Jtdylktuy

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.

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

Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$

For $nin Bbb N,$ 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 $iin Bbb N$ such that $q_inot in K(n).$ 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 $S=cup_{nin Bbb N}K(n).$

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

We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$

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 ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$

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