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

  • 2
    $\begingroup$ The statement $I \subset J$ is not correct. $\endgroup$ – littleO Sep 11 at 8:52
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.