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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 $b_i < \infty$.

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  • $\begingroup$ 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. $\endgroup$
    – learner
    Jul 9, 2020 at 15:39
  • $\begingroup$ 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 $\varnothing = \mathbb{R}^c \in A$. $\endgroup$ Jul 9, 2020 at 23:56

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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 $A$.

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