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I read some proofs that show that the outer measure $m^*(I)$ of an interval is equal to its length $l(I)$, i.e. $m^*(I)=l(I)$, where for an interval $I=[a,b]$, we have $l(I)=b-a=m^*(I)$.

I understand the part that $m^*(I) \leq l(I)$, but for the other direction $m^*(I) \geq l(I)$, I could not see why the proofs really wanted to use the compactness property of $I$ (being bounded and closed). From what I read, outer measure of an interval $I$ is:

$$ m^*(I) = \inf \bigg\{\sum_{j\in J} l(j) \bigg\} $$

where $J$ forms an open covering of $I$, and $j$ refers to any open interval inside the open covering $J$ - so that the outer measure gets the infimum of the sum above for all open coverings of $I$.

Since we have (for sure) that $I \subseteq J$, shouldn't it hold trivially that $m^*(I) \geq l(I)$ ? given that whether $J$ is finite or infinite countable, it should be able to cover all elements of $I$.

So why do we need to guarantee (using compactness and the Heine Borel theorem) that there is a $J$ with finite cardinality $|J| \neq \infty$ that covers $I$ to show that $m^*(I) \geq l(I)$ ?

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    $\begingroup$ The statement $I \subset J$ is not correct. $\endgroup$ – littleO Sep 11 at 8:52
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    $\begingroup$ Yes $(I)$ is a cover of $I$ by intervals. But the definition of outer measure is the $\inf$ over all such covers, so what follows trivially from this is $m_*(I)\le|I|$, not the other way around. The non-trivial part is showing that for any cover of $I$ by inntervals the sum of the lengths is $\ge|I|$. This is easy to see for a cover by finitely many intervals. Compactness is needed to reduce to that case. $\endgroup$ – David C. Ullrich Sep 11 at 15:30
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    $\begingroup$ I lied.. We're talking about covers of $I$ by open intervals, so if $I=[a,b]$ then $J=(I)$ is not such a cover. But for any $\epsilon>0$, $J=((a-\epsilon,b+\epsilon))$ is a cover of $I$ by open intervals; hence $m_*(I)\le|I|+2\epsilon$, hence $m_*(I)\le|I|$. $\endgroup$ – David C. Ullrich Sep 11 at 15:36
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    $\begingroup$ How are you getting from "whether $J$ is finite or infinite countable, it should be able to cover all elements of $I$" to $m^*(I)\geq l(I)$? Yes, every element of $I$ is contained in some element of $J$, but why would that imply $\sum_{j\in J}l(j)\geq l(I)$? $\endgroup$ – Eric Wofsey Sep 11 at 16:09
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    $\begingroup$ You're thinking about this wrong. If you're claiming it's trivial, the burden is on you to prove it! I think you are assuming that "length" has some properties that seem intuitively obvious, but those properties need to be proved and their proofs are far from obvious! $\endgroup$ – Eric Wofsey Sep 12 at 1:30
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Let $\epsilon>0$. By definition of $inf$ there exists an open covering $J$ such that

$$m^*(I)+\epsilon\geq \sum_{j\in J}l(j).$$

But $I$ is compact, so without lost of generality you can choose $J$ finite. However $J$ is an open finite cover of $I$, so it is clear that

$$\sum_{j\in J}l(j)\geq b-a=l(I).$$

Thus for each $\epsilon>0$ you have that

$$m^*(I)\geq l(I)-\epsilon\to_{\epsilon\to 0^+} l(I);$$

this means $m^*(I)\geq l(I)$

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