New Outer Measure Question Let $m^*(A)$ be the outer measure of a set $A$ $\in$ $\mathbb{R}$. This is defined by
$m^∗(A)$= $inf(Z_A)$ where $Z_A$={ $\sum_{n=1}^\infty l(I_n)$ : $I_n$ are intervals, $A\subseteq$ $\bigcup_{n=1}^\infty l(I_n)$}. 
We want to define a new outer measure. First we choose $\epsilon > 0$. Now we define $n^*_{\epsilon}(A)$ as the outer measure $m^*(A)$. The only difference is that intervals in the covering at least have length $\epsilon$. We now let $\epsilon$ go to $0$ en define 
$n^*(A)=\lim_{\epsilon \downarrow 0} n^*_{\epsilon}(A)$. 
Show that $n^*(A)=m^*(A)$ if A is compact.
I don't know how to start. I know that a compact set A has a finite covering. We can use that but i dont have an idea how to too. Can u help me to solve this?
 A: Note: throughout this proof, I assume that by interval, you mean an open interval, in the spirit of the standard definition of the Lebesgue outer measure.
First, note $m^*(A)$ is an infimum over a set of numbers larger than is the set defining $n_{\varepsilon}^*(A)$ for any $\varepsilon>0$ (given that in the former definition the lengths of the intervals are unrestricted), so that $$m^*(A)\leq n^*_{\varepsilon}(A)\quad\forall\varepsilon>0.$$ Therefore, $m^*(A)\leq n^*(A)$ is preserved in the limit.
[Note: The limit $\lim_{\varepsilon\to 0}n_{\varepsilon}^*(A)$ actually exists, since as you take smaller $\varepsilon$'s, the values of $n_{\varepsilon}^*(A)$ will get smaller. This is because as you relax the lower bound on the lengths of the intervals you take infima over larger sets. Therefore, $n_{\varepsilon}^*(A)$ is decreasing in $\varepsilon$ and is bounded below by $m^*(A)$, so that the limit exists.]
To obtain a contradiction, suppose that $m^*(A)<n^*(A)$. Since $n^*(A)\leq n_{\varepsilon}^*(A)$ (see the argument above in brackets), it follows from the definitions of the infima defining $m^*(A)$ and $n_{\varepsilon}^*(A)$ that there exist intervals $\{I_j\}_{j=1}^{\infty}$ such that $A\subseteq \bigcup_{j=1}^{\infty}I_j$ and, for all $\varepsilon>0$,
$$(\spadesuit)\quad \sum_{j=1}^{\infty}\ell(I_j)<\sum_{j=1}^{\infty}\ell(J_j)$$ for any collection of intervals $\{J_j\}_{j=1}^{\infty}$ such that $A\subseteq \bigcup_{j=1}^{\infty} J_j$ and $\ell(J_j)\geq\varepsilon$ for all $j\in\mathbb{Z}_+$.
Given that $A$ is compact and $A\subseteq \bigcup_{j=1}^{\infty}I_j$, there exists a finite subcollection of these intervals (I label them without loss of generality as $\{I_1,\ldots,I_m\}$) such that $A\subseteq \bigcup_{j=1}^{m}I_j$ for some $m\in\mathbb{Z}_+$. (Note: here's when I exploit the assumption that the intervals in question are open.) Let $$\varepsilon\equiv\min_{j\in\{1,\ldots,m\}}\ell (I_j)>0.$$ Then, $\ell(I_j)\geq\varepsilon$ for all $j\in\{1,\ldots,m\}$. But this contradicts $(\spadesuit)$, since we have found a covering of $A$ by open intervals of lengths of at least $\varepsilon$, but $$\sum_{j=1}^m \ell(I_j)\leq\sum_{j=1}^{\infty} \ell(I_j).$$
