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