The law of a continuous stochastic process and its canonical realization

Question

Let

• $T>0$
• $d\in\mathbb N$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $X:\Omega\times[0,T]\to\mathbb R^d$ be a continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$ and $$\tilde X:\Omega\to C^0([0,T],\mathbb R^d)\;,\;\;\;\omega\mapsto X(\omega,\;\cdot\;)$$

We can Show that $\tilde X$ is a random variable on $(\Omega,\mathcal A,\operatorname P)$, i.e. $\tilde X$ is measurable with respect to $\mathcal A$-$\mathcal B(C^0([0,T],\mathbb R^d))$. Moreover, the distribution $\operatorname P_{\tilde X}$ of $\tilde X$ is called distribution of $X$.

Now, let $$Y:C^0([0,T],\mathbb R^d)\times[0,T]\to\mathbb R^d\;,\;\;\;(\omega,t)\mapsto\omega(t)\;.\tag 1$$

If we could show that $Y$ is a stochastic process on $\left(C^0([0,T],\mathbb R^d),\mathcal B(C^0([0,T],\mathbb R^d)),\operatorname P_{\tilde X}\right)$, then it would make sense to call $Y$ canonical realization of $X$.

All we need to show is that $Y_t$ is measurable with respect to $\mathcal B(C^0([0,T],\mathbb R^d))$-$\mathcal B(\mathbb R^d)$, for all $t\in[0,T]$, isn't it?

However, in a book that I'm reading, the author writes that $Y$ is a stochastic process on $\left(C^0([0,T],\mathbb R^d),\mathcal B(C^0([0,T],\mathbb R^d)),\operatorname P_{\tilde X}\right)$, cause $(1)$ is (jointly) continuous and hence $Y$ is measurable with respect to $\mathcal B(C^0([0,T],\mathbb R^d))\otimes\mathcal B([0,T])$-$\mathcal B(\mathbb R^d)$.

The former measurability is often called product measurability. I don't see why we need it here. A stochastic process doesn't need to be product measurable. Moreover, I don't think that the stated argumentation is correct. As far as I know, the only sufficient condition of this kind is that if $Y$ is a stochastic process (i.e. each $Y_t$ is measurable) with one-side continuous paths, then $Y$ is product measurable.

Is there anything I'm missing?

Answer 1

You don't need it, but it does imply that $Y$ is a stochastic process. As you note the map $Y$ is continuous (from one metric space to another) and so Borel measurable. $Y$ is therefore product measurable (this is the bonus provided by this kind of argument; notice that the Borel $\sigma$-algebra on $C^0([0,T])\times[0,T]$ coincides with the product $\sigma$-algebra because $C^0([0,T])$ and $[0,T]$ are separable metric spaces); by Fubini, the "section" $\omega\mapsto Y_t(\omega)$ is measurable for each $t\in[0,1]$. This latter measurability just says that each $Y_t$ is a random variable.