I am reading Stochastic Processes by Richard F. Bass on my own. At one place the author says the following:
Let $X$ and $Y$ be two stochastic processes with continuous paths. Define new random variables $\bar{X}, \bar{Y}:\Omega\to \mathcal{C}[0,\infty)$ defined by $\omega\mapsto (t\to X(t,\omega))$ (and similarly for $Y$). Saying that $X$ and $Y$ has same law means $\mathbb{P}(\bar{X}\in A)=\mathbb{P}(\bar{Y}\in A)$ for every Borel subset of $C[0,\infty).$
I have the following question. We know two things about $X$ (and $Y.$)
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$X_t$ is measurable for each $t\ge 0.$ (This is part of the definition of a stochastic process.)
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For each fixed $\omega$, the map $t\to X(t,\omega)$ is continuous.
Is it enough to guarantee that $\bar{X}:\Omega\to \mathcal{C}[0,\infty)$ is measurable? Or, when Bass writes the above definition he is assuming that the $\bar{X}$ is measurable? To elaborate, will $\bar{X}$ always be measurable or the correct way to read above definition is that
Suppose $X, Y$ are stochastic processes with continuous path such that $\bar{X}:\Omega\to \mathcal{C}[0,\infty)$ defined by $\omega \mapsto (t\to X(t,\omega))$ is measurable. Then, we say that $X$ and $Y$ have the same law if $\mathbb{P}(\bar{X}\in A)=\mathbb{P}(\bar{Y}\in A)$ for all $A.$
Best Answer
The measurability is automatically satisfied.
It can be shown that the Borel $\sigma$-algebra on $C[0,\infty)$ is the smallest $\sigma$-algebra such that all projections $$\pi_t: C[0,\infty) \to \mathbb{R}, f \mapsto f(t)$$ are measurable. This implies that a mapping $$Y: \Omega \to C[0,\infty)$$ is measurable if, and only if, $$\omega \mapsto \pi_t(Y)(\omega) = (Y(\omega))(t)$$ is measurable for each $t \geq 0.$
Now let $(X_t)_{t \geq 0}$ be a stochastic process with continuous sample paths, and set $$\tilde{X}: \Omega \to C[0,\infty), \omega \mapsto \tilde{X}(\omega) :=(t \mapsto X_t(\omega)).$$ Since $\pi_t(\tilde{X})= X_t$, it follows from the definition of the stochastic process that $\pi_t(\tilde{X})$ is measurable for each $t \geq 0$. Hence, by our previous consideration, $\tilde{X}$ is measurable.