I'm reading this question by @KennethNg, i.e.,
I am reading the Dubins-Schwarz theoem on Brownian motion and stochastic calculus. It says, given a continuous local martingale $M$ such that $\lim_{n\to\infty}[M]_t = \infty$ a.s., where $[M]_t$ denotes the quadratic variation of $M$. For each $s \geq 0$, the book defines a stopping time $$T(s)=\inf\{t \geq 0: [M]_t > s\}$$ and says the stopped process $M_{t\wedge T(s)}$ is a martingale.
I would like to ask how can we see the stopped process is indeed a martingale?
The accepted answer by @saz is
Let $(\sigma_n)_{n \in \mathbb{N}}$ be a localizing sequence of the local martingale $M$, i.e. an increasing sequence of stopping times such that $\sigma_n \to \infty$ and $(M_{t \wedge \sigma_n})_{t \geq 0}$ is a martingale for each $n \in \mathbb{N}$.
Applying the optional stopping theorem, we find that $(M_{t \wedge \sigma_n \wedge T(s)})_{t \geq 0}$ is a martingale, i.e.
$$\mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = M_{u \wedge \sigma_n \wedge T(s)} \tag{1}$$
for all $u \leq t$. Since $M$ has continuous sample paths, the right-hand side converges to $M_{u \wedge T(s)}$ as $n \to \infty$. If we can show that
$$\lim_{n \to \infty} \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = \mathbb{E}(M_{t \wedge T(s)} \mid \mathcal{F}_u)\tag{2}$$
this proves that $(M_{t \wedge T(s)})_{t \geq 0}$ is a martingale. To prove $(2)$, we note that by Doob's maximal inequality
$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge \sigma_n \wedge T(s)}|^2 \right) &\leq 4 \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)}^2) = 4 \mathbb{E} \big( [M]_{t \wedge \sigma_n \wedge T(s)} \big) \leq \mathbb{E}([M]_{T(s)}) \leq 4s. \end{align*}$$
By the monotone convergence theorem this implies
$$\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right) \leq 4s < \infty.$$
Hence,
$$\mathbb{E} \left( \sup_{n \in \mathbb{N}} |M_{t \wedge \sigma_n \wedge T(s)}| \right) \leq \sqrt{\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right)} < \infty.$$
Now $(2)$ follows from the dominated convergence theorem.
My question
@jpv asked in a comment below @saz's answer that
In (1), shouldn't the conditioning be on $\mathcal{F}_{u \wedge T(s)}$ ?
@saz answered
Note that there are different version of the OST. The version I'm using is the following: Let $(M_t, \mathcal{F}_t)_{t \geq 0}$ be a martingale with continuous sample paths and $\sigma$ an $\mathcal{F}_t$-stopping time, then $\left(M_{t \wedge \sigma}, \mathcal{F}_t\right)_{t \geq 0}$ is a martingale.
So @saz applied the following OST version, i.e.,
OST theorem Let $(M_t, \mathcal{F}_t)_{t \geq 0}$ be a martingale with continuous sample paths and $\sigma$ an $\mathcal{F}_t$-stopping time, then $(M_{t \wedge \sigma}, \mathcal{F}_t)_{t \geq 0}$ is a martingale.
Now come back to @KennethNg's question. We have $T(s)$ is a stopping time and $M$ a martingale, so by above OST theorem we get $(M_{t\wedge T(s)})_{t\ge 0}$ is a martingale.
Could you confirm if my above alternative reasoning is fine?