0
$\begingroup$

The outer measure is defined as $\mu^*(S) = \inf \sum\limits_{n=1}^{\infty} \phi (E_n)$, where $E_n$ is an open elementary set, and $S \subseteq \bigcup\limits_{n=1}^{\infty} E_n$. It doesn't say what kind of function $\phi: F \to [0, +\infty]$ (where $F$ is a ring) is.

So $\mu^*(\{0\}) = \inf \sum\limits_{n=1}^{\infty} \phi (E_n)$. I have to show that $\inf \sum\limits_{n=1}^{\infty} \phi (E_n) = 0$. So I have to figure out what $\phi (E_n)$ is, but I don't know how to find it.

$\endgroup$
1
$\begingroup$

If the set function $\phi$ is not given, then it is not necessarily true that the outer measure $\mu^*$ is zero.

For example, take $\phi$ on the natural numbers which counts the number of elements in the set $E \subset \mathbb{N}$. Then $\mu^*(\{0\}) = 1$.

If we are dealing with the lebesgue measure (which assigns the value $b-a$ to intervals of the form $(a,b), (a, b], [a, b), [a, b]$), we can see that the infimum is zero since we just take the interval $(-\varepsilon, \varepsilon)$ which covers $\{0\}$ for any $\varepsilon > 0$.

$\endgroup$
  • 1
    $\begingroup$ Ah, thank you. I think Lesbesgue measure was implied here, which I didn't realize. $\endgroup$ – mr eyeglasses Dec 10 '15 at 13:10

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.