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I was working through Carothers' "Real Analysis" with my students today, and we came across an interesting question related to Lebesgue measure.

I realize that the first question I have will most likely have some answer along these lines, and that the second answer will probably involve a Cantor-like set.

The question is stated as:

Suppose that $E$ is measurable with $m(E)=1$. Show that:

1) There is a closed set $F$, consisting entirely of irrationals, such that $F\subset E$ and $m(F)=1/2$.

2) There is a compact set $F$ with empty interior such that $F\subset E$ and $m(F)=1/2$.

  • With the argument I mentioned above, it's clear that we should be able to find a closed subset, $F\subset E$, of only irrationals, with $m(F)>0$. However I am stuck when it comes to proving that such a set must exist with measure equal to exactly $1/2$. Since the open intervals we construct around the rationals inside of $E$ might overlap, I'm having trouble seeing how to control the size of the final closed set. Any hints as to what I'm missing?

  • Similarly, it's easy to construct a compact, Cantor-like set with measure $1/2$. However, when we require that set to be the subset of an arbitrary set $E$, with $m(E)$, I'm at a loss. Since $E$ might not even be bounded, I'm not even sure where to begin.

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  • The "all irrationals" part is easy. Forget the open intervals. Just remove the rationals that are in E. The new set is still measurable with measure 1, and now has all irrationals.The next part is to find a subset of that set with measure 1/2. – quasi Mar 04 '17 at 05:07
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    @quasi: but the set has to be closed. If I remove all the rationals from $(0,1)$ I don't get a closed set. I can find a sequence of irrationals that converges to $\frac 12$ – Ross Millikan Mar 04 '17 at 05:13
  • @Ross Millikan: I read it too quickly. I should have realized it wasn't that easy. So you need the open intervals after all. – quasi Mar 04 '17 at 05:14

2 Answers2

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Suppose that $E\subseteq\mathbb R$ is measurable with $m(E)\ge1.$

Find an interval $[a,b]$ such that $m(E\cap[a,b])\gt\frac12.$

Find a closed set $F_0$ consisting entirely of irrationals such that $F_0\subset E\cap[a,b]$ and $m(F_0)\gt\frac12.$

Since $\varphi(x)=m(F_0\cap[a,x])$ is a continuous function with $\varphi(a)=0$ and $\varphi(b)\gt\frac12,$ we can find $c\in(a,b)$ such that $\varphi(c)=m(F_0\cap[a,c])=\frac12.$

Let $F=F_0\cap[a,c].$ Then $F$ is a closed subset of $E$ consisting entirely of irrationals, and $m(F)=\frac12.$ Since $F$ is closed and bounded, it's compact; since $F$ is closed and contains no rational numbers, it's nowhere dense.

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Lebegue measure is Radon-inner-regular: If $E$ is measurable and $c(E)$ is the set of compact subsets of $E$ then $m(E)=\sup \{m(F): F\in c(E)\}.$

If $E$ is measurable with $m(E)=1$:

Let $E\supset F$ where $F$ is compact and $m(F)>3/4.$

Let $ G$ be an open set with $G\supset \mathbb Q$ and $m(G)<1/4.$ The set $G$ exists because $m(\mathbb Q)=0$ because $\mathbb Q$ is countable .

Now $F$ \ $G$ is compact, is disjoint from $\mathbb Q$, and is a subset of $E$, and we have:

$m(F$ \ $G)\geq m(F)-m(G)>1/2.$

And since the complement of $F$ \ $G$ is dense (because $\mathbb R$ \ $(F$ \ $G)\supset \mathbb Q $) we have $Int (F$ \ $G)=\phi. $

DanielWainfleet
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