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.

• For finite unions $A$ and $B$ of half open intervals it must be shown that $A\cup B$ and the complement of $A$ are finite unions of half open intervals as well. That proves that the collection is an algebra. If it would be a sigma-algebra then $(a,b)$ should also be a finite union of half open intervals, which is not the case. Apr 1, 2014 at 19:59
• A more direct proof I think is to consider the union $\cup_n A_ n = (1,2]\cup (3,4] \cup \dots$. This set and its complement are composed of infinite unions and, as @drhab points, are not in the class. Hence the class is not a $\sigma$-algebra. Mar 21, 2022 at 12:30
• Equivalently, $\cup_n A_n$ does not have a maximum and hence the class is not a $\sigma$-algebra... Mar 21, 2022 at 13:14

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.

• Is in general 'difference of two' enough? Should it not be 'complement of one'? If I remember well then each algebra contains the whole set. Apr 1, 2014 at 20:09
• Yes, you are right. Apr 1, 2014 at 20:31

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.

1. You must prove that for every $A\in\mathcal{A}$, the complement of $A$ is also in $\mathcal{A}$.
2. You must show that if $A_{1},\ldots A_{k}$ are all elements of $\mathcal{A}$, then $$\left(\bigcup_{i=1}^{k}A_{i}\right)\in\mathcal{A}.$$
3. 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!