A stochastic process is a collection $(X_t)_{t\in T}$ of random variables from a prob. space $(\Omega,\mathcal{F},P)$ to some measurable space $(E,\mathcal{E})$. Now, in order to understand the whole process as a 'single' random variable one look at it as a map $X:\Omega\to E^T$, where $E^T=\prod_{t\in T}E$ is the set of all functions from $T$ to $E$.

$E^T$ is usually equipped with the product sigma algebra $\mathcal{E}^T$ which is produced by the one-dimensional cylinder sets, i.e. $\mathcal{E}^T=\sigma(\pi^{-^1}(A): A\in\mathcal{E}, t\in T)$ (is is the same as the sigma algebra produced by all finite-dimensional cylinders?), where $\pi_t$ is the projection map from $E^T$ to $E$. Now, if we are interested just in a subspace $S\subset E^T$, then it is often the case that $S\notin\mathcal{E}^T$.

For example take $T=[0,1]$, $E=\mathbb{R}$, $\mathcal{E}=\mathcal{B}_{\mathbb{R}}$ and $C[0,1]$ as a subspace of $\mathbb{R}^{[0,1]}$, then $C[0,1]\notin\mathcal{E}^T$.

By looking at $\mathcal{S}=S\cap \mathcal{E}^T=\{S\cap A:A\in\mathcal{E}^T\}$ one can avoid this problem. Now my main question: If our subspace $S$ is a metric space it is natural to look at the Borel $\sigma$-algebra $\mathcal{B}_S$. Is it always the case that $\mathcal{S}=\mathcal{B}_S$? I think that we need some additional property like separability, but I cannot prove it.

  • $\begingroup$ Take $S=E^T$, endowed with the discrete metric. $\endgroup$ – Michael Greinecker Apr 15 '18 at 9:20
  • $\begingroup$ And if we assume that $E$ is separable with Borel sigma algebra $\mathcal{B}_E$? $\endgroup$ – aaaaaaaa Apr 16 '18 at 13:04
  • $\begingroup$ The question is rather what $S$ should be. It works for $S=C[0,1]$, but there is no reason it should work for every metric space. $\endgroup$ – Michael Greinecker Apr 16 '18 at 14:57
  • $\begingroup$ OK, thank you. Does it also work for the space of cadlag functions $D[0,1]$ if it is equipped with a "right" metric? $\endgroup$ – aaaaaaaa Apr 16 '18 at 17:47
  • $\begingroup$ I'm not sure. The reason why it works for $C[0,1]$ is that every function in $C[0,1]$ is determined by countably many coordinates. I don't know if this argument transfers to cadlag functions. $\endgroup$ – Michael Greinecker Apr 16 '18 at 17:55

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