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Theorem: Given $\{ E_\alpha \}_{\alpha \in \mathbb{R}}\subset \mathcal{M}$ such that $E_\alpha \subset E_\beta$ for $\alpha < \beta$. We have also that $\bigcup_{\alpha \in \mathbb{R}}E_\alpha=X$ and $\bigcap_{\alpha \in \mathbb{R}}E_\alpha=\emptyset$. Then there exists a measurable function $f:X\rightarrow \mathbb{R}$ such that $f(x)\le \alpha$ if $x\in E_\alpha$ and $f(x)\ge \alpha$ if $x\notin E_\alpha$.

I proved this by defining, $$f(x) = \inf \{ q\in \mathbb{Q} : x\in E_q \}$$

Then it is easy to show that $f$ has the desired properties and $f(x) = \inf_{q \in \mathbb{Q}} f_q(x)$. Where $f_q(x)$ is $q$ if $x\in E_q$ and is $\infty$ otherwise. Then since each $f_q$ is measurable we have that $f$ is measurable.

I don't feel very confident about this proof because I don't use the fact that $\bigcap_{\alpha \in \mathbb{R}}E_\alpha=\emptyset$ and the book hints that I should use the fact that $g:X\rightarrow \overline{\mathbb{R}}$ is measurable if $g^{-1} \left( (r,\infty] \right)\in\mathcal{M}$ for $r\in \mathbb{Q}$. And I don't use this fact, so I'd appreciate if someone could tell me if I'm making a bad mistake here.

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How do you know that there are no $x\in E_\alpha$ such that $x\notin E_\beta$ for all $\beta<\alpha$? Your function only looks at $E_q$ for rational $q$, but never at $E_\alpha$ itself. (wait, is the $\alpha$ in the claim "there exists a function such that" universally quantified, or is it fixed?) –  Mario Carneiro May 26 '13 at 7:02
    
You used $\bigcap_{\alpha \in \mathbb{R}}E_\alpha=\emptyset$ to make sure that $f$ is not $+\infty$ for any $x$. –  Mario Carneiro May 26 '13 at 7:06
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@MarioCarneiro you don't need it to show that $f$ is not $+\infty$, you need it to ensure that $f$ is not $-\infty$. As for your first comment, you are taking the infimum over the whole of $\mathbb{Q}$. –  Amudhan May 26 '13 at 7:31
    
@Amudhan You're right, my mistake. –  Mario Carneiro May 26 '13 at 7:34
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up vote 2 down vote accepted

Stuart, since your question calls for a $f:X\rightarrow \mathbb{R}$, you need to ensure that $f$, as you have defined it, is never $\pm \infty$. This is where your two conditions come in. ( $\cup_{\alpha} E_\alpha=X , \cap_{\alpha} E_\alpha = \emptyset$)

To use the fact about $g^{-1}((r,\infty])$ being measurable for $r \in \mathbb{Q}$, just show that $f$ satisfies that condition, and then show that $f$ takes only real values.

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I don't follow why showing that $f$ satisfies the condition and $f$ taking real values implies that $f$ is measurable. Are you saying that $f^{-1}((r,\infty])=E_r^c$? –  Stuart May 27 '13 at 1:56
    
You would need to show that $x\notin E_r$ and $x\in E_q$ for $q>r$ cannot occur. But it isn't clear to me how to show this. –  Stuart May 27 '13 at 2:04
    
$f^{-1}((r,\infty])=cup_{q \in \mathbb{Q}, q>r} f^{-1} ([q,\infty]) = cup_{q \in \mathbb{Q}, q>r} E_q ^c $. It is not, in general, $E_r^c$. [Look at $\mathbb{R}$, with $E_r = (-\infty, r)$]. –  Amudhan May 27 '13 at 13:18
    
You are saying that $f^{-1}([q,\infty]) = E_q^c$? I think I'm misreading you because this isn't true. –  Stuart May 28 '13 at 18:00
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@Stuart sorry, my mistake. $f^{-1}([q,\infty]) = X\setminus (\cup_{r<q,r\in \mathbb{Q}} E_r) = E_q^c \cup (E_q\setminus \cup_{r<q,r\in \mathbb{Q}} E_r ) $ –  Amudhan May 29 '13 at 8:25
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