2
$\begingroup$

From example 1.1.3 on page 3 of Durrett's Probability

$S_d$ = the empty set plus all sets of the form

$(a_1, b_1]$ $\times$ $...$ $\times$ $(a_d, b_d] \subset R^d $ where $-\infty \le a_i < b_i \le \infty$

Durrett then claims that $S_1$ is a semialgebra.

I see how $S_1$ is closed under finite intersection, but I do not see how the complement of a set in $S_1$ is the finite union of disjoint sets in $S_1$. For example, if a $\ne -\infty$, then $(a, b]^C$ = $[-\infty, a] \cup (b, \infty]$, which is not the finite disjoint union of sets in $S_1$ since $[-\infty, a]$ is not in $S_1$.

  • 1
    No. In fact, $(a,b]^c = (-\infty,a]\cup (b,\infty)$. And in the definition of $S^d$, $(a,\infty]$ should be understood as $(a,\infty)$. – 2017-01-14
  • 0
    by understood, do you mean (a, $\infty$) = (a, $\infty$] or that the sets (a, $\infty$) are just understood to be apart of $S_1$ as well? – 2017-01-14
  • 1
    To clarify: Durrett considers $\mathbb{R}$, not its completion $\mathbb{R}\cup\{\infty\}$ or something similar. So $(a,\infty]$ just makes no sense. But it would be very inconvenient to write something like: "$S_d$ consists of sets of the form $A_1\times\dots\times A_d$ where $A_i$ is either $(a,b]$ with $-\infty\le a$(a,\infty)$ with $a\ge -\infty$". – 2017-01-14

0 Answers 0