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Let $O = [0, \infty)$ and $F_1$ the class of all intervals of the type $[a, b)$ or $[a, \infty)$, where $0 \le a < b < \infty$. Let $F_2$ be the class of all finite disjoint unions of intervals of $F_1$. Show that $F_1$ is not a field and $F_2$ is a field but not a sigma field.

What does "finite disjoint unions of intervals" mean in this context ? Does that mean $F_2$ is empty should the word disjoint be in there ?

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    Yuval’s reading is really the only possible one; $[a,b)$ and $[a,\infty)$ in the definition of $F_1$ are used to describe the **types** of intervals that included in that collection; there is no implication that $a$ or $b$ is a fixed constant $-$ quite the opposite, actually.2012-02-11

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$F_2$ is the set of all finite disjoint unions of intervals, that is $[a_1,b_1) \cup \cdots \cup [a_n,b_n),$ where $b_i \leq a_{i+1}$ and $b_n$ could be $\infty$. The restriction $b_i \leq a_{i+1}$ ensures that the intervals are disjoint. Some examples: $ [1,2) \cup [3,4), \quad [5,6), \quad \emptyset, \quad [7,8) \cup [8,9) \cup [9,\infty). $

But in fact, $F_2$ is also the set of all finite (unrestricted) unions of intervals. They don't have to be disjoint. Do you see why?

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    @HennoBrandsma Thanks mate,$a$little thought along with re-reading Yuval Filmus does show that F2 has complements and finite intersections.2012-02-11