Generated $\sigma$-algebras with cylinder set doesn't contain the space of continuous functions

Consider $\mathbb R^{[0,1]}$ the space of all functions from $[0,1]$ to $\mathbb R$ and the cylindrical sigma algebra $\mathcal B$ on it. The question is: how to prove that $C[0,1]\notin \mathcal B$.

Solution 1:

Hint For any set $A \in \mathcal{B}$ there exists a countable set $S \subseteq [0,1]$ such

$$\forall f \in \mathbb{R}^{[0,1]}, w \in A: w|_S = f|_S \Rightarrow f \in A \tag{1}$$

To prove this, define

$$\mathcal{A} := \{A \in \mathcal{B}; \exists S \subseteq [0,1] \, \text{countable}: (1) \, \text{holds}\}$$

Show that $\mathcal{A}$ is a $\sigma$-algebra and contains a generator of the cylindrical $\sigma$-algebra $\mathcal{B}$.

The claim follows by showing that $C[0,1]$ does not satisfy $(1)$.

(cf. René L. Schilling, Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 4.)

Solution 2:

Using saz's lemma (sounds great btw) and supposing that the set of continuous functions $\mathcal{C}$ is a measurable set of $\mathcal{B}$, let's us be given $S_C$ the countable set that characterizes it in the lemma.

Now let's pick up one $\omega\in \mathcal{C}$ and take $f=\omega+ 1_{u}$ where $u\not \in S_C$.

As $\omega|_{S_C}=f|_{S_C}$ we must have $f\in\mathcal{C}$ but by construction $f$ isn't continuous. This shows that there's no such countable set $S_C$ in $[0,1]$ and so $\mathcal{C}$ cannot be a measurable set of $\mathcal{B}$.

Sorry to saz that deserves all the credit.

Regards