Why L is not a σ-algebra?

Question

Why $L$ is not a $σ$-algebra?

Let $L$ be the collection of all finite disjoint unions of all intervals of the form:

$(−∞, a], (a, b], (b, ∞), ∅, R.$

Then L is an algebra over R, but not a σ-algebra because

union of sets $$\bigcup _{i=1}^{\infty} {(0,(i − 1)/i]} = (0,1)$$ which doesn't belong to L

Why (0, 1) doesn't belong to L ?

Any help?

Answer 1

Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+\infty)$ or $(a,b]$, neither of which fits $(0,1)$.

If you want a more rigorous proof, suppose $(0,1)\in L$, then there is a finite set $\mathcal{A} = \{I_1,\dots, I_n\}$ of intervals of the given form such that $I := \bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, \dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + \frac{1-b}{2}\in (0,1)$ but $b+ \frac{1-b}{2} \not\in I$; or $b\geq 1$, so $1\in I$ but $1\not\in (0,1)$. Both contradicts $(0,1) = I$.

Answer 2

Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.