I have a question regarding the differential form of a local martingale and its quadratic variation (the source of the question is p. 136-137 in https://galton.uchicago.edu/~mykland/paperlinks/I.A.1-Econometrics_of_High_Frequency_Data.pdf)
Given a local martingale on the form $X_t = X_0 + \int_0^t \sigma^2_s \, dW_s$. Suppose we have a grid of time points $\mathcal{G} = \{t_1 = 0, \ldots, t_n = T \}$ on the interval $[0,T]$ and for $t \in [0,T]$ we define $t^{\ast} = \max \{ t_i \in \mathcal{G} \colon t_i \leq t \}$ and the local martingale \begin{align} M_t &= 2\sum_{t_{i+1} \leq t} \int_{t_i}^{t_{i+1}} (X_s - X_{t_i}) \, dX_s + 2 \int^t_{t^{\ast}} (X_s - X_{t^{\ast}}) \, dX_s. \tag{1}\label{1} \end{align} Then \begin{align} d M_t = 2(X_t - X_{t^{\ast}}) dX_t. \tag{2}\label{2} \end{align} I agree with this, because it is what I obtain by considering $M_{t + \Delta t} - M_t$ and then taking limits as $\Delta t \to 0$. My problem is the following, consider the quadratic variation of $M_t$ \begin{align*} [M, M]_t &= 4\sum_{t_{i+1} \leq t} \int_{t_i}^{t_{i+1}} (X_s - X_{t_i})^2 \, d[X, X]_s + 4 \int^t_{t^{\ast}} (X_s - X_{t^{\ast}})^2 \, d[X,X]_s, \end{align*} then supposedly \begin{align} d[M,M]_t = 4(X_t - X_{t_i})^2 d[X,X]_t. \tag{3}\label{3} \end{align} Using that $d[X,X]_t = \sigma^2_t \, dt$ I am able to (at least I think I am) to show that \begin{align} d[M,M]_t = 4(X_t - X_{t^{\ast}})^2 d[X,X]_t. \tag{4}\label{4} \end{align} (note that (4) differs from (3) by including $t^{\ast}$ and not $t_i$) by considering $[M,M]_{t + \Delta t} - [M, M]_t$ and taking limits. Obviously, I must be misunderstanding something, as (3) is the correct answer. What way is the correct way to show (3)?
Instead of $t^*$ which the authors denote by $t_*\,$ I'd rather define the left continuous step function $$ T_t=\max\{t_i\in{\cal G}:t_i\color{red}{<}t\} $$ by which $$\tag{1} M_t=2\sum_{t_{i+1}\le t}\int_{t_i}^{t_{i+1}}(X_s-X_{t_i})\,dX_s+2\int_{T_t}^t(X_s-X_{\textstyle T_\color{red}{t}})\,dX_s $$ can be written as $$\tag{2} M_t=2\int_{0}^{t}(X_s-X_{\textstyle T_\color{red}{s}})\,dX_s\,. $$ This is possible because
for $t_i<s\le t_{i+1}$ we have $t_i=T_s$ and
for $T_t<s\le t$ we have $T_s=T_t\,.$
Then we get as usual $$\tag{3} [M,M]_t=4\int_0^t(X_s-X_{\textstyle T_\color{red}{s}})^2\,d[X,X]_s\,. $$ If we want we could now use the bullet points to switch back to the clumsier formula $$\tag{4} [M,M]_t=4\sum_{t_{i+1}\le t}\int_{t_i}^{t_{i+1}}(X_s-X_{t_i})^2\,d[X,X]_s +4\int_{T_t}^t(X_s-X_{\textstyle T_\color{red}{t}})^2\,d[X,X]_s\,. $$