I'm attending a course of Measure and Integration and have some homework to do. We don't have a specific book to follow, neither for exercise.

I'm asked to proof that every set $A \in R$ with strictly positive outer measure contains a set nonmeasurable (Lebesgue).

I thought about the Cantor set. Thinking that A is omeomorphus to Cantor set and about sharing $[0,1]$ with a numerable family of copies disjointed of A. But honestly I'm not sure how to do exactly, and I would like to get a precise answer 'cause I'll have to explain in a foreigner language so the better my notes are the easier it will be.

I checked other questions on this website before asking wihtout finding anything, just some references to this property as known; probably it's my fault, so if somebody knows where to find the answer please let me know.

Update: May be is not clear but A is strictly positive in the outer Lebesgue measure and I think that the subset has to be strictly included.

  • $\begingroup$ Do you require this subset stricly contained in $A$? If not, we just deal with the case $A$ measurable: then $A=B\cup N$, where $B$ is a Borel set and $N$ of measure $0$. $B$ has positive measure. $\endgroup$ – Davide Giraudo Oct 3 '12 at 14:56
  • $\begingroup$ I have the same doubt,the questions is exactly as I wrote it. But I think the inclusion is strict. On the opposite it's going to be too easy compared to rest of the sheet I have to face with. $\endgroup$ – Laura Oct 3 '12 at 15:00
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    $\begingroup$ See Rudin, Real and Complex Analysis, Theorem 2.22, p.53. $\endgroup$ – commenter Oct 3 '12 at 15:06
  • $\begingroup$ Is it really a problem, considering $A_x:=A\cap (-x,x)$? And is the exercise too easy even with this (even if it's compared to the rest of teh sheet? $\endgroup$ – Davide Giraudo Oct 3 '12 at 15:08
  • $\begingroup$ I think this time it's going to work better, I'm going to check properly the whole proof to be sure. I'm just wondering why our teachers suggested as to think about Cantor strictly.Anyway, thanks for the help $\endgroup$ – Laura Oct 3 '12 at 15:20

Theorem 2.22 on page 53 of Real and Complex Analysis Rudin states:

If $A \subset \mathbb{R}$ is a subset with the property that every subset is measurable then $A$ is a null set.

If $A$ has positive outer measure it must be uncountable. Suppose $A$ is not measurable. Removing any point $a \in A$ yields a non-measurable proper subset $B = A \setminus \{a\}$, for otherwise $A = B \cup \{a\}$ would be measurable as a union of two measurable sets.

If $A$ is measurable and has positive outer measure, it has positive measure, so it has a subset $B$ which isn't measurable by the theorem. Of course $B$ must be a proper subset.

Thus, in both possible cases a set of positive outer measure has a non-measurable proper subset, as you wanted.

I don't know why your teachers insisted on the Cantor set, I don't think that helps a lot here. In the context of non-measurable sets, Cantor sets are used to show that not every Lebesgue measurable set is Borel measurable: The collection of Borel sets has cardinality continuum while there are $2^{\# \mathbb{R}}$ subsets of the Cantor set. Since the ternary Cantor set is a Lebesgue null set, all of its subsets are Lebesgue measurable, but only continuum many of them are Borel measurable, so most subsets of the Cantor set are Lebesgue but not Borel measurable.

If you want to see some non-Borel subsets of $\mathbb{R}$, see also:

The proof of the theorem in Rudin is the usual Vitali argument, using inner regularity of Lebesgue measure: take a set of representatives $E$ of $\mathbb{Q}$-cosets so that $\mathbb{R} = \bigcup_{q \in \mathbb{Q}} (E + q)$ and hence $$A = \bigcup_{q \in \mathbb{Q}} [\underbrace{(E + q) \cap A}_{=: A_q}].$$ It suffices to show that $\mu(A_q) = 0$ because then $A$ is a null set as a countable union of null sets. Since $A_q \subseteq A$, it follows from the hypothesis on $A$ that $A_q$ is measurable, so $\mu(A_q) = \sup\{\mu(K) : K \subseteq A_q \text{ is compact}\}$ and it suffices to show that $\mu(K) = 0$ for every compact $K \subseteq A_q$. Note that for rational $s \neq r$ the sets $K+r$ and $K+s$ are disjoint, and the set $F = \bigcup_{r \in \mathbb{Q} \cap [0,1]} K + r$ is bounded, so $F$ has finite measure. By pairwise disjointness of the sets $K + r$ $$ \mu(F) = \sum_{r \in \mathbb{Q} \cap [0,1]} \mu(K+r) = \sum_{r \in \mathbb{Q} \cap [0,1]} \mu(K) $$ since Lebesgue measure is translation invariant. It follows that $K$ must be a null set, and hence $K$ is a null set.

Replacing $\mathbb{Q}$ by $\mathbb{Q}^n$ and $[0,1]$ by $[0,1]^n$, the same argument yields that the theorem isn't special to $\mathbb{R}$, it also holds in $\mathbb{R}^n$.

  • $\begingroup$ Im working a similar problem and was needing some input. Can you take a look @commenter? math.stackexchange.com/questions/599061/… $\endgroup$ – cele Dec 9 '13 at 16:34
  • $\begingroup$ @commenter Why the set $F$ is Lebesgue measurable? You, as rudin, say it is measurable because it is bounded. I can't follow why. $\endgroup$ – jouge Jun 13 '14 at 10:53
  • $\begingroup$ @jouge: $K$ is compact, hence measurable. Thus, so is each $K + r$. It follows that $F$ is measurable as a countable union of measurable sets. The measure of $F$ is finite, because it is bounded. It is bounded, because of $F \subset \overline{B_{R + 1}} (0)$ if $K \subset \overline{B_R}(0)$. $\endgroup$ – PhoemueX Nov 4 '14 at 9:19
  • $\begingroup$ I would like to add that assuming Rudin's theorem, the result follows immediately: Assume $m(A)>0$ and that every subset of $A$ is measurable, then by the theorem $m(A)=0$! $\endgroup$ – hjhjhj57 Feb 25 '15 at 7:13

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