Measurable Spaces – Problem (45/365)

Posted on October 29, 2016 by Naren Sundar

Show that the following is not a Borel set in \mathcal{B}(R^{[0,1]}) (this is the \sigma-algebra of functions over the domain [0,1] unlike the previous onces we looked at where the domain was over the natural numbers).

\displaystyle A = \{ x : x_t = 0 \text{ for at least one } t \in [0,1] \}

If A is a Borel set, then so is its complement \bar{A}. We know that all sets in \mathcal{B}(R^{[0,1]}) must have this form \{ x : (x_{t_1}, x_{t_2}, \dots) \in B \} for some B \in \mathcal{B}(R^\infty) and t_1, t_2, \dots \in [0,1] (proved in book and is simple). The function y_t = 1 belongs to \bar{A} and therefore (y_{t_1}, \dots) \in B. Consider

\displaystyle z_t = y_t \text{ if } t \in (t_1, t_2, \dots) \\ z_t = 0 \text{ otherwise}

Then, since (z_{t_1}, \dots) = (y_{t_1}, \dots) the function z_t belongs to \{ x : (x_{t_1}, x_{t_2}, \dots) \in B \}. But z_t clearly does not belong to \bar{A}. Hence \bar{A} is not a Borel set and neither is A.

This entry was posted in Uncategorized and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s