Example of an algebra which is not a σ-algebra. I have troubles with constructing an example of an algebra of sets which is not a σ-algebra. Could you please help me with this?
 A: Let $X$ be an infinite set, and $\mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $\mathcal A$ is an algebra of sets which is not a $\sigma$-algebra.
A: Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−\infty, a], (a, b], (b, \infty), \emptyset, \mathbf{R}$.
Then $L$ is an algebra over $\mathbf{R}$, but not a σ-algebra because
union of sets $\left\{(0,\frac{i-1}{i}]\right\}$ for all $i \ge 1 = (0, 1) \notin L $.
A: For example, let $\mathbb{R}^{[0,+\infty)}$ denote the set of all the real value functions on $[0,+\infty)$. Consider the so called n-dimensional cylinder set, defined as
\begin{equation*}
C_{t_1,t_2,\ldots, t_n,B}=\{f\in \mathbb{R}^{[0,+\infty)}:(f(t_1),f(t_2),\ldots, f(t_n))\in B\},
\end{equation*}
where $B\in\mathcal{B}^d$ is a Borel set and $C\subseteq \mathbb{R}^{[0,+\infty)}$. Then the set $\mathcal{C}'$ of all the cylinders $C$, that is,
\begin{equation*}
\mathcal{C}'=\{C_{t_1,t_2,\ldots t_n,B}, \ \forall \ n\geq 1, \ \forall\ (t,t_1,t_2,\ldots,t_n), \ \forall \ B\in \mathcal{B}^n  \} 
\end{equation*}
is an algebra but not  $\sigma$ - algebra.
Anyway it is well known that there exists the smallest $\sigma$ - algebra containing $\mathcal{C}'$, but this is another story. 
A: As another example which supports the first answer above, let $\Omega $ $=$ $(0,1]$, and let $\mathcal{F}$
contain $\varnothing $, all $(a,b]$ with $a,b$ $\in $ $\mathbb{Q}$, $a,b \in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $\mathcal{F}$ is a
field by straightforward verification. It is not a $\sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}\pi ]$, hence $A_{n}$ $\in $ $\mathcal{F}$ but $\cup
_{n=1}^{\infty }A_{n}$ $=$ $(\pi ,1]$ $\notin $ $\mathcal{F}$.
