Question on Construction of Brownian motion with Kolmogorov-Centsov Theorem I am studying the construction of Brownian motion with Kolmogorovs Extension Theorem.
In the attached section from Karatzas and Shrive (Brownian Motion and Stochastic Calculus) it says that $C([0,\infty) \not\subset \mathscr{B}(\mathbb{R}^{[0, \infty)})$, doesn't it? But I thought that every continuous function is Borel measurable, which would contradict this claim, so what is my mistake?

 A: No, the statement is $C[0,\infty) \notin \mathcal{B}(\mathbb{R}^{[0,\infty)})$.  Note that's a $\notin$, not $\not\subset$.
Let's unpack this.  $\mathbb{R}^{[0,\infty)}$ is the set of all functions from $[0,\infty)$ to $\mathbb{R}$.  $\mathcal{B}(\mathbb{R}^{[0,\infty)})$ is a $\sigma$-field on this set; a set of functions $A$ which is in $\mathcal{B}(\mathbb{R}^{[0,\infty)})$ could be said to be "measurable".  (I don't have the book in front of me, but I think it's the so-called cylindrical $\sigma$-field, not the Borel $\sigma$-field)  But this is not the same as saying that "every element of $A$ is a measurable function"!
Now, $C[0,\infty)$ is a set of functions, i.e. a subset of $\mathbb{R}^{[0,\infty)}$.  The question is whether it's a measurable subset, i.e. whether it's one of the sets that's in $\mathcal{B}(\mathbb{R}^{[0,\infty)})$.  The claim is that it's not.  But this has nothing to do with the fact that every continuous function is Borel measurable.  It's true that every element of $C[0,\infty)$ is a Borel measurable function on $[0,\infty)$, but that isn't relevant in determining whether or not $C[0,\infty)$ is a measurable set.  
