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I am trying to prove Blumenthal's zero-one law using Kolmogorov's zero-one law. I use that $B_t$ Brownian $\iff$ $tB_{1/t}$ Brownian. Can I change the intersection and union as follows?

Start of my proof

If $(B_t)_{t\geq 0}$ is a Brownian motion then $(B_t^*)_{t\geq 0}:=(tB_\frac{1}{t})_{t\geq 0}$ is a Brownian motion. We have $$\mathcal{F}_{0+}^B=\cap_{t>0}\mathcal{F}_t^B=\cap_{t>0}\sigma(B_s:s\leq t) .$$ For any event in this $\sigma$-algebra there must also exist an event in $\mathcal{F}_{0+}^{B^*}$ with the same probability. But \begin{align*}\mathcal{F}_{0+}^{B^*}&=\cap_{t>0}\mathcal{F}_t^{B^*}=\cap_{t>0}\sigma(B_s^*:s\leq t)=\cap_{t>0}\sigma(sB_\frac{1}{s}:s\leq t)=\cap_{t>0}\sigma(B_s:s>t)\\ &= \cap_{t\in\mathbb{N}}\sigma(B_s:s\geq t) = \cap_{t\in\mathbb{N}}\cup_{n\in\mathbb{N}} \sigma(\sigma(B_t),\sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N}))\\ &= \cap_{t\in\mathbb{N}}\cup_{n\in\mathbb{N}} \sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N})\\ &=\cup_{n\in\mathbb{N}}\cap_{t\in\mathbb{N}} \sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N}) \end{align*} By application of Komogorov's zero-one law we see that all events in the latter set must have either probability one or zero. QED.

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