Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $\mathcal G = (\mathcal G_t, t \ge 0)$ a completed filtration. Let $M$ be a real-valued continuous local martingale w.r.t. $\mathcal G$. I'm reading below lemma from Le Gall's Brownian Motion, Martingales, and Stochastic Calculus, i.e.,
Lemma 5.14 We have a.s. for every $0 \leq a<b$, $$ M_t=M_a, \forall t \in[a, b] \iff\langle M, M\rangle_b=\langle M, M\rangle_a . $$
Proof of Lemma 5.14 Thanks to the continuity of sample paths of $M$ and $\langle M, M\rangle$, it is enough to verify that for any fixed $a$ and $b$ such that $0 \leq a<b$, we have $$ \left\{M_t=M_a, \forall t \in[a, b]\right\} =: B = C := \left\{\langle M, M\rangle_b=\langle M, M\rangle_a\right\}, \quad \text { a.s. } $$
The fact that the event in the left-hand side is (a.s.) contained in the event in the right-hand side is easy from the approximations of $\langle M, M\rangle$ in Theorem 4.9.
Let us prove the converse. Consider the continuous local martingale $N_t=M_t-$ $M_{t \wedge a}$ and note that $$ \langle N, N\rangle_t=\langle M, M\rangle_t-\langle M, M\rangle_{t \wedge a} . $$
For every $\varepsilon>0$, introduce the stopping time $$ T_{\varepsilon}=\inf \left\{t \geq 0:\langle N, N\rangle_t \geq \varepsilon\right\} . $$
Then $N^{T_{\varepsilon}}$ is a martingale in $\mathbb{H}^2$ (since $\left\langle N^{T_{\varepsilon}}, N^{T_{\varepsilon}}\right\rangle_{\infty} \leq \varepsilon$ ). Fix $t \in[a, b]$. We have $$ E\left[N_{t \wedge T_{\varepsilon}}^2\right]=E\left[\langle N, N\rangle_{t \wedge T_{\varepsilon}}\right] \leq \varepsilon . $$ Hence, considering the event $A:=\left\{\langle M, M\rangle_b=\langle M, M\rangle_a\right\} \subset\left\{T_{\varepsilon} \geq b\right\}$, $$ E\left[\mathbf{1}_A N_t^2\right]=E\left[\mathbf{1}_A N_{t \wedge T_{\varepsilon}}^2\right] \leq E\left[N_{t \wedge T_{\varepsilon}}^2\right] \leq \varepsilon . $$ By letting $\varepsilon$ go to 0 , we get $E\left[1_A N_t^2\right]=0$ and thus $N_t=0$ a.s. on $A$, which completes the proof.
I could not understand the reasoning behind
The fact that the event in the left-hand side is (a.s.) contained in the event in the right-hand side is easy from the approximations of $\langle M, M\rangle$ in Theorem 4.9.
The quoted result is
Theorem 4.9 Let $M=\left(M_t\right)_{t \geq 0}$ be a continuous local martingale. There exists an increasing process denoted by $\left(\langle M, M\rangle_t\right)_{t \geq 0}$, which is unique up to indistinguishability, such that $M_t^2-\langle M, M\rangle_t$ is a continuous local martingale. Furthermore, for every fixed $t>0$, if $0=t_0^n<t_1^n<\cdots<t_{p_n}^n=t$ is an increasing sequence of subdivisions of $[0, t]$ with mesh tending to 0 , we have $$ \langle M, M\rangle_t=\lim _{n \rightarrow \infty} \sum_{i=1}^{p_n}\left(M_{t_i^n}-M_{t_{i-1}^n}\right)^2 $$ in probability. The process $\langle M, M\rangle$ is called the quadratic variation of $M$.
My attempt Let $0=t_0^n<\cdots<t_{p_n}^n=b$ be a sequence of sub-divisions of $[0, b]$ whose mesh tends to $0$ as $n \to \infty$. Let $$ X^{(n)}_t := \sum_{i=1}^{p_n} (M_{ t \wedge t_i^n} - M_{t \wedge t_{i-1}^n})^2 \quad \forall t \in [0, b] \quad \forall n \in \mathbb N. $$
For each $t \in [0, b]$, we have $X^{(n)}_t \to \langle M \rangle_t$ in probability as $n \to \infty$. We have $B \subset \{X^{(n)}_a = X^{(n)}_b\}$ for all $n$.
Could you explain how to get $B \subset C$ a.s.?