# Using hint to prove a result about the Lebesgue outer measure.

I'm trying to solve this but without success.

The Question

Prove that: if $F_1$ $F_2$ are bounded closed sets in $\mathbb{R}$, so $m^*(F_1\cup F_2)=m^*(F_1)+m^*(F_2)$ where $m^*$ denote the outer measure of Lebesgue.

The Hint

Show that two bounded closed sets that $F_1 \cap F_2 = \emptyset$ are at a distance $\delta>0$. Show that there is a covering of $F_1 \cup F_2$ by a countable family of open intervals of lengths smaller than $\delta$.

What I thought

I know that using this hint the exercise becomes easy (simply apply the definition of the outer Lebesgue measure). However my problem is in using it.

How can I show that every pair of closed can be covered by a countable family of open? What (I get is all the closed are the countable intersection of open sets.)

This $\delta$ exists by the Hausdorff property, but why cover this union with these open intervals of length smaller than $\delta$ help?

Note that the definition of Lebesgue outer measure is that given any subset $A$ of $\mathbb{R}$, its outer measure is defined by $$m*(A)=\inf\{\sum_{n=1}^\infty m(I_n)\mid A\subset \bigcup_{n=1}^\infty I_n\}$$ Where the $I_n$ are open intervals and $m(I_n)$ denotes its length. This definition requires that your covers be all countable, so you can always assume that your covers are countable. I don't know very much about the Hausdorff property that you say, but another approach to show that there's such $\delta$ is the next: since $F_1$ and $F_2$ are closed and bounded, by the Heine-Borel property, they are compact. You can consider then the continuous function $\rho_{F_1}:F_2\to \mathbb{R}$ given by $\rho_{F_1}(x)=\inf\{|x-y|\mid y\in F_1\}$ and deduce that such a $\delta$ is the minimum of the function.

Once we have that the distance between the sets are positive, note usually, to show that $m^*(F_1\cup F_2)=m^*(F_1)+m^*(F_2)$ we show that the inequalities $$m^*(F_1\cup F_2)\leq m^*(F_1)+m^*(F_2)\\ m^*(F_1\cup F_2)\geq m^*(F_1)+m^*(F_2)$$ Both hold. The first one is obvious by monocity of the outer measure, so the problem is showing that the second holds. Note, however, that given any cover of a subset $A$ by open sets, $\{I_n\}_{n\in\mathbb{N}}$ by the archimedean property of $\mathbb{R}$, we can assume that each $I_n$ has length less than $\delta$. To solve the problem, you can take an $\varepsilon>0$ and a cover $\{I_n\}$ of $F_1\cup F_2$ such that $$m^*(F_1\cup F_2)>\sum_{n=1}^\infty m(I_n)-\varepsilon$$ Again, assume that each $I_n$ has length less that $\delta$ and separe the cover in the next way:

The set $\{J_j\}$ will be formed by the intervals that share at least one point with $F_1$.

The set $\{H_h\}$ will be formed by the intervals that share at least one point with $F_2$.

The set $\{K_k\}$ will be formed by the intervals that don't share points with $F_1$ and $F_2$.

Just note that by the assumptions, we have that $F_1\subset \bigcup_j J_j$, $F_2\subset \bigcup_h H_h$ , that for every $j$ and for every $h$ we have $F_1\cap H_h=F_2\cap J_j=\emptyset$ and that $$\sum_{n=1}^\infty m(I_n)= \sum_j m(J_j)+\sum_h m(H_h)+\sum_k m(K_k)$$

Edit:

To conclude your result, note that since $m^*$ is defined as an infimum, we will have $$m^*(F_1)\leq \sum_j m(J_j)\\ m^*(F_2)\leq \sum_h m(H_h)$$ Since $\sum_k m(K_k)\geq 0$, then $$m^*(F_1\cup F_2)+\varepsilon> \sum_{n=1}^\infty m(I_n)\geq \sum_j m(J_j) +\sum_h m(H_h)\geq m^*(F_1)+m^*(F_2)$$ Then, $m^*(F_1)+m^*(F_2)<m^*(F_1\cup F_2)+\varepsilon$. Since this is valid for every $\varepsilon>0$ we conclude that $m^*(F_1)+m^*(F_2)\leq m^*(F_1\cup F_2)$ as we wanted.

• Can you edit this: \m^*(F_1\cup F_2)>\sum_{n=1} m{I_n)-\varepsilon ? – Felipe Feb 1 '14 at 16:35
• Sorry, I have edited it – Brandon Feb 1 '14 at 17:28
• Can you assume that each $I_n$ has length less than $\delta$ and assume both $m^*(F_1\cup F_2)>\sum^\infty_{n=1} m(I_n)-\varepsilon$? What follows for the last part? $\sum_h m(H_h)=0$?! – Felipe Feb 2 '14 at 11:57
• Yes, we can assume both statements. The second one is essentially the definition of Lebesgue outer measure as an infimum. I have edited again my answer to show how to finish the problem. – Brandon Feb 2 '14 at 15:00
• I am not satisfied with this statement. You can argue anything more about it in detail? The impression I have is that if we fix intervals shorter than $\delta$ can guarantee anything beyond that (we can not guarantee that this inequality holds because we are fixing a priori a size to those intervals). I must be wrong but you can convince me? – Felipe Feb 2 '14 at 17:00