I am trying to show rigorously that $m^{*}(\emptyset)$ is zero. I know intuitively why it makes sense, but intuitively isn't going to cut it - I need to show it rigorously where $m^{*}(A) = \inf \left\{ \sum_{k=1}^{n}l(I_{k}) | I_{1},\cdots,I_{n}\,\text{open intervals,}\,A \subseteq \cup_{k=1}^{n}I_{k}\right\}$. So, it's NOT THE USUAL DEFINITION OF AN OUTER MEASURE; IN THIS CASE, ONLY FINITE OPEN COVERS ARE ALLOWED.
I actually have to show that $m^{*}$ is monotone and finitely subadditive as well, but right now, I'm just working on this part.
My idea is that any collection of open intervals covers $\emptyset$, since it is a subset of every set. So, we have some leniency in choosing our $I_{k}$'s. I thought $\left(-\frac{1}{k},\frac{1}{k} \right)$ would be a good choice for the $I_{k}$'s, since its length $l(I_{k})=\frac{2}{k}$ for every $k$.
The problem is that I need to show that $\inf \left\{\sum_{k=1}^{n} \frac{2}{k} \right\}$ either equals $0$ or is $\leq 0$, and I am not sure how to do that. As essentially the $k$-th partial sum of the divergent series $\sum_{k=1}^{\infty}\frac{2}{k}$, its is not getting any smaller as $k$ gets larger, so perhaps I picked the wrong series of intervals? Could somebody please help me 1) choose a good (finite) series of intervals and 2) show that the inf of the sum of the lengths of those intervals is what I need it to be in order to show that $m^{*}(\emptyset)=0$ for this finite outer measure I have defined.
(If you had hints on the other two parts, that would be great as well!)