# Every subset of $\mathbb{R}$ with finite measure is the disjoint union of a finite number of measurable sets

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?

• 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)? – Jonas Meyer Jan 25 '11 at 1:58
• (On second thought, only one of my approaches uses measurability.) – Jonas Meyer Jan 25 '11 at 2:06
• 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 $E\setminus[-M,M]\subseteq U$. – Jonas Meyer Jan 25 '11 at 2:12
• @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. – ragrigg Jan 25 '11 at 2:19
• 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. – 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.

• 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. – Jonas Meyer Jan 25 '11 at 2:23