# Measurability of C([0,1]) for the completion of the Wiener measure

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 ?

• In the conventional way to construct Wiener measure, you show that $C[0,1]$ has outer measure $1$, so if you restrict the cylinder outer measure to the Borel sets of $C[0,1]$ to get a genuine countably-additive measure there. – Gerald Edgar Nov 28 '18 at 22:18

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 $$\mathcal{B}_0$$.