Prove: $m^*({x_0}) = 0 = m^*(\emptyset)$

Question

Prove: $m^*({x_0}) = 0 = m^*(\emptyset)$ for any $x_0\in \mathbb{R}$

By definition for $E\subset \mathbb{R}$

$$m^*(E)=\inf\left\{\sum_{n=1}^{\infty}|I_n|\,\middle|\,E\subset\cup_{n=1}^{\infty} I_n\right\}$$

Where $I_n$ are open intervals.

let $I_n=(x_0-\epsilon_n,x_0+\epsilon_n)$ for an arbitrary small $\epsilon$ there for $|l_n|=2\epsilon$ and $m^*(\{x_0\})=0$

As for $m^*(\emptyset)$ we can say that any cover including $0$ will cover the empty set?

Answer 1

In your definition of $I_n$ there is a $\varepsilon_n$ thar vanishes after that.

Just take $\varepsilon>0$ and define $I_n=(x_0-\varepsilon,x_0+\varepsilon)$ for each $n\in\mathbb N$. Then your argument works. And since $\emptyset\subset I_n$, the same argument proves that $m^*(\emptyset)=0$.