I'm trying to prove that if $E \subset \mathbb{R}$ has finite measure and $\varepsilon \in \mathbb{R}$ such that $\varepsilon>0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\varepsilon$.

Here is what I've done:

Let $\varepsilon \in \mathbb{R}$ such that $\varepsilon>0$. Since $m^{*}(E) < +\infty$, there exists a countable collection $\{I_{k}\}_{k \in \mathbb{Z}^{+}}$ of open and bounded intervals (measurable sets) covering $E$ and such that $\sum \limits_{k=1}^{\infty } \ell ( I_{k})<m^{*}( E) +\varepsilon /2$. Define $G_{1}:=I_{1}$, and for each $k\in \mathbb{Z}^{+}$ such that $k\geq 2$ define $G_{k}:=I_{k}\setminus\cup_{n=1}^{k-1} I_{n}$. Therefore ${\{G_{k}\}}_{k\in \mathbb{Z}^{+}}$ is a countable disjoint collection of bounded and measurable sets such that $G:=\cup _{k\in \mathbb{Z}^{+}}G_{k}=\cup _{k\in \mathbb{Z}^{+}}I_{k}$. So, $m^{*}( G) <m^{*}( E) +\varepsilon /2$ and $E\subseteq G$. Now define $F_{k}:=\overline{G}_{k}$ (the closure of $G_k$). Since $G_{k}\subseteq I_{k}$, we know that $F_{k}\subseteq \overline{I_{k}}$; i.e., each $F_{k}$ is a closed and bounded subset of $\mathbb{R}$. By the Heine-Borel Theorem we conclude that each $F_{k}$ is compact.

If $x\in \mathbb{R}$, then we define $N( x,\varepsilon) := (x-\varepsilon/2,x+\varepsilon/2)$. For each $k\in \mathbb{Z}^{+}$ the collection $\{N( x,\varepsilon ) \colon x\in F_{k}\}$ is an open covering of $F_{k}$. The compactness of each $F_{k}$ implies that there exist $x_{1}^{k},\ldots ,x_{n_{k}}^{k}\in F_{k}$, with $n_{k}\in \mathbb{Z}^{+}$, such that $F_{k}\subseteq \bigcup _{i=1}^{n_{k}}N(x_{i}^{k},\varepsilon ) $. Thus $E\subseteq \cup _{k\in \mathbb{Z}^{+}}G_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}F_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}(\cup_{i=1}^{n_{k}}N( x_{i}^{k},\varepsilon ))$ ...

I can't figure out how to make this last union over $k$ finite. I suppose I've to use the fact that $m^*(G)<m^*(E)+\varepsilon/2$. I can't see how! If that union is finite, then the problem is solved; one would just have to make some intersections and take some complements.

On the other hand, I think that if we remove the hypothesis of finite outer measure on $E$, then the conclusion is the same but with a countably infinite union. Am I right?

Thank you in advance.

  • $\begingroup$ I edited my answer and added a disclaimer because 2 of my approaches had assumed that $E$ is measurable. However, the problem is false as stated if $E$ is nonmeasurable. A finite (or countable) union of measurable sets is measurable. Is it assumed that $E$ is measurable, or are the sets only required to have outer measure less than $\varepsilon$ (with no assumption of measurability)? $\endgroup$ – Jonas Meyer Jan 25 '11 at 1:58
  • $\begingroup$ (On second thought, only one of my approaches uses measurability.) $\endgroup$ – Jonas Meyer Jan 25 '11 at 2:06
  • 1
    $\begingroup$ Another possible correction of the problem is that $E$ is only supposed to be contained in the disjoint union, rather than equal to it. In that case, my first approach would still work; Just take $I_1,\ldots,I_n$ together with an open set $U$ of measure less than $\varepsilon$ such that $U\subset(\mathbb{R}\setminus[-M,M])$ and $E\setminus[-M,M]\subseteq U$. $\endgroup$ – Jonas Meyer Jan 25 '11 at 2:12
  • $\begingroup$ @Jonas: First of all, thank you for your help! I'll work on your idea. On the other hand, the actual statement of the problem says "$E$ has finite measure". So, I think I've made a mistake by adding the word "outer". I'm a little confused about that, but I suppose that "finite measure" means that $E$ is actually measurable. Sorry! I'll make the corresponding changes. Also, I'll update my approach if I can make it work. $\endgroup$ – ragrigg Jan 25 '11 at 2:19
  • $\begingroup$ OK, thanks! In that case, the paragraph in my answer that mentions Littlewood's 3 principles is the one that I thought aligns most closely with your approach. $\endgroup$ – Jonas Meyer Jan 25 '11 at 2:22

Note: The problem is false as stated. Either $E$ has to be measurable, or some of the sets in the disjoint union must be nonmeasurable.

An easier approach is to first show that there is an $M$ such that $m^*(E\setminus[-M,M])<\varepsilon$, and break up $[-M,M]$ into a finite number of disjoint intervals $I_1,\ldots,I_n$ of length less than $\varepsilon$. Then $E\setminus[-M,M],E\cap I_1,\ldots,E\cap I_n$ will do the trick. (This gives sets of outer measure less than $\varepsilon$, which cannot all be measurable unless $E$ is.)

If you only want to have $E$ contained in the union rather than equal to it, then you can take the sets to be measurable. Let $U$ be an open set containing $E\setminus{[-M,M]}$ of measure less than $\varepsilon$, and intersect it with $\mathbb{R}\setminus[-M,M]$ if necessary to make it disjoint with $[-M,M]$. Then $U,I_1,\ldots,I_n$ will do the trick.

Here's another approach (for the case of equality of the union, where the sets may not be measurable). Let $n$ be such that $n\varepsilon\gt m^*(E)$ and $(n-1)\varepsilon\leq m^*(E)$, and proceed by induction on $n$. The base case is that the outer measure is less than $\varepsilon$ already, and for the inductive step you can apply the intermediate value theorem to the function $f(t)=m^*(E\cap[-t,t])$ to find and remove a subset of outer measure $\varepsilon$. For what it's worth, this would give you the smallest possible number of sets.

The answer to your question about countable unions is "yes". You can intersect $E$ with intervals of the form $[n\varepsilon,(n+1)\varepsilon)$. (Again, you can't get the sets to be measurable unless $E$ is.)

I'm not sure how to salvage your approach.

Edit: The following approach works if $E$ is measurable. I hastily missed the fact that that is not assumed.

If $E$ is measurable, then here is an approach more in line with yours. You could prove the useful fact that there is a finite collection of intervals $I_1,\ldots I_n$ such that the measure of the symmetric difference of $E$ with $\cup_k I_k$ is less than $\varepsilon$. This is (one version of) one of Littlewood's 3 principles. One way to start the proof would be to find an $M$ similar to above, and then to use a compactness argument along with the definition of measure. After you have this, you already have control on the stuff outside of $\cup_k I_k$, and the stuff inside $\cup_k I_k$ could be handled by a method similar to yours.

  • 1
    $\begingroup$ I'm not going to edit any more right now. I just learned from the OP that in all likelihood $E$ is assumed to be measurable, so all the warnings above can safely be ignored. $\endgroup$ – Jonas Meyer Jan 25 '11 at 2:23

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.