## Measurable Spaces – Problem (45/365)

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 $A$.

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