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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Gerald Edgar Nov 28 '18 at 22:18
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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 $\mathcal{B}_0$.

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