Proof of Finite Additivity of Lebesgue Outer Measure for Separated Sets

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 $$\sum_{n=0}^{\infty}l(I_n) and note that we can write each $$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) 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 $$\sum_{k=0}^{\infty}l(S_k) and then show that $$\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.

• 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 $l(J_i ^n)<\alpha$) Jun 17 '20 at 4:04
• @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. Jun 17 '20 at 12:51
• 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. Jun 17 '20 at 17:30
• @Masacroso How specifically can I control the overlap when I create my finite collection of subintervals? Jun 17 '20 at 17:41

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 $$\epsilon >0$$.$$\Box$$