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!)
Best Answer
Firstly, note that $\varnothing \subseteq I , \forall I\in O(\mathbb{R})$ where $O(\mathbb{R})$ denotes the collection of all open intervals in $\mathbb{R}$ . Secondly, since the measure $l:O(\mathbb{R}) \rightarrow [0,+\infty)$ is nonnegative, we know that $0 \leq \sum_{i \in \mathbb{N}}l(I_i)$ for all $I_i \in O(\mathbb{R}), i \in \mathbb{N}$.
Now fix $\epsilon > 0$ and set $I_i=(-\epsilon/2^{i+1},\epsilon/2^{i+1}), i=1,\dotsc,n$, which is in $O(\mathbb{R})$. Then $$0 \leq m^*(\varnothing)=\inf\{\sum_{i=1}^n l(R_i): \varnothing \subseteq \bigcup_{i=1}^n R_i, R_i \in O(\mathbb{R}) \} \leq \sum_{i=1}^n l(I_i) \leq \sum_{i=1}^\infty l(I_i)=\epsilon.$$ Since $\epsilon>0$ is arbitrary, we conclude that $m^*(\varnothing)=0$ for your "finite" outer measure definition. It may be worth mentioning that the countable finite covering has serious implications in some relevant examples. Take $E=\mathbb{Q}\cap [0,1]$ and find $m^*(E)$ in comparison with the countable infinite version $\mu^*(E)$ where $\mu^*(E)=\inf\{\sum_{i=1}^\infty l(R_i): E \subseteq \bigcup_{i=1}^\infty R_i, R_i \in O(\mathbb{R}) \} $, taking into account the denseness of $\mathbb{Q}$ in $[0,1]$.
Extending your question to higher dimensions, if we work with intervals in $O(\mathbb{R^d})$, and then with the product measure $\mu$, we can put instead $I_i=(a_1,b_1)\times \cdots \times (a_d,b_d)$ where $a_k=-\epsilon^{(1/d)}/2^{i/d +1}$ and $b_k=\epsilon^{(1/d)}/2^{i/d +1}, k=1,\dotsc,d$, for some given $\epsilon >0$. Then $\mu(I_i)=(a_1,b_1)\dotsc (a_d,b_d)=(\epsilon^{(1/d)}/2^{i/d})^d=\epsilon/2^{i}$. Therefore, analogously,
$$0\leq m^*(\varnothing)=\inf\{\sum_{i=1}^n \mu(R_i): \varnothing \subseteq \bigcup_{i=1}^n R_i, R_i \in O(\mathbb{R^d}) \} \leq \sum_{i=1}^n \mu(I_i) \leq \sum_{i=1}^\infty \mu(I_i)=\epsilon. $$
From this we have that $m^*(\varnothing)=0$.