# Prob. 15, Chap. 2, in Royden's REAL ANALYSIS: For any set of positive measure and $\epsilon > 0$, there are finitely many disjoint measurable sets ...

Here is Prob. 15, Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

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 $$\epsilon$$.

Here $$E \subset \mathbb{R}$$ of course, and $$m^*(E) < \infty$$, where $$m^*(E)$$ is the infimum of the the set of all the sums of the form $$\sum_{k = 1}^\infty l \left( I_k \right)$$, where $$\left\{ I_k \right\}_{k=1}^\infty$$ is a countable collection of non-empty, bounded open intervals which cover $$E$$ and, for each $$k$$, $$l \left( I_k \right)$$ denotes the length of the interval $$I_k$$.

How to proceed from here?

For each $$n\in \Bbb Z$$, let $$S_n := E\cap \big[ n\epsilon, (n+1)\epsilon \big) .$$ Then $$+\infty > m^*(E)=\sum_{n\in \Bbb Z} m^* \left( S_n \right)= \lim_{p\to \infty} \sum_{n \in \mathbb{Z}, \lvert n \rvert \leq p} m^* \left( S_n \right).$$ So take $$p\in \Bbb N$$ large enough that $$m^*(E)-\epsilon \leq \sum_{ n \in \mathbb{Z}, \lvert n \rvert \leq p} m^* \left( S_n \right) \leq m^*(E).$$

Let $$\mathscr{B} := \left\{ S_n \colon n \in \mathbb{Z}, \lvert n \rvert \leq p \right\},$$ and let $$\mathscr{A} := \mathscr{B} \bigcup \left\{ E\setminus \cup_{B \in \mathscr{B} } B \right\}.$$

Then $$\mathscr{A}$$ is a finite pairwise-disjoint measurable family with $$\bigcup_{A \in \mathscr{A}} A =E ,$$ and $$A \in \mathscr{A} \implies m(A) \le \epsilon.$$

• thank you so much for such a wonderful answer. I've made some edits to your post in the interest of readability. Do you agree with these edits? Commented Mar 27, 2021 at 10:06
• Your edits are fine............ Commented Mar 28, 2021 at 12:59
• how are the sets $S_n$ measurable if $E$ is not necessarily measurable? And, how do we know that $$m^*(E) = \sum_{n \in \mathbb{Z} } m^* \left( S_n \right)?$$ Commented May 19, 2021 at 8:34
• The textbook says "if $E$ has finite measure" which means $E$ IS measurable, which implies each $S_n$ is measurable. If $E$ was not measurable it could not be a union of a finite or countable family of measurable sets. Exercise: If $T$ is a finite or countably infinite family of pairwise-disjoint real intervals and $F\subset \cup T$ then $m^*(F)=\sum_{t\in T}m^*(F\cap t).$ Commented May 19, 2021 at 11:50

The statement is trivial if $$E$$ is an interval. For example, if $$E = (a, b),$$ then we may write $$E = \bigcup_{0 \leq k \leq \lfloor \frac{b - a}{\varepsilon} - 1 \rfloor} (a + k \varepsilon, a + (k + 1) \varepsilon] \cup (a + \lfloor \frac{b - a}{\varepsilon}\rfloor \varepsilon, b).$$ Thus the statement is true for intervals. The same can be done for any finite union of intervals, since this can be written as a finite union of disjoint intervals.

Assume now that $$E$$ is a general measurable set with finite measure and let $$\varepsilon > 0$$. Then let $$\bigcup_{1 \leq n \leq N} I_n$$ such that $$m(E \Delta \bigcup_{1 \leq n \leq N} I_n) < \varepsilon.$$ Cover the union of intervals as discussed in the previous paragraph. Note that $$E \Delta \bigcup_{1 \leq n \leq N} I_n = (E \setminus \bigcup_{1 \leq n \leq N} I_n) \cup(\bigcup_{1 \leq n \leq N} I_n \setminus E)$$ and this has measure smaller than $$\varepsilon.$$ Then so does $$E \setminus \bigcup_{1 \leq n \leq N} I_n.$$ Then a covering such as the one you desire is given by the sets covering $$\bigcup_{1 \leq n \leq N} I_n$$ intersected with $$E$$ afterwards (of course, the measure is still less than $$\varepsilon$$) and $$E \setminus \bigcup_{1 \leq n \leq N} I_n.$$ I hope this helps. :)

• thank you for taking the time answering my question in such depth and detail, but I would be reallly grateful if you could limit your proof to only the machinery that Royden has put at the reader's disposal up to this particular point in his book. Commented Mar 26, 2021 at 19:58
• I don't know what machinery Royden developed up to that point, but if you were talking about the outer measure in the statement of the problem, then this answer only relies on those concepts. If you read the post I linked in my answer, then you would have seen that what is used in the proof of the fact that any measurable set can be approximated by finite unions of intervals is the definition of the outer measure. Hence the only theoretical concept used here is the definition of outer measure. Commented Mar 26, 2021 at 20:07