Borel algebra on the postive real line I´m considering the borel sigma algebra on the positive real line, $ \mathcal{B} (\mathbb{R}_+ ) $ and  I would like to show that intervals given by $\{ [0,t] : t \in  \mathbb{R}_+ \}$ satisfy that 
$ \sigma ( \{ [0,t] : t \in  \mathbb{R}_+ \} ) = \mathcal{B} (\mathbb{R}_+ ) $
is there a simple (obvious) way to proof this claim?   
 A: Normally the Borel sigma algebra is defined as the sigma algebra generated
by the open sets.
Defining $\mathcal{W}:=\left\{ \left[0,t\right]:t\in\mathbb{R}_{+}\right\} $
and $\mathcal{V}:=\left\{ V:V\text{ open}\right\} $ it is to be shown
that $\sigma\left(\mathcal{W}\right)=\sigma\left(\mathcal{V}\right)$.
If $W\in \mathcal{W}$ then $W^{c}\in\mathcal{V}\subset\sigma\left(\mathcal{V}\right)$
and consequently $W=\left(W^{c}\right)^{c}\in\sigma\left(\mathcal{V}\right)$.
So $\mathcal{W}\subseteq\sigma\left(\mathcal{V}\right)$ and consequently
$\sigma\left(\mathcal{W}\right)\subseteq\sigma\left(\mathcal{V}\right)$.
An open interval in $\mathbb{R}_{+}$ of the form $\left[0,b\right)$
can be written as $\left[0,b\right)=\bigcup_{r\in\left[0,b\right)\cap\mathbb{Q}}\left[0,r\right]$.
This is a countable union of elements from $\mathcal{W}$ so $\left[0,b\right)\in\sigma\left(\mathcal{W}\right)$.
An open interval in $\mathbb{R}_{+}$ of the form $\left(a,b\right)$
can be written as $\left[0,b\right)-\left[0,a\right]$ so $\left(a,b\right)\in\sigma\left(\mathcal{W}\right)$.
Every $V\in\mathcal{V}$ can be written as a countable union of open
intervals in $\mathbb{R}_{+}$ so $V\in\sigma\left(\mathcal{W}\right)$.
So $\mathcal{V}\subseteq\sigma\left(\mathcal{W}\right)$ and consequently
$\sigma\left(\mathcal{V}\right)\subseteq\sigma\left(\mathcal{W}\right)$.
