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A previous (already submitted) exercise from my coursework was,

Prove that there does not exist a probability measure $ \mathbb{P} $ on the infinite product space $ (\mathbb{R}^{[0,\infty]}, > \mathcal{B}^{[0, \infty)}) $ with the property that a.s $ \xi \in \mathbb{R}^{[0, \infty]} $, the function $ t \rightarrow \xi_t $ is continous.

I could not come up with a good starting approach to solve this one, after revising and understanding what each of these terms mean. I was thinking about applying $ \pi - \lambda $ type metatheorem, because that can be used usually to prove that some property holds a.s. However, here the converse needs to be proved, which I find not straightforward.

I would really appreciate, some solution or hints to this question.

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Let $C=\{\omega\in\mathbb{R}^{[0,\infty)}:\omega\text{ is continuous}\}$. The only $\mathcal{B}^{[0,\infty)}$-measurable subset of $C$ is $\emptyset$. To get started proving this, first show that if $E\in\mathcal{B}^{[0,\infty)}$, then there exists a strictly increasing sequence $\{t_n\}_{n=1}^\infty$ in $[0,\infty)$ and a set $A\in\mathcal{B}(\mathbb{R}^\infty)$ such that $$ E = \{\omega: (\omega(t_1), \omega(t_2), \ldots) \in A\}. $$ (Show that the collection of such sets is a $\sigma$-algebra that contains the cylinder sets, and use the fact that the cylinder sets generate $\mathcal{B}^{[0,\infty)}$.)

With this fact in hand, now suppose there exists $P$ such that $t\mapsto\omega(t)$ is continuous a.s. This means there exists $N\in\mathcal{B}^{[0,\infty)}$ such that $P(N)=0$ and, for all $\omega\in N^c$, we have that $t\mapsto\omega(t)$ is continuous. This implies that $N^c\subset C$. By the above, $N^c=\emptyset$, so $P(N^c)=0$, which gives $P(N)=1$, a contradiction.

EDIT:

This answer assumes that $\mathcal{B}^{[0,\infty)}$ is meant to denote the product $\sigma$-algebra, $$ \mathcal{B}^{[0,\infty)} = \bigotimes_{t\in[0,\infty)} \mathcal{B}, $$ where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.

An alternative way to define a $\sigma$-algebra on $\mathbb{R}^{[0,\infty)}$ is to endow $\mathbb{R}^{[0,\infty)}$ with the product topology (or the topology of pointwise convergence) and then use the Borel $\sigma$-algebra corresponding to this topology (i.e. the smallest $\sigma$-algebra on $\mathbb{R}^{[0,\infty)}$ that contains the sets which are open in the product topology.) This latter approach produces a strictly larger $\sigma$-algebra. For example, this latter $\sigma$-algebra contains singletons, whereas $\mathcal{B}^{[0,\infty)}$ does not. See https://math.stackexchange.com/a/248587/11867, for example. As indicated in a now-deleted answer, if we take the latter approach, then the result is not true.

EDIT 2:

Here are some clarifications. A cylinder set is a set of the form $$ F = \{\omega\in\mathbb{R}: (\omega(t_1),\ldots,\omega(t_n)) \in B\}, $$ for some $n\in\mathbb{N}$ and some $B\in\mathcal{B}(\mathbb{R}^n)$. Also $$ \mathbb{R}^\infty = \{(t_1,t_2,\ldots):t_j\in\mathbb{R}\}, $$ and $$ \mathcal{B}(\mathbb{R}^\infty) = \bigotimes_{n\in\mathbb{N}}\mathcal{B}, $$ which is the $\sigma$-algebra generated by sets of the form $$ A_1\times A_2\times \cdots, $$ where $A_n\in\mathcal{B}$ for each $n$. If $A\in\mathcal{B}(\mathbb{R}^n)$, then $$ A\times\mathbb{R}\times\mathbb{R}\times\cdots = \{(t_1,t_2,\ldots):(t_1,\ldots,t_n)\in A\} \in \mathcal{B}(\mathbb{R}^\infty). $$

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  • $\begingroup$ Thank you for your answer. I'm confused when you define E, and say that is a $\sigma$-algebra, because to my understanding E are the cylinder sets, which collection can be used to generate $\mathcal{B}^{[0,\infty)}$, If E would be a $ \sigma$ algebra, then the $\sigma$-algebra generated would be itself again, which implies E is Borel. I'm also a bit confused about the difference in $ \mathcal{B}(R^{\infty})$ and $\mathcal{B}^{[0,\infty)}$. $\endgroup$ – boomkin May 2 at 9:51
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    $\begingroup$ A cylinder set uses only finitely many $t$ values, whereas $E$ uses a countably infinite number of $t$ values. Regarding the difference between those two $\sigma$-algebras, it is too complicated to explain further in a comment, but it is not relevant to your question. I only included it because of another answer and comment which is now deleted. If you want to know more, you should probably post a separate question about it. $\endgroup$ – Jason Swanson May 2 at 11:43
  • $\begingroup$ It is used in your proof stating $ A \in \mathcal{B}(\mathbb{R}^{\infty}) $, that is the only reason I'm worried about it. $\endgroup$ – boomkin May 2 at 12:32
  • $\begingroup$ Sorry, I was reading your comment on my phone, so had to parse the tex mentally and misread it. I have added an edit in order to clarify this for you. $\endgroup$ – Jason Swanson May 2 at 12:49

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