Algebra that is not a $\sigma$-algebra Let $X=\Bbb R\ $ and$\ $ $\mathcal A=\{\text{finite disjoint unions of}\ (-\infty,b],\ (a,b]\land(a,\infty)\}$. So the exercise says to prove that $\mathcal A$ is an algebra but is not a $\sigma$-algebra.
So I started it:
i) $\Bbb R\in\mathcal A$,because $\Bbb R=(-\infty,a_1]\cup (a_2,a_3]\cup (a_4,\infty)$, where $a_1\le a_2\le a_3\le a_4$
ii) Let $E,F\in\mathcal A$, so
$\qquad E$ and $F$ are finite disjoint unions of $(-\infty,b],\ (a,b]\land(a,\infty)\}$  But what does this really mean? How write them as actual finite disjoint unions? And from here I don't see a way how to prove that $E-F\in \mathcal A$ what do you suggest or any ideas?
 A: We must show that $\mathcal{A}$ contains $\mathbb{R}$ and is closed under pairwise unions and complements, or equivalently (which is what we will do here) under pairwise intersections and complements.
Choose any $a \in \mathbb{R}$. Then $\mathbb{R} = (-\infty,a] \cup (a,\infty)$, which shows that $\mathbb{R} \in \mathcal{A}$.
To show that $\mathcal{A}$ is closed under pairwise intersections, suppose that $A,B \in \mathcal{A}$. Then we can write $A = \bigcup_{n=1}^N I_n$ and $B = \bigcup_{m=1}^{M}J_m$ where the $I_n$ and $J_m$ are intervals of the indicated types, and the unions are disjoint. Then $A \cap B = \left(\bigcup_{n=1}^{N} I_n\right) \cap \left(\bigcup_{m=1}^{M} B_m\right) = \bigcup_{n=1}^{N} \bigcup_{m=1}^{M} (I_n \cap B_m)$. The last expression is a disjoint union, and you can verify that each $I_n \cap B_m$ is again an interval of one of the three types. This shows that $A \cap B \in \mathcal{A}$.
To check that $\mathcal{A}$ is closed under complements, we write $A^c = \bigcap_{n=1}^N I_n^c$. Note that each $I_n^c$ is in $\mathcal{A}$, which you can easily verify by checking the complement of each of the three types of intervals. We already showed that $\mathcal{A}$ is closed under pairwise intersections, so by a simple induction it is also closed under finite intersections, and therefore $A^c \in \mathcal{A}$.
The above shows that $\mathcal{A}$ is an algebra. To show that it is not a $\sigma$-algebra, fix $a < b$ and note that $(a,b) = \bigcup_{n=1}^{\infty}(a,b-1/n]$. Therefore, if $\mathcal{A}$ were a $\sigma$-algebra, it would have to contain $(a,b)$. But there is no way to express $(a,b)$ as a finite union of intervals of the three types, so $(a,b)$ is in fact not in $\mathcal{A}$.
To answer your last question, if $E,F \in \mathcal{A}$, then assuming by $E-F$ you mean the relative complement $E \setminus F = E \cap F^c$, this is in $\mathcal{A}$ because $\mathcal{A}$ is closed under intersections and complements.
