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I don't understand that $\{I_x\}_{x\in O}$ is disjoint. For example, $O = \{(1 , 2)\}$. Let $x = 1.5$. Then, $a_x = 1$, and $b_x = 2$. Therefore, $I_x = (1, 2)$. Let $y = 1.6$. Then, similarly, $I_y = (1, 2)$. That is, $I_x$ and $I_y$ are not disjoint, but equal.

Could you explain how $\{I_x\}_{x \in O}$ can be disjoint?

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    $\begingroup$ Royden Fitzpatrick? $\endgroup$
    – BCLC
    Jul 28, 2018 at 9:34

1 Answer 1

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The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x \cap I_y \not = \emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)

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