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 $$ \sum_{n=1}^\infty m(I_n)= \sum_j m(J_j)+\sum_h m(H_h)+\sum_k m(K_k) $$


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.

  • $\begingroup$ Can you edit this: \m^*(F_1\cup F_2)>\sum_{n=1} m{I_n)-\varepsilon ? $\endgroup$ – Felipe Feb 1 '14 at 16:35
  • $\begingroup$ Sorry, I have edited it $\endgroup$ – Brandon Feb 1 '14 at 17:28
  • $\begingroup$ 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? $\sum_h m(H_h)=0$?! $\endgroup$ – Felipe Feb 2 '14 at 11:57
  • $\begingroup$ 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. $\endgroup$ – Brandon Feb 2 '14 at 15:00
  • $\begingroup$ 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? $\endgroup$ – Felipe Feb 2 '14 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.