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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?

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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 $r\le m^*(A)$.

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  • $\begingroup$ (note that I left some work for you to do, but this should get you started and should, in particular, address your question about the "surplus", $\varepsilon$ in this case) $\endgroup$
    – Thomas
    Oct 15 '15 at 5:14
  • $\begingroup$ In the other direction, is what I did okay?: Suppose $A \subseteq \sum_{k=1}^{\infty}I_{k}$, where $I_{i}=(a_{i},b_{i})$. Then, $\forall I_{i}$, we have $(a_{i},b_{i})\cup\{ a_{i}\} \cup \{b_{i} \} = [a_{i},b_{i}]$, and $m^{*}(\{a_{i}\})=m^{*}(\{b_{i}\}) = 0$. So, $m^{*}(I_{i})=m^{*}(\overline{I}_{i}) \to l(I_{i})=l(\overline{I}_{i})$, $\forall i$. Thus, $(a_{i},b_{i})\subseteq [a_{i},b_{i}]\, \forall i$. Now, $m^{*}(A)=\{ \inf\sum_{k=1}^{\infty}l(I_{k})\} \leq \sum_{k=1}^{\infty}l(I_{k}) = \sum_{k=1}^{\infty}l(\overline{I}_{k})$. (Continued in next comment) $\endgroup$
    – user100463
    Oct 15 '15 at 16:42
  • $\begingroup$ (continued from above) Now, since $\inf \{\sum_{k=1}^{\infty}l(\overline{I}_{k})\}=c^{*}(A)$, $c^{*}(A) + \epsilon$ cannot be a lower bound for $\sum_{k=1}^{\infty}l(\overline{I_{k}})$. Therefore, $\sum_{k=1}^{\infty}l(\overline{I}_{k}) \leq c^{*} + \epsilon$, $\forall \epsilon > 0$. So, $m^{*}(A) \leq c^{*}(A)$. $\endgroup$
    – user100463
    Oct 15 '15 at 16:44
  • $\begingroup$ @JessyCat I added a note to the answer addressing your question. $\endgroup$
    – Thomas
    Oct 15 '15 at 16:51
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    $\begingroup$ @tangentbundle The sum of the volumes of the cover given by the closed $I_k$ is shown to have a value $V$ which is at most $\varepsilon >0$ smaller than the one of the open cover constructed from it. Since the $\inf$ over all open covers is certainly smaller than the value for the one I have constructed I have shown that $V > m^*(A) - \varepsilon$. Now $V$ is arbritrarily close to $c^*(A)$ (you need one more, similar, $\varepsilon$ reasoning to make this precise.) Since $\varepsilon$ is arbitrary I have shown the inequality you want to see. $\endgroup$
    – Thomas
    Sep 26 '19 at 6:56

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