# If $(X_t,t\in I)$ is a process with values in $(E,\mathcal{E})$, are $\sigma(X_t,t\in I)$ and $\sigma(X)=X^{-1}(\mathcal{E}^{\otimes I})$ equal?

Let

• $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
• $E$ be a Polish space and $\mathcal{E}$ be the Borel $\sigma$-algebra on $E$
• $I$ be an index set
• $X_t$ be a random variable on $(\Omega,\mathcal{A},\operatorname{P})$ with values in $(E,\mathcal{E})$

$X:=(X_t,t\in I)$ is called a stochastic process. We may consider $X$ as a mapping $$\Omega\times I\to E\;,\;\;\;(\omega,t)\mapsto X_t(\omega)$$ or even as a mapping $$\Omega\to E^I\;,\;\;\;\omega\mapsto X(\omega,\;\cdot\;)\;\tag{1}$$ From $(1)$, the question arises if $X$ can be considered as a random variable on $(\Omega,\mathcal{A},\operatorname{P})$ with values in $(E^I,\mathcal{E}^{\otimes I})$. So, does the $\mathcal{A}$-$\mathcal{E}$-measurability of each $X_t$ imply the $\mathcal{A}$-$\mathcal{E}^{\otimes I}$-measurability of $X$?

Many peopole write $\sigma(X)$ for the $\sigma$-algebra $$\sigma(X_t,t\in I):=\sigma\left(\bigcup_{t\in I}\sigma(X_t)\right)=\sigma\left(\bigcup_{t\in I}X_t^{-1}(\mathcal{E})\right)\tag{2}$$ generated by the family $(X_t,t\in I)$. However, if we consider $X$ in the sense of $(1)$, we've got $$\sigma(X)=X^{-1}(\mathcal{E}^{\otimes I})\tag{3}$$ Can we prove that $(2)$ equals $(3)$?

• @oxbadfood Does ${\cal E}^{\otimes I}$ refer to the product $\sigma$ algebra?? – user940 Jul 1 '15 at 17:32

The result is true, and has nothing to do with topology or countability. Let ${\cal C}$ be the collection of subsets of $E^I$ of the form $\pi_i^{-1}(E)$ for some $i\in I$ and $E\in{\cal E}$. Then use the fact that $X^{-1}(\sigma({\cal C}))=\sigma(X^{-1}({\cal C}))$.
• So $\mathcal{E}^{\otimes I}$ does not denote the Borel $\sigma$-algebra? That was my viewpoint. – drhab Jul 1 '15 at 17:28
• You are right, Byron. $\mathcal{E}^{\otimes I}$ is the product $\sigma$-algebra. – 0xbadf00d Jul 1 '15 at 19:37