Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$.
We know that if $(M_n)_{n\in\mathbb N_0}$ is a square-integrable $(\mathcal F_n)_{n\in\mathbb N_0}$-martingale on $(\Omega,\mathcal A,\operatorname P)$, $$Y_n:=\left.\begin{cases}M_0&\text{, if }n=0;\\M_n-M_{n-1}&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }n\in\mathbb N$$ is stationary and $$\left\|\frac1n\sum_{i=1}^nY_i^2-\sigma^2\right\|_{L^1(\operatorname P)}\xrightarrow{n\to\infty}0\tag1$$ for some $\sigma>0$, then $$\frac1{\sqrt n}M_n\xrightarrow{n\to\infty}\mathcal N_{0,\:\sigma^2}\tag2$$ in distribution.
Now assume $(M_t)_{t\ge0}$ is a square-integrable $(\mathcal F_n)_{n\in\mathbb N_0}$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $$M_{s+t}-M_s\sim M_t-M_0\tag3\;\;\;\text{for all }s,t\ge0$$ and $$\left\|\frac1t[M]_t-\sigma^2\right\|_{L^1(\operatorname P)}\xrightarrow{t\to\infty}0\tag1$$ for some $\sigma>0$. How can we show that again $$\frac1{\sqrt t}M_t\xrightarrow{t\to\infty}\mathcal N_{0,\:\sigma^2}\tag2$$ in distribution.
We clearly want to utilize the discrete-time result, but how can we do that? My idea is to define $$X^n_t:=\sum_{i=1}^n1_{(i-1,\:i]}(t)\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$ and consider the stochastic integral process $$(X^n\cdot M)_t=M_n-M_0\;\;\;\text{for all }t\ge n\in\mathbb N\tag4.$$ However, we need to clarify how the quadratic variation $[M]$ is defined (since $M$ is not assumed to be continuous) and utilize the Kunita-Watanabe identity $[X\cdot M,N]=X\cdot[M,N]$ to proceed ...