# Show that the collection of the subsets $\cup_{i=1}^k(a_i, b_i], \quad -\infty \leq a_i < b_i < \infty$ for some $0\leq k < \infty$ is an algebra.

Definition (Algebra) Let $$\Omega$$ denote a universal set. A collection $$S$$ of subsets of $$\Omega$$ is called an algebra or field if

1. $$\Omega \in S$$
2. If $$A \in S$$, then $$A^c \in S$$, where $$A^c$$ denotes the complement of $$A$$.
3. If $$A\in S$$ and $$B\in S$$ then $$A \cup B \in S$$.

Let $$\Omega = \mathbb{R}$$ and let $$A$$ denote the collection of subsets on the form

\begin{align} \cup_{i=1}^k(a_i, b_i], \quad -\infty \leq a_i < b_i < \infty \end{align}

for some $$0 \leq k < \infty$$. This is clearly an algebra, but it is not a sigma algebra. ...

I don't understand how to show the fact that it is an algebra. I would need to show that
$$\mathbb{R} \in A$$
$$A$$ is closed under complement.
$$A$$ is closed under union.

The first thing that causes problems is that I don't understand the definition of $$A$$. Is $$k$$ fixed or does $$A$$ contain all subsets on the form $$\cup_{i=1}^k(a_i, b_i], \quad -\infty \leq a_i < b_i < \infty, \quad k \in \mathbb{Z}_+$$, that is \begin{align} A = \{ \cup_{i=1}^0(a_i, b_i], \cup_{i=1}^1(a_i, b_i], \cup_{i=1}^2(a_i, b_i]...,\}? \end{align}

In order to show that $$A$$ is closed under complement it seems to me like I would need to show that $$(a_i, b_i]^c = (-\infty, a_i] \cup (b_i, \infty) \in A$$, but I don't see how such a disjoint union could equal any $$(a_j, b_j] \in A$$? Also, how do we know there is some $$(a_j, b_j]\in A$$ with $$a_j = -\infty$$?

I also wonder how to prove that $$\mathbb{R} \in A$$, I thought it wouldn't be the case since $$b_i < \infty$$.

• A few points to note. The set of reals does not contain $\infty$. $a_i$'s and $b_i$'s are of your choosing. You can fix $k$ or vary it to prove the properties. Jul 9, 2020 at 15:39
• I'd guess the author just meant to allow $b_i = \infty$, where e.g. $(0, \infty] = \{x \in \mathbb{R} : 0 < x \leq \infty\} = (0, \infty)$. Then it is an algebra but not a sigma algebra. Also, to be clear you need to allow $k=0$ (as opposed to "$k \in \mathbb{Z}_+$") so that $\varnothing = \mathbb{R}^c \in A$. Jul 9, 2020 at 23:56

1)) I guess that $$k$$ is not fixed, but varies on non-negative integers.
2)) The family $$A$$ is not an algebra, because all members of $$A$$ are bounded from above subsets of $$\Bbb R$$, whereas $$\Bbb R$$ is not bounded from above. By the same reason $$A$$ is not closed with respect to complement. So I guess there is a misprint in the definition of $$A$$.