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How does Axler know he has found the infimum?

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I am reading the following example from Measure, Integration & Real Analysis by Sheldon Axler about the outer measure:

Suppose $\displaystyle A=\{ a_1,a_2,...,a_n \}$ is a finite set ofreal numbers. Suppose $\varepsilon > 0$. Define a sequence$\displaystyle I_1,I_2,...$ of open intervals by:

$\displaystyle \displaystyle I_k=\left\{\begin{array}{ll}(a_k-\varepsilon,a_k+\varepsilon)&\text{if }k\le n\\\emptyset&\text{if }k>n\end{array}\right.$

Then $I_1,I_2,...$ is a sequence of open intervals whose union contain$A$. Clearly $\sum_{k=1}^{\infty}\ell(I_k)=2\varepsilon n$. Hence$\left| A\right|\le2\varepsilon n$. Because $\varepsilon$ is anarbitrary positive number, this implies that $\left| A\right| = 0.$

I am wondering how Axler knows he has found the infimum of all sums of lengths of open intervals. As far as I can tell, the outer measure of a set can never be less than $0$. Obviously, $\left| A\right|\le \sum_{k=1}^{\infty}\ell(I_k)$, for any sequence $(I_k)$. So since Axler can find one sequence, such that $\left| A\right|\le2\varepsilon n$, $\varepsilon$ can be made arbitrarily small, and $0\le\left| A\right|$, we know the infimum has to be $0$. Therefore $\left| A\right|=0$.

Have I understood his example correctly?


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