Ito Integral Continuous Semimartingale Left Endpoint Riemann Sum For Brownian motion, we can show that the left endpoint Riemann sum converges to the Ito integral:
$$\sum\limits_{i=0}^{n-1} W_{t_i}(W_{t_{i+1}} - W_{t_i}) \to \int_0^t W_s dW_s$$
However, I am wondering if this is true more generally for any $X$ that is continuous and square-integrable martingale. I first tried to show that
$$X_t^{(n)} := \sum\limits_{i=0}^{n-1} X_{t_i} \mathbb{1}_{(t_i, t_{i+1}]}(t)$$
on $[0,T]$ is Cauchy in $L^2$ under the norm $|| X ||_T := E\int_0^T  X_t^2 d\langle X\rangle_t$ for every $T$. Then since $X_t^{(n)} \to X_t$ a.s. as $n\to \infty$ by continuity, we have that $X_t^{(n)} \to X_t$ in $|| \cdot ||_T$ for every T. If we assume that $E\int_0^T X_s^2 d\langle X\rangle_s < \infty$ for every $T$, I believe the stochastic integral $\int_0^TX_s dX_s$ does exist. Then we should get that by definition $\int_0^T X_s^{(n)} dX_s \to \int_0^T X_sdX_s$ in $L^2(P)$. However, I am already stuck showing the first step simply because I cannot use Fubini like in Brownian case where $\langle X \rangle_s = s$ deterministically. Is there a counter-example to my statement or is there a more involved proof?
 A: I believe that, in general, $L^2$ convergence fails to hold. If we pose stronger assumptions on the integrability of $(X_t)_{t \geq 0}$, then $L^2$-convergence holds, but unfortunately I'm not aware of a result on the optimal integrability condition. In my answer I will show that the Riemann sums $$\sum_{i=0}^{n-1} X_{t_i} (X_{t_{i+1}}-X_{t_i})$$ converge in probability.
For $R>0$ define stopping times by
$$\tau_R := \inf\{t \geq 0; |X_t| \geq R\}.$$
Since $(X_t)_{t \geq 0}$ has continuous sample paths, we habe $\tau_R \uparrow \infty$ as $R \to \infty$. For fixed $\epsilon>0$ and $T>0$ we can choose $R>0$ such that $\mathbb{P}(\tau_R \leq T) \leq \epsilon$. If we denote by $M^{(n)}$ the stochastic integral of $X^{(n)}$ with respect to $X$ we find
$$\begin{align*} \mathbb{P} \left( \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) &= \mathbb{P} \left( \tau_R \leq T, \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) \\ &\quad + \mathbb{P} \left( \tau_R>T, \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) \\ &\leq \mathbb{P}(\tau_R \leq T) + \mathbb{P} \left( \sup_{t \leq T} |M^{(n)}_{t \wedge \tau_R}-M^{(m)}_{t \wedge \tau_R}| \geq \delta \right) \\ &=: I_1+I_2 \tag{1} \end{align*}$$
By our choice of $R$ we know that $I_1 \leq \epsilon$. For the second term we use Markov's inequality, the maximal inequality and Itô's isometry to conclude that
$$\begin{align*} I_2 &\leq  \frac{4}{\delta^2} \mathbb{E} \left( \int_0^T |X_{t \wedge \tau_R}^{(n)}-X^{(m)}_{t \wedge \tau_R}|^2 \, d\langle X \rangle_t \right) \\ &\leq \frac{4}{\delta^2} \mathbb{E} \left( \sup_{t \leq T} |X_{t \wedge \tau_R}^{(n)}-X_{t \wedge \tau_R}^{(m)}|^2 \langle X \rangle_T \right).  \tag{2} \end{align*}$$
Since the sample paths of $(X_t)_{t \geq 0}$ are uniformly continuous on compact sets, it is not difficult to see that
$$\sup_{t \leq T} |X_{t \wedge \tau_R}^{(n)}-X_{t \wedge \tau_R}^{(m)}|^2 \to 0$$
as $m,n \to \infty$. On the other hand, the expression is bounded by $4R^2$, and therefore we may apply the dominated convergence theorem in $(2)$ to conlucde that $I_2 \to 0$ as $m,n \to \infty$. Consequently we have shown that $M^{(n)}$ is a ("uniform") Cauchy sequence. On the other hand, $X^{(n)} \to X$, and therefore
$$\int_0^T X_s \, dX_s \tag{3}$$
exists and
$$\int_0^T X_s \, dX_s = \mathbb{P}-\lim_{n \to \infty} M^{(n)}_T.$$
Remark: The existence of the stochastic integral $(3)$ is to be understood in the sense of convergence in probability and not in $L^2$-convergence. In fact, defining stochastic integrals as a limit in probability theory is a standard way to extend stochastic integrals to a larger class of integrators and integrands.
