Let $P_{t}$ be a price process of an asset. Assume the log-price $p_{t}=\ln P_{t}$ follows a generalized Ito process $$ d p_{t}=\mu_{t} d t+\sigma_{t} d w_{t}, \quad t \in R, \quad (1) $$ where $\mu_{t}$ is a continuous and locally bounded function, $\sigma_{t}$ is a càdlàg function, and $w_{t}$ is the standard Brownian motion. The integrated variance and the integrated volatility of the asset for day $t$ are $$ \int_{t-1}^{t} \sigma_{s}^{2} d s, \sqrt{\int_{t-1}^{t} \sigma_{s}^{2} d s}, t=1,2, \ldots $$ A simple version of realized variance, $R V$, is the sum of squares of intra-day log-returns. For a day $t$, assume a set of equally spaced intra-day price observations $P_{t-1+h j}, j=0,1, \ldots, N, h=1 / N$ is available. Then the $h$-spaced realized variance for day $t$ is given by $$ R V_{t}^{(h)}=\sum_{j=1}^{N} r_{t j}^{(h) 2}, t=1,2, \ldots, h=1 / N $$ where $$ r_{t j}^{(h)}=p_{t-1+h j}-p_{t-1+h(j-1)} $$ is the log-return for the $h$-spaced interval $[t-1+h(j-1), t-1+h j]$. The notation $R V_{t}$ indicates both realized variance and realized volatility and $I V_{t}$ indicates both integrated variance and integrated volatility.
I read here and here that, since $r_{t j}^{(h)}, j=1,2, \ldots, N$ in Eq. $(1)$ are uncorrelated, $R V_{t}^{(h)}$ converges in probability to $I V_{t}$ as $h \rightarrow 0$. How could I prove this? I apologize if this question may seem trivial but I am not very familiar with convergence in probability.
The question is a bit old but I find the topic interesting and so I give it a try...
Let $\Pi_N := \max_{j = 1, \cdots, N-1}(t_{j+1} - t_j)$ be a partition of $[0,T]$. The quadratic variation of the return process is
$$ QV_t = [r]_t := L^2 -\lim_{||\Pi|| \to 0} \sum_{j = 0}^{N-1}(p(t_{j+1}) - p(t_j))^2, \quad N \to \infty$$
If $p$ is an Ito process (i.e. it is a diffusion process) then $QV_t = IV_t$. Thus, if you take the $L^2$-limit of $RV_t$ you get exactly the definition of $QV_t$. The consistency of the estimator (i.e. the convergence in probabbility) follows from the fact that convergence in $L^2$ always implies convergence in probability.