\sum_{j}v(a_j)c_{a_{j}}^{-1/2}\left(\int_{a_{j-1}}^{a_{j}}\sqrt{c_{s}}dB_{s}-c_{a_{j}}^{1/2}\Delta B_{a_{j}}\right)=0.$$
We pull the increment
$$\sum_{j}v(a_j)\left(\frac{1}{\Delta B_{a_{j}}}\int_{a_{j-1}}^{a_{j}}\sqrt{\frac{c_{s}}{c_{a_{j}}}}dB_{s}-1\right)\Delta B_{a_{j}}=:\sum_{j}v(a_j)g_{j}\Delta B_{a_{j}}.$$
Here we use that the difference in the paranthesis goes to zero (https://math.stackexchange.com/questions/3962360/derivative-of-a-stochastic-integral-with-respect-to-limit-with-respect-to-inte)

If $H$ is adapted, right-continuous, and bounded then $$\lim_{t \rightarrow 0} \frac{1}{W_t} \int_0^t H_s dW_s = H_0$$ in probability.

