# Two stochastic processes with the same distribution inducing different measures

I am currently reading Strook's $\textit{Probability Theory: An Analytic View}$, and I am confused by the following statement on page 156:

"I take for $D(\mathbb{R}^N)$ the measurable structure given by the $\sigma$-algebra $\mathcal{F}:=\sigma(\psi(t): t\in[0,\infty))$ generated by the maps $\psi\mapsto\psi(t)$ as $t$ runs over $[0,\infty)$. The reason for insisting on this choice is that I want two $D(\mathbb{R}^N)$-valued stochastic processes $X_t$ and $Y_t$ to induce the same measure on $D(\mathbb{R}^N)$ if they same distribution. Seeing as $\mathcal{F}\subset\mathcal{B}(D(\mathbb{R}^N)$, this would not be true if I were to choose the Borel structure."

Any help clarifying this is greatly appreciated.

Suppose that $X$ and $Y$ are equal in distribution in some space $(\Omega,\mathcal{F},\mathbb{P}$). Stroock would like to choose $\mathcal{F}$ such that $\mathbb{P} (X \in A) = \mathbb{P} (Y \in A)$ for any $A \in \mathcal{F}$. This does not hold when you equip the set $\Omega$ with its natural Borel structure $\mathcal{F} := \sigma$ say. Instead, he defines $\mathcal{F} := \mathfrak{B}(D(\mathbf{R}^N))$ to allow this.
There are many problems with this choice of $\mathcal{F}$. In the same page he discusses how $\mathcal{F}$ measurability no longer follows from continuity. Determining which (infinite dimensional) stochastic processes are now measurable is.. considerably more difficult.
On page 160 he asks us to show that $\mathfrak{B}(D(\mathbf{R}^N))$ is not $(\mathbb{R}^N)^{[0,\infty[}$ measurable nor is a special subset of it (he calls it $C$). The set $\mathfrak{B}(D(\mathbf{R}^N))$ is rather bizarre.
• Thanks for your answer Arbias. But why specifically does there exist an $A\in\mathcal{F}$ such that $\mathbb{P}(X\in A)\neq\mathbb{P}(Y\in A)$ even when $X\overset{\mathcal{D}}{=}Y$ when $\mathcal{F}$ is the Borel $\sigma$-algebra? – Mike Oct 6 '14 at 13:02