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

Solution 1:

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.