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This question is about Durrett Edition 5 Example 1.1.8, in which he claims that

Let $\Omega=\mathbb{R}$, and $\mathcal{S}=\mathcal{S}_{1}$ then $\overline{\mathcal{S}}_{1}=$the empty set put all sets of the form $\bigcup_{k=1}^{n}(a_{k}, b_{k}]$, where $-\infty\leq a_{i}<b_{i}\leq\infty$ is an algebra.

He did not specify what is $\mathcal{S}_{1}$, so I assumed it is a set containing empty set and the intervals of the form $(a,b]$ where $-\infty\leq a<b\leq\infty.$

It is clear that $\overline{\mathcal{S}}_{1}$ is the set of finite disjoint union of sets in $\mathcal{S}_{1}$. Thus, $\overline{\mathcal{S}}_{1}$ being an algebra follows immediately if we prove that $\mathcal{S}_{1}$ is a semi-algebra.

Indeed, for $(a_{1}, b_{1}], (a_{2}, b_{2}]\in\mathcal{S}_{1}$, the intersection of them is of the form $(x_{1},y_{2}]$ where $x_{1}\in \{a_{1}, a_{2}\}$ and $y_{2}\in\{b_{1}, b_{2}\}$, and thus $\mathcal{S}_{1}$ is closed under finite intersection.

However, I don't know how to show for any $(a,b]\in\mathcal{S}_{1}$, $(a,b]^{c}$ can be expressed as a finite disjoint union of elements in $\mathcal{S}_{1}$. I don't even this this is possible, so the definition of $\mathcal{S}_{1}$ must be something else.

Any idea? I think perhaps I assumed the wrong definition. Thank you!

Edit:

Okay I figured it out. The definition of $\mathcal{S}_{1}$ is not wrong, Durrett actually gave a general example in example 1.1.5 of $\mathcal{S}_{d}$, with respect to $\mathcal{R}^{d}$.

The point here is that the end points of the interval can be $\pm\infty$.

I will answer my own question.

JacobsonRadical
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1 Answers1

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If for $(a, b]\in\mathcal{S}_{1}$, then $(a,b]^{c}=(-\infty, a]\sqcup (b,+\infty]$ which is a finite disjoint union of sets in $\mathcal{S}_{1}$.

As I said in my edit, the point here is that you can have $\pm\infty$, so it is immediate.

JacobsonRadical
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  • Your answer is correct. But I believe it should be (-infinity, a] u (b, +infinity) since these intervals are defined as subsets of R. At least if this is the case we need not to think about extended real line. Do you agree? – georgy_dunaev Nov 26 '22 at 19:17
  • @georgy_d This is what I don't like about Durret's book. He used $\mathbb{R}$, and then he used the interval $(a,b]$ where $b$ can be infinity, which means that $(a,\infty]$ exists in its algebra, but $(a,\infty]$ does not lie in $\mathbb{R}$. I can't remember what he meant by $\mathcal{S}{1}$ and $\overline{\mathcal{S}{1}}$, but I guess here whether it is $(b,+\infty)$ or $(a,+\infty]$ is not important. It depends on the definition of $\mathcal{S}_{1}$. – JacobsonRadical Nov 28 '22 at 20:52
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    Also if we fix a set and then intersect all elements of algebra with it, then we also obtain algebra. So he may also be understood as constructing algebra on (-inf;+inf] and then taking "subset algebra". – georgy_dunaev Nov 29 '22 at 21:27