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I am working out a book on Lebesgue measure by Bartle, and would like to see the steps that go into the construction of a proof for the following:

Show that any $\sigma$-algebra of subsets of $\mathbb{R}$ which contains all open intervals also contains all closed intervals.

That is, $[a,b]=\cap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$

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  • $\begingroup$ What part are you struggling with? For example, do you see why the last statement is true? Using the assumptions in the problem statement, what can you conclude from it (even if you don't yet understand why it's true)? $\endgroup$
    – cardinal
    May 24, 2012 at 12:34
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    $\begingroup$ I can see that the RHS is an intersection of intervals that are a part of the $\sigma-$ algebra and hence, $[a,b]$ is also a part of a $\sigma-$ algebra. The statement looks trivial on a softer note. I am interested in the constructive elements of the proof, that make it seem precise and formal. Am trying to learn from the way, the proof is constructed in showing the equality to be true. $\endgroup$
    – hearse
    May 24, 2012 at 12:45
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    $\begingroup$ That's a good start! Here are some thoughts on how to continue: For each $x \in [a,b]$, is $x$ in every set on the RHS? Why or why not? Suppose $x > b$. Write $x = b + \delta$ for some $\delta > 0$. Recall the Archimedean property of the reals. Now, do something analogous for $x < a$. Conclusion? $\endgroup$
    – cardinal
    May 24, 2012 at 13:09
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    $\begingroup$ $x$ can be between $a \leq x \leq b$ for all $n$ from $1$ to $\infty$ in the RHS, where the intersection of these infinite subsets would be $[a,b]$ as the length of the interval is getting larger. But in a case where $x$ is $<a$ or $>b$, I would have to continue the proof with this case based on the Archimedean property for the reals. That I will go through here, forth for the second case. Is this another step in the right direction w.r.t the first case? $\endgroup$
    – hearse
    May 25, 2012 at 2:20

1 Answer 1

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  • We have for each positive integer $n$ that $a-1/n \lt a\lt b\lt b+1/n$, hence $[a,b]\subset (a-1/n,b+1/n)$; in particular $[a,b]$ is contained in the intersection. If $x\in (a-1/n,b+1/n)$ for each positive $n$, then $a-1/n\lt x\lt b+1/n$. Taking the limit $n\to \infty$, we get $x\in [a,b]$.

  • Let $\mathcal A$ be a $\sigma$-algebra on $\mathbb R$ which contains all the open intervals. Then $\mathcal A$ being stable under countable intersections, this $\sigma$-algebra contains the countable intersections of open intervals. As a closed interval can be expressed like that, we are done.

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