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I'm taking a course in Real Analysis covering Measure Theory. We are using Real Analysis by Royden/Fitzpatrick as the text. I'm going through the book and I'm struggling to understand the approach of Proposition 1 which states: The outer measure of an interval is it's length. This makes me think I'm misunderstanding the definition of the outer measure.

I understand the first part which tells us that if we have an interval $[a,b]$, let $\epsilon > 0$ be given. Then $ [a,b] \subseteq (a- \epsilon, b+ \epsilon)$ so $m^*([a,b]) \leq \ell((a- \epsilon, b+ \epsilon) = b - a + 2\epsilon$. Since this holds for any $\epsilon > 0$ then $m^*([a,b]) \leq b-a$.

My question is: Why can't we use the same trick to show the other inequality? We have that $(a - \epsilon, b + \epsilon)$ is a cover of $[a,b]$. Then $\ell(a-\epsilon ,b+\epsilon) = b - a + 2\epsilon \geq b-a$. Since the outer measure, in this case, is defined as $$m^*([a,b]) = \inf \left\{ \sum_{k=1}^{\infty} \ell (I_k) : [a,b] \subseteq \cup_{k=1}^{\infty} I_k \right\},$$ then why can't we say $b-a + 2\epsilon \geq m^*([a,b]) \geq b-a$, by virtue of $m^*([a,b])$ being defined as the infimum?

We have an open cover that is bounded below by $b-a$, so can't we say that the the infimum over all open covers is bounded below by $b-a$? The textbook takes a different approach using the compactness of the interval. I understand the proof of that. I just don't understand we we have to take that approach. This makes me think I'm misunderstanding something subtle about the definition of the outer measure. If someone could help clarify the definition that would be greatly appreciated!

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  • $\begingroup$ You have to prove that $m^*([a,b]) \geq b-a$. The above doesn't prove it. $\endgroup$
    – copper.hat
    Sep 16 '20 at 16:25
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By definition of the infimum: the infimum of a subset $S$ of a partially ordered set $T$, denoted $\inf S$ is the greatest element in $T$ that is less than or equal to all elements of $S$.

Applying that, you have no way to select a set of interval $I_k$ with $[a,b] \subseteq \cup_{k=1}^{\infty} I_k$ and conclude that $m^*([a,b]) \ge b-a$.

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  • $\begingroup$ Okay I think I see what you mean. I posted an answer myself and I think it aligns with what you are saying. Thank you! $\endgroup$
    – MGeoSnyder
    Sep 16 '20 at 20:48
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So I think the issue is that just because we found an open cover whose sum is bounded below by $b-a$ it doesn't mean that every open cover's sum is bounded below by $b-a$. We could possibly still find an open cover that is less that $b-a$. This is why the text approaches it in this way. You start with an arbitrary open cover, reduce it to a finite cover and show that it is larger than $b-a$. Since the given open cover is arbitrary, it holds for all open covers and thus the measure is also greater than or equal to $b-a$.

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