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I'm currently looking through a proof that Lebesgue outer measure, $m^*$, is finitely additive for bounded sets $A$ and $B$ if we have $$\forall a \in A\space\forall b \in B\space(|a-b|≥\alpha)$$ with $\alpha>0$.

Since we have countable subadditivity, all that must be shown is $m^*(A)+m^*(B) ≤ m^*(A\cup B)$.

In the proof that I am looking at, we take a countable cover of $A\cup B$ of nonempty bounded, open intervals $\{I_{n}:n≥0\}$ such that $$\sum_{n=0}^{\infty}l(I_n)<m^*(A\cup B)+\frac{\epsilon}{2}$$ and note that we can write each $I_{n}$ as a finite union of open subintervals $\{J_i^n:i≤K(n)\}$ with $l(J_i^n)<\alpha$ for each $i≤K(n)$. This part is clear, but the author also says that in addition to each $l(J_i^n)<\alpha$, we also have $$\sum_{i=0}^{K(n)}l(J_i^n)<l(I_n)+\frac{\epsilon}{2^{n+2}}$$ It's this last inequality that has me confused.

I've tried all sorts of things to convince myself this is true. I tried to invoke a cover of nonempty bounded, open intervals of $I_n$, say $\{S_k:k≥0\}$, such that $$\sum_{k=0}^{\infty}l(S_k)<l(I_n)+\frac{\epsilon}{2^{n+2}}$$ and then show that $$\sum_{i=0}^{K(n)}l(J_i^n)≤\sum_{k=0}^{\infty}l(S_k)$$ but couldn't get it to work. I feel like there should be a constructive way to show this. I'd greatly appreciate any clarification.

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    $\begingroup$ the inequality $\sum_{i=0}^{K(n)}l(J_i^n)<l(I_n)+\frac{\epsilon}{2^{n+2}}$ just mean that the overlap of the intervals $J_i ^n$ can be chosen arbitrarily small (regardless of the condition of $l(J_i ^n)<\alpha $) $\endgroup$
    – Masacroso
    Jun 17 '20 at 4:04
  • $\begingroup$ @Masacroso I was thinking about that, but why is there any reason to think that when I write this finite union that there won't be a lot of overlap with the subintervals? It seems like I only have control over their length, not their position with respect to one another. $\endgroup$
    – Newman
    Jun 17 '20 at 12:51
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    $\begingroup$ for the real line this is easy to see because any interval have the form $(a,b)$. In $\mathbb{R}^n$, where the intervals are replaced by boxes, is a bit more complicate to see it. $\endgroup$
    – Masacroso
    Jun 17 '20 at 17:30
  • $\begingroup$ @Masacroso How specifically can I control the overlap when I create my finite collection of subintervals? $\endgroup$
    – Newman
    Jun 17 '20 at 17:41
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Let $I:=(a,b)$ any bounded interval in $\mathbb{R}$. Then for any chosen $\alpha ,\epsilon >0$ there are intervals $I_1,\ldots ,I_n$ with $l(I_k)<\alpha $ for each $k$ and $\sum_{k=1}^n l(I_k)< l(I)+\epsilon $.

Proof: we have that $l(I)=b-a$, then for large enough $n$ we have that $r:=\frac{b-a}{n}<\alpha $, and setting $I_k:=(a+(k-1)r,a+k(r+\delta ))$ we find that $l(I_k)=r+k\delta $. Choosing $0<\delta <\min\{\frac{\alpha -r}n,\frac rn\}$ we have that $I_j \cap I_k=\emptyset $ when $|j-k|>1$, $l(I_k\cap I_{k+1})=k\delta $ for each $k$ and of course $I\subset \bigcup_{k=1}^nI_k$.

And $$ \sum_{k=1}^n l(I_k)-l\left(\bigcup_{k=1}^nI_k\right)=\sum_{k=1}^{n-1}l(I_k \cap I_{k+1})=\delta \sum_{k=1}^{n-1}k=\delta \frac{n(n-1)}{2}\\ \text{ and }\quad l\left(\bigcup_{k=1}^nI_k\right)-l(I)=a+n(r+\delta )-b=n\delta \\ \therefore\quad \sum_{k=1}^n l(I_k)-l(I)=\delta \left(n+\frac{n(n-1)}{2}\right)=\delta \frac{(n+1)n}{2} $$

So choosing $\delta $ enough small we find that $\sum_{k=1}^n l(I_k)-l(I)<\epsilon $ for arbitrarily chosen $\epsilon >0$.$\Box$

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