Every set of finite measure can be written as the disjoint union of finitely many measurable sets with measure $\le \epsilon$

Question

Let $E\subseteq \mathbb{R}$. Show that if $E$ has finite measure and $\epsilon>0,$ then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\epsilon$.

I want to use the Caratheodory condition of measurability. Let $\{I_x\}$ be a finite family of intervals such that length of each $I_{x}<\epsilon$, then $L(I_x)=m^*(E\cap I_x)+m^*(E^c\cap I_x).$ We can see $m^*(E\cap I_x)＜\epsilon$, but then I can't go on. This problem comes from Royden's Real Analysis (page 40).

The method from How can I find a subset of a set with “half the size” of the original? allows you to produce a subset of any given measure that is smaller than the set's measure. You can then produce your disjoint union by induction, chipping off subsets of given size $\leq\epsilon$ from $E$. — Conifold, Sep 20 '20 at 03:21

score 5 · Accepted Answer · answered Sep 20 '20 at 04:58

5

Let $E_n = E \cap [n\epsilon,(n+1)\epsilon)$, we have $\sum_k m E_k < \infty$.

Choose $K$ such that $\sum_{|k| >K} m E_k < \epsilon$. Let $E'=\cup_{|k|>K} E_k$.

Then $E_{-K},E_{-K+1},...,E_K, E'$ are disjoint, measurable and each have measure $\le \epsilon$.

answered Sep 20 '20 at 04:58

copper.hat

172,524

How to choose Such K – user158796 Sep 20 '20 at 05:49
You are given that the measure is finite so the sequence converges. – copper.hat Sep 20 '20 at 05:51
My understand is :since m(E) is finite then when – user158796 Sep 20 '20 at 05:56
My understand is: since m(E) is finite then when – user158796 Sep 20 '20 at 05:57
My understand is: since m(E) is finite the when |k| is a large number ∪En is a small number – user158796 Sep 20 '20 at 06:00
Review convergence of series. – copper.hat Sep 20 '20 at 06:01

Every set of finite measure can be written as the disjoint union of finitely many measurable sets with measure $\le \epsilon$

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