Let $\mu$ be the Lebesgue measure and, $E$ and $F$ compact subsets of $\mathbb{R}$ such that $E\subset F$ and $\mu(E)=1$ and $\mu(F)=3$. Prove there is a compact subset $K$ of $\mathbb{R}$ such that $E \subseteq K \subseteq F$ and $\mu(K)=2$. First and foremost, as $\mu(E)=1$ and $\mu(F)=3$ are compact subsets of $\mathbb{R}$ there are open subsets of $\mathbb{R}$, $E'$ and $F'$, such as $E \subset E'$ and $F' \subset F$ and $\mu^{\ast}(E'- E) < \epsilon$ and $\mu^{\ast}(F'- F) < \epsilon$. Where $\mu(E')=\mu(E)=1$ and $\mu(F')=\mu(F)=3$. Obviously, $\mu^{\ast}$ stands for Lebesgue outer measure. So far, I´ve been run out of ideas and aproaches to solve this. I have googled properties about Lebesgue measure of compact sets and there is nothing about them that I have found helpful, for instance, all compact sets are Lebesgue measurable and $\mu(E)< \infty$ for every compact set in $\mathbb{R}$. But I dont know how I´m supposed to construct this compact set $K$ such as $E \subseteq K \subseteq F$ and $\mu(K)=2$. Thanks!
$E$ and $F$ measurable compact sets of $\mathbb{R}$ such $\mu(E)=1$ and $\mu(F)=3$. There exist $K$ compact such $\mu(K)=2$.
lebesgue-measuremeasure-theoryreal-analysis
Related Solutions
Assuming you're talking about Lebesgue measure on $\mathbb{R}$ or $\mathbb{R^n}$, then here's one approach.
Without loss of generality, we may assume that $E$ is contained in a bounded subset of $\mathbb{R^n}$ (by splitting it up into countably parts if need be).
Recall the method of defining the Lebesgue measure of a set by considering outer measure, where we try to find a minimal covering by your (open) generating sets. I think you can see this by looking at a proof of Caratheodory's Extension theorem. From this it follows that you can approximate from outside by open sets. In a bounded set, we may take complements without worry about infinite measure cropping up, so we can use the same logic to show that we can approximate from inside by closed sets. Recalling that closed, bounded sets are compact, this shows the result for bounded E, which we can then extend by countable additivity.
You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that $$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$ And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set. Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).
Best Answer
Hint:
Consider the sets $[-r,r] \subseteq \mathbb{R}$.
Can you show $r \mapsto \mu(E \cup [-r,r] \cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us?
I hope this helps ^_^