Something is weird from a proof that I am reading:
The well-known theorem of characterization of quadratic variation states that:
Suppose $X$ is a continuous local martingale and $A$ is a continuous adapted process of bounded variation. Then the process $X^2-A$ is a local martingale if and only if $A = \langle X \rangle + A_0$.
The ''if'' direction of the proof starts by claiming that
There exists a sequence of stopping times $(T_n)$ that reduces $X$ to a continuous martingale such that $\sup_{t \geq 0} \mathbb{E} ((X^{T_n}_t)^2) < +\infty$, so $(X^{T_n})^2 - \langle X^{T_n} \rangle$ is a martingale.
My concern is: Why is it true that $\sup_{t \geq 0} \mathbb{E} ((X^{T_n}_t)^2) < +\infty$ ?
There is a theorem that states that
If $X$ is a continuous local martingale such that $X_0$ is bounded, then the sequence of stopping times defined by $S_n = \inf \{ t \geq 0 : |X_t | >n \}$ reduces $X$ to a bounded martingale, for each $n$.
However, in this case, $X_0$ is not assumed to be bounded. Hence, how does the claim follow?
If $X_0$ is not bounded, then there does not necessarily exist a sequence of stopping times $(T_n)_n$ such that
$$\sup_{t \geq 0} \mathbb{E}((X_t^{T_n})^2) < \infty. \tag{1}$$
Indeed: For $X_0 \in L^1 \backslash L^2$ the process $X_t := X_0$ defines a continuous martingale, but obviosuly there does not a exist a sequence of stopping times $(T_n)_n$ such that $(1)$ holds.
However, in the original statement (i.e. the one about the uniqueness of the quadratic variation) we may assume without loss of generality that $X_0 =0$. Indeed: If $X_0 \neq 0$, then can consider the martingale
$$Y_t := X_t-X_0$$
instead. (Note that $\langle X \rangle = \langle Y \rangle$.)