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I'm looking for a dense set in $\mathbb{R}$ with outer measure $1$. There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$.. Also I've tried with Bernstein sets but I don't get it.

Any hint is appreciated.

Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf In the proof of theorem 4.13, he said:

...A set of outer measure $1$ is clearly dense...

In the Density Topology, $\mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $\mathbb{R}$ with outer measure $1$, I would appreciate it.

guchihe
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  • $(0,1)$ works. Also $[0,1]$. Also $[0,1]\setminus\Bbb Q$. – Asaf Karagila Jan 09 '19 at 08:16
  • But I need it to be dense in $\mathbb{R}$. – guchihe Jan 09 '19 at 08:18
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    I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$). – Asaf Karagila Jan 09 '19 at 08:50
  • No, the space is defined in $\mathbb{R}$ and use the Lebesgue measure. – guchihe Jan 09 '19 at 08:55
  • I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies – DanielWainfleet Jan 10 '19 at 00:17
  • That's the thing. I think he means that exist one such that has outer measure $1$ and is dense, but we need such set of minimum cardinality for yield the inequality that I need. – guchihe Jan 10 '19 at 00:34

3 Answers3

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Take $\mathbb{Q}\cup[0,1]$, for instance.

José Carlos Santos
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In the usual (Euclidean) topology, $\Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$

Let $\Bbb Q=\{q_i: i\in \Bbb N\}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1\in K(1).$

For $n\in \Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:

$K(n)$ is bounded so $\Bbb Q \not \subset K(n).$ Let $i(n)$ be the least $i\in \Bbb N$ such that $q_i\not \in K(n).$ Since $q_{i(n)} \not \in \overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}\in U(n)\subset \Bbb R \setminus K(n).$ Since $K(n)\cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)\cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$

Let $K(n+1)=K(n)\cup U(n)\cup V(n).$

Let $S=\cup_{n\in \Bbb N}K(n).$

It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_n\in S$ by induction on $n,$ as follows:

We have $q_1\in S.$ For $n\in \Bbb N,$ if $\{q_i:i\le n\}\subset S,$ then for each $ i\le n$ let $f(i)$ be the least (or any) $j$ such that $q_i\in K(j),$ and let $g(n)=\max \{f(i):i\le n\}.$

Now if $q_{n+1}\in K(g(n))$ then $q_{n+1}\in S.$

But if $q_{n+1}\not \in K(g(n)),$ then, since $\{q_i:i\le n\}\subset \cup_{i\le n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}\in U(g(n))\subset K(g(n)+1)\subset S.$

DanielWainfleet
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  • By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $T\subset \Bbb R$ such that $m(T)=1,$ and $ m(T\cap (n,n+1))\ne 0$ for every $n\in \Bbb Z.$ Although this is not the kind of set in the Q. – DanielWainfleet Jan 10 '19 at 00:26
  • Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer. – guchihe Jan 10 '19 at 02:19
  • By the Lebesgue Density Lemma, any measurable $E\subset \Bbb R$ has a measurable subset $E^$ with $m(E)=m(E^)$ and such that, in the notation of F.D. Tall, $\phi(E^)\supseteq E^,$ which makes $E^$ open in the Density Topology. And if $m(E)\ne 0$ then $E^$ is not empty because $m(E^*)=m(E)$.... In particular let $E=\Bbb R\setminus S.$ – DanielWainfleet Jan 10 '19 at 09:14
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    Oh it's true. In fact, just by that, I need a non-measurable set. – guchihe Jan 10 '19 at 19:57
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Let $\{a_1, a_2, \cdots, a_n, \cdots\}$ be the set of all rational numbers. For each $i$, take a neighborhood of length $\frac{1}{2^{i+1}}$ near $a_i$ and denote it by $B_i$. Let $U=\cup B_i$. Then $|U| \le \Sigma |B_i| \le \frac{1}{2}$. Theore, $U$ is an open dense subset of measure less than $1$. Scale it to any measure you want.

If you need a non-measurable set, just take the union of this set and the Vatali set in the unit interval.

gaoqiang
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