# Prove that a right-continuous stochastic process is product measurable

Let

• $X=(X_t,t\ge 0$ be a real-valued stochastic process on a measurable space $(\Omega,\mathcal{A})$ with almost surely right-continuous paths
• $\mathbb{F}:=(\mathcal{F}_t,t\ge 0)$ be a filtraiton on $(\Omega,\mathcal{A})$ and $X$ be adapted to $\mathbb{F}$

I want to show, that for all $t\ge 0$ $$\Omega\times [0,t]\to\mathbb{R}\;,\;\;\;(\omega,s)\mapsto X_s(\omega)\tag{1}$$ is $\mathcal{F}_t\otimes\mathcal{B}\left([0,t]\right)$-measurable.

Let $$A_k^{(n)}:=\left[\frac t{2^n}(k-1),\frac t{2^n}k\right]\;\;\;\text{and}\;\;\; t_k^{(n)}:=\frac t{2^n}k\;\;\;\;\;\text{for }k=0,\ldots,2^n$$ Moreover, let $$X^{(n)}(\omega,s):=\sum_{k=0}^{2^n}1_{A_k^{(n)}}(s)X_{t_k^{(n)}}(\omega)\;\;\;\text{for }(\omega,s)\in\Omega\times [0,t]\tag{2}$$ Clearly, $X^{(n)}$ is $\mathcal{F}_t\otimes\mathcal{B}\left([0,t]\right)$-measurable and we've got $$X^{(n)}(\omega,s)=X_{t_k^{(n)}}(\omega)\Leftrightarrow s\in A_k^{(n)}\tag{3}$$ How do we need to use the continuity from the right to show $$\lim_{n\to\infty}X^{(n)}(\omega,s)=X_s(\omega)\;\;\;\text{almost surely}\;?\tag{4}$$

First a remark concerning the definition of $X^{(n)}$: Instead of the closed interval $A_k^{(n)} := [t_{k-1}^{(n)},t_k^{(n)}]$ you should use the half-open interval $A_k^{(n)} := (t_{k-1}^{(n)},t_k^{(n)}]$ (otherwise you might run into trouble at the boundary points $\{t_k^{(n)}\}$). Then

$$X^{(n)}(s,\omega) := X_0(\omega) 1_{\{0\}}(s) + \sum_{k=1}^{2^n} 1_{A_k^{(n)}}(s) X_{t_k^{(n)}}(\omega)$$

does the job.

Fix $s \in (0,t]$. For any $n \geq 1$, there exists a unique number $k_n \in \{1,\ldots,2^n\}$ such that

$$s \in \bigg( t_{k_n-1}^{(n)}, t_{k_n}^{(n)} \bigg].$$

By definition,

$$X^{(n)}(s,\omega) = X(t_{k_n}^{(n)},\omega).$$

Thus, since $t_{k_n}^{(n)} \to s$ as $n \to \infty$ and $t_{k_n}^{(n)} \geq s$ for all $n \in \mathbb{N}$, it follows from the right-continuity that

$$\lim_{n \to \infty} X^{(n)}(s,\omega) = \lim_{n \to \infty} X(t_{k_n}^{(n)},\omega) = X_s(\omega).$$

• Is $t_{k_n}^{(n)}\downarrow s$, especially $$t_{k_{n+1}}^{(n+1)}\le t_{k_n}^{(n)}\;,$$ really true? Commented May 2, 2015 at 21:55
• @0xbadf00d What I mean by $t_k^{(n)} \downarrow s$ is that $t_k^{(n)} \to s$ as $n \to \infty$ and $t_k^{(n)} \geq s$ (i.e. the sequence converges "from the right" to $s$).
– saz
Commented May 3, 2015 at 5:29
• There are some errors in your answer. The sum in the first equation should run from $1$ to $2^n$ and you don't use $k_n$ in your second equation. Commented May 4, 2015 at 10:37
• Thinking about that: Why can't we use $$t_k^{(n)}:=\frac knt$$ instead of $t_k^{(n)}:=\frac k{2^n}t$? Commented May 4, 2015 at 10:40
• @user32423 I'm sorry, but I can't follow your thoughts. In the second part of my answer, I proved that the approximation does the job... so you can either a) why you think that the proof is wrong or b) give me a specific counterexample where the approximations fails.
– saz
Commented May 20, 2018 at 16:08