The book "Discretization of Processes" starts with a simple example. Suppose that $X=\sigma\,W$ where $W$ is a Brownian motion and $\sigma>0$. Suppose that $X$ is sampled on the equispaced partition $t_{j,n}=j\,\Delta_n$ of the interval $[0,t]$, with $j=1,\dots,\lfloor t/\Delta_n\rfloor$. Define $\Delta_j^nX=X_{j\,\Delta_n}-X_{(j-1)\,\Delta_n}$. The problem is to derive a law of large numbers (LLNs) for the quantity $$ V^{\prime n}(f,X)_n=\Delta_n\,\sum_{j=1}^{\lfloor t/\Delta_n\rfloor}f\left(\Delta_j^nX/\sqrt{\Delta_n}\right),\quad (1) $$ where $f$ is a regular enough function. The authors say that it is enough to consider a sequence of iid random variables $\xi_j$ that are distributed as $f(X_1)$. In fact, the LLNs implies $$ Z_n\doteq\frac{t}{\lfloor t/\Delta_n\rfloor}\sum_{j=1}^{\lfloor t/\Delta_n\rfloor}\xi_j\stackrel{a.s.}{\rightarrow}t\,\mathsf{E}\{\xi_1\} =t\,\int_{\mathbb{R}}f(x)\exp(-x^2/(2\,\sigma^2))\,(2\,\pi\,\sigma^2)^{-1/2}\,dx\quad(2). $$ My problem is the following. They say that $V^{\prime n}(f,X)_n$ in $(1)$ is distributed as $Z_n$ in $(2)$, but I do not see how this is possible (from this observation, is not difficult to find a LNNs for $V^{\prime n}(f,X)_n$) .
Some attempts
They claim that $$ \Delta_n\,\sum_{j=1}^{\lfloor t/\Delta_n\rfloor}f\left(\Delta_j^nX/\sqrt{\Delta_n}\right) \stackrel{d}{=}\frac{t}{\lfloor t/\Delta_n\rfloor}\sum_{j=1}^{\lfloor t/\Delta_n\rfloor}f(X_1) $$ Since the function floor is involved, I need help understanding how to proceed.
Proof
I think I have figured out how to prove it. I use the same notation of the book, so $\rho_c(f)=\int f(x)\,\rho_c(dx)$ where $\rho_c$ is the centered Gaussian law $\mathcal{N}(0,c)$.
