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Let $A,B$ be bounded sets in $\Bbb R$ for which $\exists \alpha>0$ such that $|a-b|>\alpha$ for any $a\in A,b\in B$ .

Prove that $m^*(A\cup B)=m^*(A)+m^*(B)$.

By countable sub-additivity of outer measure $m^*(A\cup B)\le m^*(A)+m^*(B)$.

To show the reverse inequality let $\epsilon>0$. Then there exists a sequence of open intervals $\{I_n\}$ such that $A\cup B\subset \cup I_n$ and $m^*(A\cup B)+\epsilon >\sum l(I_n)$.

If I can show that for each sequence of intervals $\{I_n^{'}\}$ and $\{I_n^{"}\}$ covering $A,B$ respectively we have $\sum l(I_n)>\sum l(I_n^{'})+\sum l(I_n^{"})$ then we are done.

But I am stuck .Any help.

1 Answers 1

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Fix $\epsilon > 0$. Then there is a countable set $\{I_n \}$ of intervals such that $$\sum_n l(I_n) < m^*(A \cup B) + \epsilon/2.$$

Write each open interval $I_n$ as a finite union of open subintervals $\{ J_{i,n}\}_{i=1}^{N(n)}$ such that $l(J_{i,n}) < \alpha$ and $\sum_{i=1}^{N(n)}l(J_{i,n}) < l(I_n) + \frac{\epsilon}{2^{n+1}}$.

Now, the sets $C(A) = \{J_{i,n} : J_{i,n} \cap A \neq \emptyset \}$ and $C(B) = \{J_{i,n} : J_{i,n} \cap B \neq \emptyset \}$ are disjoint open covers of $A$ and $B$, respectively. (Why?)

Hence, we have $$m^*(A \cup B) + \epsilon > \sum_n l(I_n) + \epsilon/2 > \sum_n \sum_{i=1}^{N(n)} l (J_{i,n}) \geq \sum_{J' \in C(A)}l(J') + \sum_{J'' \in C(B)}l(J'') \geq m^*(A) + m^*(B).$$

As $\epsilon$ is arbitrary, we are done.

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    How can each $I_n$ be expressed in terms of $J_i$ as shown – 2017-01-23
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    Maybe you should try to check that part on your own. The idea is to "decompose" $I_n$ into finitely many (not necessarily disjoint) open subintervals $J_{i,n}$, each with length less than $\alpha$. This can be done in such a way that the total overlap of the $J_{i,n}$ is less than $\epsilon/(2^{n+1})$. – 2017-01-23
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    Why don't you try checking a concrete example, say $I_n = (0,1)$ and $\alpha = 1/4$? If you can work that out, you will see that it can be generalized without much difficulty. – 2017-01-23