Let $\xi_1, \xi_2, \dots$ be an i.i.d. sequence of zero-mean, unit-variance random variables defined on some probability space $(\Omega, \mathcal{A}, \mathbb{P})$. Consider the random walk $(S_n)_{n\geq 0}$, where
[math]S_n = \begin{cases} \sum_{i=1}^{n} \xi_i, & \text{if } n \geq 1, \\ 0, & \text{if } n = 0. \end{cases}[/math]For $\omega \in \Omega$, define the continuous function $X_n(\cdot, \omega) : [0,1] \to \mathbb{R}$ as follows: Set
[math]X_n(j/n, \omega) := \frac{S_j(\omega)}{\sqrt{n}}, \quad 0 \leq j \leq n,[/math]and define $X_n(t, \omega)$ for $t \in (j/n, (j+1)/n)$, $0 \leq j < n$, by linear interpolation. It follows that
[math]X_n(t, \omega) = \frac{S_{\lfloor nt \rfloor}(\omega)}{\sqrt{n}} + (nt - \lfloor nt \rfloor) \frac{\xi_{\lfloor nt \rfloor + 1}(\omega)}{\sqrt{n}}, \quad t \in [0,1].[/math]Prove that the map
[math]X_n : (\Omega, \mathcal{A}, \mathbb{P}) \to (C[0,1], \mathcal{B}_{C[0,1]}),[/math]taking $\omega$ to $X_n(\cdot, \omega)$, is measurable.
To show that $X_n$ is measurable as a map into $(C[0,1], \mathcal{B}_{C[0,1]})$, we use the fact that the Borel $\sigma$-algebra on $C[0,1]$ is generated by finite-dimensional evaluations of functions.
For any fixed $t \in [0,1]$, we express $X_n(t, \omega)$ explicitly as
[math]X_n(t, \omega) = \frac{S_{\lfloor nt \rfloor}(\omega)}{\sqrt{n}} + (nt - \lfloor nt \rfloor) \frac{\xi_{\lfloor nt \rfloor + 1}(\omega)}{\sqrt{n}}.[/math]Since $S_k(\omega) = \sum_{i=1}^{k} \xi_i(\omega)$ is a sum of measurable random variables and $\xi_{\lfloor nt \rfloor + 1}(\omega)$ is also measurable, it follows that $X_n(t, \omega)$ is a measurable function of $\omega$ for each fixed $t$.
For any finite set of time points $t_1, \dots, t_k \in [0,1]$, consider the map
[math]\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega)).[/math]Since each $X_n(t_i, \omega)$ is measurable in $\omega$, the map $\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega))$ is measurable as a function from $(\Omega, \mathcal{A})$ to $(\mathbb{R}^k, \mathcal{B}(\mathbb{R}^k))$.
The Borel $\sigma$-algebra $\mathcal{B}_{C[0,1]}$ on $C[0,1]$ is generated by sets of the form
[math]\{ f \in C[0,1] : (f(t_1), \dots, f(t_k)) \in U \},[/math]where $t_1, \dots, t_k \in [0,1]$ and $U \subset \mathbb{R}^k$ is an open set. Since we have shown that the finite-dimensional evaluations $\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega))$ are measurable, it follows that $\omega \mapsto X_n(\cdot, \omega)$ is measurable as a map into $C[0,1]$.
Thus, we conclude that $X_n$ is a random element of $C[0,1]$, i.e.,
[math]X_n : \Omega \to C[0,1][/math]is measurable.
Can someone check this proof?
[math]S_n = \begin{cases} \sum_{i=1}^{n} \xi_i, & \text{if } n \geq 1, \\ 0, & \text{if } n = 0. \end{cases}[/math]For $\omega \in \Omega$, define the continuous function $X_n(\cdot, \omega) : [0,1] \to \mathbb{R}$ as follows: Set
[math]X_n(j/n, \omega) := \frac{S_j(\omega)}{\sqrt{n}}, \quad 0 \leq j \leq n,[/math]and define $X_n(t, \omega)$ for $t \in (j/n, (j+1)/n)$, $0 \leq j < n$, by linear interpolation. It follows that
[math]X_n(t, \omega) = \frac{S_{\lfloor nt \rfloor}(\omega)}{\sqrt{n}} + (nt - \lfloor nt \rfloor) \frac{\xi_{\lfloor nt \rfloor + 1}(\omega)}{\sqrt{n}}, \quad t \in [0,1].[/math]Prove that the map
[math]X_n : (\Omega, \mathcal{A}, \mathbb{P}) \to (C[0,1], \mathcal{B}_{C[0,1]}),[/math]taking $\omega$ to $X_n(\cdot, \omega)$, is measurable.
To show that $X_n$ is measurable as a map into $(C[0,1], \mathcal{B}_{C[0,1]})$, we use the fact that the Borel $\sigma$-algebra on $C[0,1]$ is generated by finite-dimensional evaluations of functions.
For any fixed $t \in [0,1]$, we express $X_n(t, \omega)$ explicitly as
[math]X_n(t, \omega) = \frac{S_{\lfloor nt \rfloor}(\omega)}{\sqrt{n}} + (nt - \lfloor nt \rfloor) \frac{\xi_{\lfloor nt \rfloor + 1}(\omega)}{\sqrt{n}}.[/math]Since $S_k(\omega) = \sum_{i=1}^{k} \xi_i(\omega)$ is a sum of measurable random variables and $\xi_{\lfloor nt \rfloor + 1}(\omega)$ is also measurable, it follows that $X_n(t, \omega)$ is a measurable function of $\omega$ for each fixed $t$.
For any finite set of time points $t_1, \dots, t_k \in [0,1]$, consider the map
[math]\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega)).[/math]Since each $X_n(t_i, \omega)$ is measurable in $\omega$, the map $\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega))$ is measurable as a function from $(\Omega, \mathcal{A})$ to $(\mathbb{R}^k, \mathcal{B}(\mathbb{R}^k))$.
The Borel $\sigma$-algebra $\mathcal{B}_{C[0,1]}$ on $C[0,1]$ is generated by sets of the form
[math]\{ f \in C[0,1] : (f(t_1), \dots, f(t_k)) \in U \},[/math]where $t_1, \dots, t_k \in [0,1]$ and $U \subset \mathbb{R}^k$ is an open set. Since we have shown that the finite-dimensional evaluations $\omega \mapsto (X_n(t_1, \omega), \dots, X_n(t_k, \omega))$ are measurable, it follows that $\omega \mapsto X_n(\cdot, \omega)$ is measurable as a map into $C[0,1]$.
Thus, we conclude that $X_n$ is a random element of $C[0,1]$, i.e.,
[math]X_n : \Omega \to C[0,1][/math]is measurable.
Can someone check this proof?
