Why is this not a $\sigma$ - algebra Why is 
$\displaystyle \{\cup_{i=1}^{\infty} (a_i, b_i] : a_i < b_i, i=1,...,n, n \in \mathbb{N_0} \}$
not a $\sigma$-algebra?
It looks ok to me, the only way I can see that this would fail is if $i \in \mathbb{R}$, which is clearly not the case.
 A: Take $a_i=-n$ and $b_i=n$ for $n\in\Bbb N$, and $n>0$. Then $\bigcup (a_n,b_n]=\Bbb R$.
However $\varnothing$ cannot be written as a union of these intervals since none of them is empty.
A: I am a little confused at the notation, but here goes:


*

*If the sets of your collection are finite unions of half-open intervals, i.e., your collection is
$$\left\lbrace \bigcup_{i = 1}^{n} (a_i, b_i], \quad a_i < b_i, \quad i = 1, 2, \dotsc, n, \quad n \in \mathbb{N}_0 \right\rbrace,$$
[note that $\mathbb{N}_0$ includes $0$ for the empty union, i.e. the empty set], then you do have an algebra (assuming you allow $a_i = -\infty$ and $b_i = \infty$ ).  Yet this is not a $\sigma$-algebra, since it is not closed under countable unions, for you can get open intervals with countable unions of such sets, e.g.
$$(0, 1) = (0, 1/2] \cup (1/2, 3/4] \cup (3/4, 7/8] \cup \dotsb.$$

*If the sets of your collection are countable unions of half-open intervals, i.e., your collection is 
$$\left\lbrace \bigcup_{i = 1}^{\infty} (a_i, b_i], \quad a_i < b_i,\quad i = 1, 2, \dotsc, \infty \right\rbrace,$$
then $(0, 1)$ is in your collection by the above construction, but its complement $(- \infty, 0] \cup [1, \infty)$ is not in the set, so it is not even an algebra.  
I think there is a typo in your notes.  In the first interpretation, any particular member of the collection should be a finite union, and in the second interpretation, it is not clear what role $n$ plays. 
