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Let $\mathcal{F}$ be the Borel $\sigma$-algebra on $(0,\infty)$ and fix $t>0$. I'm trying to show that

$$\sigma(\{[s,\infty):0<s\leq t\})=\{B\in\mathcal{F}:B\subset (0,t)\text{ or } B^c\subset (0,t)\}.$$

The intervals generating the $\sigma$-algebra on the left are contained in the set on the right. We may also check directly that the righthand set is a $\sigma$-algebra. So it suffices to show that

$$\sigma(\{[s,\infty):0<s\leq t\})\supset\{B\in\mathcal{F}:B\subset (0,t)\text{ or } B^c\subset (0,t)\}.$$

How to proceed from here?

Thank you.

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up vote 1 down vote accepted

Let $\mathcal B$ a $\sigma$-algebra containing $\{[s,+\infty): 0<s\leq t\}$. We have to show that $\mathcal B$ contains $\{B\in \mathcal F,B\subset (0,t)\mbox{ or }B^c\subset (0,t)\}$.

  • $\mathcal B$ contains each interval of the form $[a,b),0<a<b\leq t$ because $[a,b)=[a,+\infty)\setminus [b,+\infty)$. Since $\{a\}=\bigcap_{n\geq 1}[a,a+n^{-1})$, we get that $\mathcal B$ contains in fact each interval of the form $(a,b), 0\leq a\leq b\leq t$. Now take $B\in \{B\in \mathcal F,B\subset (0,t)\mbox{ or }B^c\subset (0,t)\}$.
  • If $B\subset (0,t)$ then $B$ is in the trace of $\mathcal F$ on $(0,t)$, and thanks to the definition of trace topology and $\sigma$-algebra, $B$ is a Borel subset of $(0,t)$, i.e. an element of $\mathcal B(0,t)$ (see here for the argument). Such an element is necessarily in the $\sigma$-algebra generated by $\{(a,b), 0\leq a\leq b\leq t\}$, hence in $B$.
  • If $B^c\subset \mathcal F$, then by the previous case $B^c\in\mathcal B$ and since $\mathcal B$ is a $\sigma$-algebra $B\in \mathcal B$.
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Thank you!${}{}{}{}{}$ –  Ben Derrett May 31 '12 at 17:51
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