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:

  1. 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?

  2. 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?

  3. Why we are sure that the series converges in the unbounded case?

  4. What is the intuition behind defining $E_{1} := E \cap \left(\bigcup_{N+1}^{\infty}I_{k}\right)$?

  5. 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.


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    $\begingroup$ I typed the solution more or less as it was in the image, but beware that there was a typo in there (which I chose to faithfully repeat): The fact that the series converges tells us that there is an $N$ such that $\sum_{k = N}^\infty \ell (I_k)<\varepsilon$, not $\sum_{k = 1}^N$. Also, the solution mixes up $\leq \varepsilon$ and $<\varepsilon$, and in some places that made it actually incorrect. $\endgroup$
    – Arthur
    Sep 9, 2019 at 12:22
  • $\begingroup$ You mean the solution has a typo?@Arthur ..... not the question ..... correct? $\endgroup$
    – Secretly
    Sep 9, 2019 at 12:28
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    $\begingroup$ Yes. The solution image had typos. $\endgroup$
    – Arthur
    Sep 9, 2019 at 12:29
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    $\begingroup$ Just use $<\varepsilon$ everywhere. If everything is $<\varepsilon$, then it is also $\leq\varepsilon$, so if we solve it using $<$, then the original problem asking for $\leq$ is satisfied. However, the converse doesn't always hold, and that's what gets the solution author into trouble here. $\endgroup$
    – Arthur
    Sep 9, 2019 at 16:53
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    $\begingroup$ That's exactly how we know that. $\endgroup$
    – Arthur
    Sep 9, 2019 at 18:48

1 Answer 1

  1. 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 $[-M,M]$.

  2. There are unbounded sets of finite measure. Consider $E=\mathbb{N}$. Then $m(E)=0,$ but $E$ is unbounded.

  3. 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?

  4. 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 $\varepsilon$.

  5. I'm not sure I follow: Would you like a second way of proving the statement?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Sep 10, 2019 at 21:40

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