We have a polish space $\mathbb{R}^{+\infty}$ with a metric $\rho(x, y) = \sum_{k=1}^{+\infty} 2^{-k} {|x_k - y_k| \over 1 + |x_k - y_k|}$. Show that in this space Borel $\sigma$-algebra coincides with cylindrical one.
Asked
Active
Viewed 89 times
1 Answers
0
You may argue that the topology induced by $\rho$ is equivalent to the product topology, $\tau$, on $\mathbb{R}^{\infty}$. Then we know that $\sigma(\tau)$ equals to the cylindrical $\sigma$-algebra, $\mathcal{C}$. To see the latter assertion, note that the coordinate functions are continuous (w.r.t. $\tau$) and, therefore, $\sigma(\tau)$-measurable ($\therefore \mathcal{C}\subseteq\sigma(\tau)$). On the other hand, $\tau$ is second countable and the sets in its base belong to $\mathcal{C}$ ($\therefore \sigma(\tau)\subseteq\mathcal{C}$).