Consider the completion $(\mathbb{R}^{[0,1]}, \mathcal{B}, \mu)$ of the Wiener measure on $\mathbb{R}^{[0,1]}$ (with the cylinder set $\sigma$-algebra). 


Is the following true : 


* $C([0,1])\in \mathcal{B}$ ?


I am aware that $\mu^\*(C([0,1]))=1$ where $\mu^\*$ is the outer measure associated to the Wiener measure, but since the outer measure is not additive, we can't conclude that $C([0,1])$ is measurable for the outer measure. Is it true nonetheless ?



Measurability of C([0,1]) for the completion of the Wiener measure
Let $\mathcal{B}\_0$ denote the cylinder $\sigma$-algebra. Since a cylinder set $A \in \mathcal{B}\_0$ only specifies the values of functions at countably many points, if it is nonempty then it contains a discontinuous function. Hence the inner measure of $C([0,1])$ is
$$\mu\_\*(C([0,1])) = \sup\{\mu(A) : A \in \mathcal{B}\_0, A \subset C([0,1])\} = \sup\{\mu(\emptyset)\} = 0.$$
Since $\mu\_\*(C([0,1])) \ne \mu^\*(C([0,1]))$ it is not $\mu$-measurable and is not in the $\mu$-completion of $\mathcal{B}\_0$.


