# Why is the outer measure of $\{0\} = 0$?

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.

## 1 Answer

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 $\varepsilon > 0$.

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