4

In Royden's text Vitali's theorem states that "Any set E of real numbers with positive outer measure contains a subset that fails to be measurable". So my question the Cantor set is a set of real numbers with outer measure zero does this set have a subset which fails to be measurable?

CLARE
  • 43
  • What do you mean by saying "fails to be measurable"? – qwertyguy Feb 11 '20 at 14:40
  • No, because any cover of the Cantor set by intervals is also a cover of any subset of the Cantor set by intervals. So, the measure of a subset of the Cantor set has an outer measure that is bounded above and below by zero. – SlipEternal Feb 11 '20 at 14:43
  • Lebesgue measure is complete: every subset of any set of measure $0$ is measurable (with measure $0$). – saulspatz Feb 11 '20 at 14:45
  • To summarize (and perhaps add to) the comments: Let "measure" refer to Lebesgue measure on $\mathbb R$ and let $E \subseteq \mathbb R.$ (1) If $E$ has positive (or infinite) outer measure, then there exists a nonmeasurable set $E'$ such that $E' \subseteq E.$ (2) If $E$ has zero outer measure, then every subset of $E$ is measurable. Corollary of (1): If $E$ has positive (or infinite) measure, then there exists a nonmeasurable set $E'$ such that $E' \subseteq E.$ Corollary of (2): If $E$ has zero measure, then every subset of $E$ is measurable. – Dave L. Renfro Feb 11 '20 at 15:29

1 Answers1

2

As the comments say, any set of outer measure zero is measurable. Consequently, every subset of a set of outer measure zero is measurable (since $\mu^*(A)\le\mu^*(B)$ whenever $A\subseteq B$).

Exactly how to prove this depends on what definition of measurability you use. If you use the "inner and outer measures agree" definition of measurability, this is basically immediate: the inner measure is at most the outer measure and is always nonnegative, so if the outer measure of $A$ is zero the inner measure of $A$ must also be zero. If you use the "differs from a Borel set by a set of outer measure zero" definition of measurability, this fact is literally trivial (since the emptyset is Borel).

If you use the Caratheodory definition of measurability - which is my personal favorite - then things aren't quite immediate, but the proof is still straightforward. A set $A$ is measurable iff for every $X$ we have $$\mu^*(X)=\mu^*(X\cap A)+\mu^*(X\cap A^c).$$

Now assume $\mu^*(A)=0$. We have $\mu^*(X\cap A)+\mu^*(X\cap A^c)=\mu^*(X\cap A^c)$, so it's enough to prove that for $A$ of outer measure zero and $X$ an arbitrary set we have $$\mu^*(X)=\mu^*(X\cap A^c).$$ One direction is trivial: since $X\cap A^c\subseteq X$ we know $\mu^*(X)\ge\mu^*(X\cap A^c)$, so all we need to prove is that $\mu^*(X)\le\mu^*(X\cap A^c).$ To do this we'll "combine covers," using the fact that $A$ has outer measure zero. Specificaly:

Given an open cover $\mathcal{C}$ of $X\cap A^c$ with total size (= sum of diameters of intervals involved) $k$ and $\epsilon>0$, let $\mathcal{D}$ be an open cover of $A$ with total size $<\epsilon$ (which exists since $\mu^*(A)=0$) and consider $\mathcal{C}\cup\mathcal{D}$. Intuitively, it never takes much more measure to cover $X$ than it does to cover $X\cap A^c$ if $A$ has outer measure zero.

(It's a good exercise to pin down why this same argument doesn't show that every set is measurable.)

Noah Schweber
  • 245,398