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

copper.hat · Accepted Answer · 2024-11-25T04:04:22.133

6

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 $\le \epsilon$.

Addendum addressing @ziabadar's question.

Note that the $E_n$ are disjoint and measurable and $\cup_{n \in I} E_n \subset E$ for any index set $I$. For any finite index set $I$ we have (by Proposition 6, p.36 Royden) $m(\cup_{n \in I} E_n) = m^*(\cup_{n \in I} E_n) = \sum_{n \in I} m^* E_n = \sum_{n \in I} m E_n \le m E$, it follows that $\sum_n m E_n \le m E$ (which is finite).

edited Nov 25 '24 at 04:04

answered Sep 20 '20 at 04:58

copper.hat

178,207

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
@copper.hat problem states $m(E)\lt \infty$ but how did you get $\sum_km(E_k) \lt \infty $, please confirm you have used countable addivity of lebesgue measure because royden did not provided proof of countable addivity before this problem. – zia badar Nov 25 '24 at 03:20
@ziabadar It follows from an earlier proposition, I have elaborated above. – copper.hat Nov 25 '24 at 04:04
@copper.hat could you please guide in the direction from where I can understand finite summation inequality $\sum_{n\in I}m(E_n)\leq m(E)$ also satisfies infinite summation inequality $\sum_{n}m(E_n)\leq m(E)$ – zia badar Nov 25 '24 at 13:47
or is it just simple induction. – zia badar Nov 25 '24 at 13:56
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@copper.hat okay I have understood, if equality satisfies for every finite index set then it also satisfies for set of natural numbers or set of integers. – zia badar Nov 25 '24 at 14:34

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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