The question and its answer are given below:
Problem: Show that if $E$ has finite measure and $\varepsilon>0$, then E is the disjoint union of a finite number of measurable sets, each of which has measure at most $\varepsilon$.
Solution First assume $E\subseteq [-M, M]$ is bounded. Then a finite number of disjoint intervals of the form $J_k := \left(\frac{k\varepsilon}2, \frac{(k+1)\varepsilon}2\right]$ will cover $[-M, M]$ (by the Archimedean property) and thus $E$. Since $E_k = M\cap J_k$ is measurable of measure $\leq \varepsilon$ we have $E$ a disjoint union of measurable sets $E_k$ with $m(E_k)<\varepsilon$.
If $E$ is unbounded cover $E$ by a countably infinite collection $(I_k)$ of bounded open intervals such that $\sum_k \ell(I_k)<m(E) + 1$. Since the series converges there exists $N$ such that $\sum_{k = 1}^N\ell(I_k)<\varepsilon$. Let $E_{1} := E \cap \left(\bigcup_{N+1}^{\infty}I_{k}\right)$. Then $E_1$ is measurable of measure $<\varepsilon$ and $E_2 := E\setminus E_1\subseteq \bigcup_{k = 1}^N$ is bounded and can be written by the above as a finite union of disjoint measurable sets with each of measure $\leq \varepsilon$.
My questions are:
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I do not understand how by Archimedian property a finite number of disjoint intervals will cover $[-M, M]$ and why we are sure that the intervals are disjoint?
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Why the author of the answer considered the case "if $E$ is unbounded", does not the question said directly that $E$ has a finite measure and hence bounded so this case must not be considered ….. am I correct?
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Why we are sure that the series converges in the unbounded case?
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What is the intuition behind defining $E_{1} := E \cap \left(\bigcup_{N+1}^{\infty}I_{k}\right)$?
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I found a solution to the problem here Every subset of $\mathbb{R}$ with finite measure is the disjoint union of a finite number of measurable sets but I am confused about a logical ordered solution for this problem as the question here seems to discuss cases that are not in the question, so a logical sequence of steps for the solution will be greatly appreciated.
Thanks!
Best Answer
For any $\varepsilon>0,$ we have $$ (\frac{k}{2}\varepsilon,\frac{k+1}{2}\varepsilon]\cap (\frac{j}{2}\varepsilon,\frac{j+1}{2}\varepsilon]=\emptyset $$ if $k\neq j$ simply by construction. The Archimedean principle tells you that there is some $n\in \mathbb{N}$ such that $n> M/(\varepsilon/2),$ in which case the intervals $(\frac{k}{2}\varepsilon,\frac{k+1}{2}\varepsilon]$ for $-n\leq k\leq n$ will cover $[-M,M]$.
There are unbounded sets of finite measure. Consider $E=\mathbb{N}$. Then $m(E)=0,$ but $E$ is unbounded.
The series converges by construction: All of its elements are positive and picked so that $\sum_{k} \ell(I_k)\leq m(E)+1,$ which is finite by assumption. Are you asking why it is possible to do this?
We want all of the sets in our disjoint union to have measure at most $\varepsilon>0,$ irrespective of what $m(E)$ is. The intuitive statement here is that "all of $E$'s measure except for, at most, $\varepsilon$, lives inside some bounded interval." The intersection considered is "the rest" - the unimportant bit that does have measure less than $\varepsilon$.
I'm not sure I follow: Would you like a second way of proving the statement?