# How to prove that such measure does not exist on an infinite product space?

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.

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 $$A\times\mathbb{R}\times\mathbb{R}\times\cdots = \{(t_1,t_2,\ldots):(t_1,\ldots,t_n)\in A\} \in \mathcal{B}(\mathbb{R}^\infty).$$

• 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 $\mathcal{B}^{[0,\infty)}$. – boomkin May 2 at 9:51
• 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. – Jason Swanson May 2 at 11:43
• It is used in your proof stating $A \in \mathcal{B}(\mathbb{R}^{\infty})$, that is the only reason I'm worried about it. – boomkin May 2 at 12:32
• 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. – Jason Swanson May 2 at 12:49