For $n \in \mathbb{N}$, let $(X^n_t)_{t \geq 0}$ be symmetric, nearest neighbour continuous time random walks taking values in $n^{-1/2} \mathbb{Z}$ which jump at rate $n$.
It is standard to prove that, as $n \to \infty$, $X^n \Rightarrow W$ where $W$ is a standard Brownian motion and $\Rightarrow$ denotes weak convergence.
Is it also the case that we also have convergence of the corresponding local times? (for simplicity taken at $0$, say) That is do we have $$\int_0^t \mathbb{1}_{0}(X^n_s) ds \Rightarrow L^0(t)$$ where $L^0$ is the local time at $0$ of $W$. If so is there a simple proof of this fact?
The answer is Yes; perhaps the first result in this direction is due to F.B. Knight [Random walks and a sojourn density process of Brownian motion, Trans. Amer. Math. Soc. 109 (1963) 56–86]. (N.B. "sojourn density" = "local time".) A more modern approach would be to use the Tanaka formula $$ |W_t|=\int_0^t \text{sign}(W_s)\,dW_s+L^0(t), $$ in combination with the analogous formula for the random walk and the appropriate martingale central limit theorem (applied to the stochastic integral term) (or invariance principle for weak convergence to the entire process $(L^0(t))_{0\le t\le T}$).