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Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $X = (X_t, t\ge 0)$ be a continuous square-integrable martingale and $\langle X, X\rangle$ its quadratic variation. Let $Y = (Y_t, t\ge 0)$ be a process with continuous trajectories. Let $f:\mathbb R \to \mathbb R$ be continuous. I'm trying to prove below result used in the proof of Itô's lemma, i.e,

Theorem For each $t>0$, there is an increasing sequence $0=t_0^n<\cdots<t_{p_n}^n=t$ of subdivisions of $[0, t]$ whose mesh tends to $0$ and that $$ \sum_{i=0}^{p_n-1} f (Y_{t_i^n}) (X_{t_{i+1}^n}-X_{t_i^n})^2 \quad \underset{n \rightarrow \infty}{\longrightarrow} \quad \int_0^t f (Y_s) \mathrm{d}\langle X, X\rangle_s \quad \text{almost surely}. $$

Could you have a check if my below attempt is fine?


Proof Clearly, there is an increasing sequence $0=t_0^n<\cdots<t_{p_n}^n=t$ of subdivisions of $[0, t]$ whose mesh tends to $0$. We note that $$ \sum_{i=0}^{p_n-1} f (Y_{t_i^n}) (X_{t_{i+1}^n}-X_{t_i^n})^2 = \int_{[0, t]} f(Y_s) \mathrm d \mu_n (s), $$ where $\mu_n$ is the random (discrete) measure on $[0, t]$ defined by $$ \mu_n := \sum_{i=0}^{p_n-1} (X_{t_{i+1}^n}-X_{t_i^n})^2 \delta_{t_i^n}. $$

Let $D$ be the set that consists of all $t_i^n$ for $n \geq 1$ and $0 \leq i \leq p_n$. Then $D$ is dense in $[0, t]$.

Lemma For every sequence $0=t_0^n<\cdots<t_{p_n}^n=t$ of sub-divisions of $[0, t]$ whose mesh tends to $0$, we have $$ \sum_{i=0}^{p_n-1} (X_{t_{i+1}^n} - X_{t_i^n})^2 \underset{n \rightarrow \infty}{\longrightarrow} \langle X, X \rangle_t \quad \text{in probability}. $$

By above Lemma, we get for every $r \in D$, $$ \mu_n([0, r]) \quad \underset{n \rightarrow \infty}{\longrightarrow} \quad \langle X, X\rangle_r \quad \text{in probability}. $$

So there is a subsequence of values of $n$ such that, along this subsequence, we have for every $r \in D$, $$ \mu_n([0, r]) \quad \underset{n \rightarrow \infty}{\longrightarrow} \quad \langle X, X\rangle_r \quad \text{almost surely}. $$

This implies that the sequence $(\mu_n, n \in \mathbb N)$ of random measures converges almost surely to a random measure $\mu$ (whose random c.d.f. is $\langle X, X\rangle_r \mathbf{1}_{[0, t]}(r)$) in the sense that $$ \mu_n (\omega) \quad \underset{n \rightarrow \infty}{\longrightarrow}\quad\mu(\omega) \quad \text{in distribution} $$ for $\mathbb P$-a.e. $\omega \in \Omega$. On the other hand, convergence in distribution is equivalent to weak convergence. Hence $$ \int_0^t f(Y_s ) \mathrm d \mu_n (s) \quad \underset{n \rightarrow \infty}{\longrightarrow} \quad \int_0^t f(Y_s) \mathrm d \mu(s) = \int_0^t f(Y_s) \mathrm{d} \langle X, X\rangle_s \quad \text{almost surely}. $$

This completes the proof.

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