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Let $B$ be a Brownian motion and consider $X:=(tB_{1/t} \Bbb{1}_{t>0})_{t\geq 0}$ the so called time inversion. I want to show that the paths of $X$ are continuous for all $t\in [0,\infty)$.

First I remark that clearly $t\mapsto X_t$ is continuous for all $t>0$. Thus it remains to check right continuity at $t=0$. Remark that by definition $X_0=0$, so we need to check that $\Bbb{P}(\lim_{t\downarrow 0}X_t=0)=1 $. $$\begin{align}\Bbb{P}\left(\lim_{t\downarrow 0}X_t=0\right)&=\Bbb{P}\left(\forall \epsilon>0~\exists \delta>0:~\forall|t|<\delta \Rightarrow |X_t|<\epsilon\right)\\&=\Bbb{P}\left(\bigcap_{\epsilon>0}\bigcup_{\delta>0}\bigcap_{t\in (0,\delta)}\|X_t\|<\epsilon\right)\\&=\Bbb{P}\left(\bigcap_{\epsilon>0}\bigcup_{\delta>0}\bigcap_{t\in (0,\delta)}\|B_t\|<\epsilon\right)=1\end{align}$$ Does this work?

Summerday
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  • You are taking intersections and union over an(or rather $3$) uncountable families . Hence the resulting set need not even lie inside the sigma algebra which you had fixed for your probability space. Instead, you have to go to a countable union and intersection. See my answer here and also read the comments under my answer. I think that should be enough to clear your confusion – Mr. Gandalf Sauron Aug 02 '23 at 19:08
  • @Mr.GandalfSauron aha so instead of $\epsilon>0$ I take $n\in \Bbb{N}$ and instead of $\delta>0$ I take $m\in \Bbb{N}$ and then I take $t\in [0,1/m]\cap \Bbb{Q}$? – Summerday Aug 02 '23 at 19:19
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    Yes. Exactly @summerday – Mr. Gandalf Sauron Aug 02 '23 at 19:22

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