Lebesgue outer measure is usually defined as a function $m^{*}: \{ \text{subsets of }\,\mathbb{R} \} \to \mathbb{R}_{\geq 0} \cup \{ \infty \} $ where $m^{*}(A)=\inf \left\{\sum_{k=1}^{\infty}l(I_{k})\vert I_{1},\cdots,\, \text{open intervals such that}\, A \subseteq \cup_{k=1}^{\infty}I_{k}\right\}$.
I am being tasked with showing that the following function, $c^{*}: \{ \text{subsets of }\,\mathbb{R} \} \to \mathbb{R}_{\geq 0} \cup \{ \infty \}$ where $c^{*}(A)=\inf \left\{\sum_{k=1}^{n}l(I_{k})\vert I_{1},\cdots,I_{n}\, \text{closed, bounded intervals such that}\, A \subseteq \cup_{k=1}^{n}I_{k}\right\} $, is equal to $m^{*}$. I have been provided with the hint to prove that every closed interval in the cover for $A$ can itself be covered by open intervals with arbitrarily small "surplus" and vice-versa, but I am really very confused and unsure how to do this (especially regarding weird subscripts).
I was thinking to cover each $I_{i}$ with the set of $\{I_{i,j} \}_{j=1}^{n}$, and then the "surplus" would be covering the points in between the n intervals covering $I_{i}$, but I don't know how to express it mathematically. Could somebody please show me how to do this?
Best Answer
if $A$ is covered by $\{I_k|k\in\mathbb{N}\}$, $\varepsilon >0$ is given and $I_k =[a_k,b_k]$ look at $(a_k-\varepsilon_k, b_k + \varepsilon_k)$ where $\varepsilon_k = \frac{\varepsilon}{2^k}$ and note that $\sum\frac{\varepsilon}{2^k}=\varepsilon$. The family covers $A$ and it's volume is only $\varepsilon$ larger than the volume of the original cover. The other direction is similar.
Edit (in response to a question in a comment): if $r := \inf\{\sum l(I_k)\}$, then, by the definition of the infimum, this implies that for any $\varepsilon_* >0$ there is a specific cover of $A$ by closed intervals, say $\{J_k\}_k$, such that $\sum l(J_k) \ge r > \sum l(J_k) - \varepsilon_*$. Let $\varepsilon = \frac{1}{2} \varepsilon_*$ and for $J_k=[a_k,b_k]$ let $J_k^* = (a_k-\varepsilon_k, b_k+\varepsilon_k)$ with $\varepsilon_k$ as in the first part of the answer.
Then a short computation shows that $\sum l(J_k^*) = \sum l(J_k) +\varepsilon_*$, so $\sum l(J_k^*) - \varepsilon_* \ge r > \sum l(J_k^*) - 2\varepsilon_*$. So $r+\varepsilon_* \le m^*(a)$ for any $\varepsilon_*>0$, which implies $r\le m^*(A)$.