# Why is compactness needed for proof that interval outer measure is its length

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 $$m^*(I) \geq l(I)$$ ?

• The statement $I \subset J$ is not correct. – littleO Sep 11 at 8:52
• 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. – David C. Ullrich Sep 11 at 15:30
• 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 $m_*(I)\le|I|$. – David C. Ullrich Sep 11 at 15:36
• 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 $\sum_{j\in J}l(j)\geq l(I)$? – Eric Wofsey Sep 11 at 16:09
• 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! – Eric Wofsey Sep 12 at 1:30

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