Collection of half open intervals is an algebra, why? I have to show, why the collection of all finite unions of such half open intervals $(a,b]$ is an algebra and not a sigma algebra. I know that $−∞≤a≤b≤∞$, and have:
$$
(a,b)=\bigcup_{n=1}^∞ \left(\right.a,b−\frac1n\left.\right]
$$
But how can I say from this, that it is an algebra? I will say that it is a sigma-algebra, since the union has the limits to infinity.
 A: The mentioned identity shows that the open interval $(a,b)$ can be written as a countably infinite union of half open intervals $(a,\ b-\frac1n]$. 
While, $(a,b)$ cannot be an element of the given set, as any finite union of half open intervals has a maximum element, while the open interval doesn't have.
So, the collection of half open intervals is not closed under countable union, i.e. it is not a $\sigma$-algebra.
This in itself doesn't show that, on the other hand, it is an algebra. But that can be easily verified: the union of two elements (finite unions of half open intervals) and the complement of one can be written as a finite union of half open intervals.
A: Let $\mathcal{A}_0$ be the collection of all subsets of $\mathbb{R}$ of sets of the form $(a,b]$ (for real $a,b$), or of the form $(a,\infty)$ or $(-\infty,b]$, and let $\mathcal{A}$ be the collection of all sets which have the form
$$\bigcup_{i=1}^{n}A_{i}$$
for some finite set $\{A_{1},\ldots A_{n}\}\subseteq\mathcal{A}_{0}$ (together with $\emptyset$, which is the union of zero intervals). Note that $\mathcal{A}_{0}\subseteq\mathcal{A}$. 
You are discussing two separate claims.  The first is that $\mathcal{A}$ is an algebra. The second is that $\mathcal{A}$ is a $\sigma$-algebra. 
The second of these claims is the easiest to resolve. You have already proven that the open interval $(0,1)$ is expressible as the union of a countable number of elements of $\mathcal{A}_{0}$.  Thus, if $\mathcal{A}$ is a $\sigma$-algebra, then $(0,1)$ must be an element of $\mathcal{A}$.  But it is obvious that any non-empty bounded element of $\mathcal{A}$ must have a maximum element.  Since $(0,1)$ does not have a maximum, it is not an element of $\mathcal{A}$.  We can therefore conclude that $\mathcal{A}$ is not a $\sigma$-algebra. 
The fact that $\mathcal{A}$ is an algebra requires a little more work to show. 


*

*You must prove that for every $A\in\mathcal{A}$, the complement of $A$ is also in $\mathcal{A}$.  

*You must show that if $A_{1},\ldots A_{k}$ are all elements of $\mathcal{A}$, then $$\left(\bigcup_{i=1}^{k}A_{i}\right)\in\mathcal{A}.$$

*You must show that $\emptyset\in\mathcal{A}$.


Condition 3 follows directly from the definition, and condition 2 follows immediately from the construction.  
For 1, let $A=\bigcup_{i=1}^{n}A_{i}$ for some finite $\{A_{i},\ldots A_{n}\}\subseteq\mathcal{A}_{0}$.  For each $i$ let $A_{i}=(a_{i},b_{i}]$ (with suitable adjustment for unbounded intervals).  You can assume without loss of generality that $b_{i}<a_{i+1}$ (why?).  So all you need to do is find a convenient form for the complement of $A$, and you're done!
