Closed nowhere dense positive measure subset of $[0,1]$ without isolated points I'm trying to prove that there exists a subset $F$ of $[0,1]$ such that:


*

*$F$ is closed

*$F$ is nowhere dense

*$F$ has a positive Lebesgue measure

*$F$ contains no isolated points.
I know that Cantor fat sets could be used but I'd like to see a different approach. 
I was thinking about using the construction that was described in this thread:
link (I mean the construction based on enumerating rationals in $[0,1]$).
Such construction leads to a set which clearly satisfies the first three conditions, but I'm having trouble with the isolated points.
Thank you.
 A: Let $F$ be any set satisfying conditions 1-3. Enumerate all open balls (or your favourite countable basis of open sets) in the interval as $(B_n)_n$, and put $\mathcal U:=\{B_n\mid B_n\cap F\textrm { is countable}\}$, and let $F':=F\setminus \bigcup \mathcal U$ (this is called the perfect (or Cantor-Bendixson) kernel  of $F$). Check that $F'$ satisfies your conditions (note that $\bigcup \mathcal U\cap F$ is a countable open subset of $F$).
This is a special case of a general theorem called the Cantor-Bendixson theorem: any closed subset of a Polish (i.e. separable completely metrisable) space can be written as a disjoint union of a perfect set (i.e. closed, without isolated points) and a countable set.
A: This is a modification of the construction described in the linked answer. The idea is to not take every ball $B(q_n,2^{-(n+1)})$ but to select them in order to avoid isolated points of the complement. To do so, we will take one ball at a time, choosing the radius small enough to our need.
So, enumerate the rationals as $(q_n)_{n\geq 1}$, and let us define inductively on a parameter $k\in\mathbb{N}$ which balls $B_k$ we will take. 
[$k=1$] The first step is to take $B_1=B(q_1,\frac14)$.
[$k\to k+1$] Now let $C_k=\bigcup_{i=1}^k B_i$ be the union of all the balls taken up to step $k$. Consider the smallest index $n\in\mathbb N$ such that $q_{n}\in \overline{ C_k}^c$, call it $n(k)$, and take the ball $B_{k+1}=B(q_{n(k)},r_k)$ where
$$r_k=\min\left\{\frac {1}{2^{k+1}}, \frac12 dist(q_{n(k)}, \overline{C_k})\right\}.$$
The algorithm is saying: at each step consider the next rational in the enumeration that's not in the closure of the balls already taken, and take the ball centered there of radius $1/2^{k+1}$, unless it falls too close to the balls already taken; in that case, choose a smaller radius. Check that this is all well defined, and let $B=\bigcup_{k=1}^\infty B_k$, and $F=[0,1]\backslash B$.
Now the key observation is that the two extreme points of every ball $B_k$ are contained in $F$ (why?). For every point $q\in F$ there are  balls among $B_k$ arbitrarily close to it, and therefore their extremes will have $q$ as a limit. The only way this could not work is if $q$ were itself an extreme of two balls simultanesously. But this is impossible, since the last ball to be chosen between the two could not lie so close to the other.
A: You already know how to construct $F\subseteq[0,1]$ with the first three properties: closed, nowhere dense, positive measure. But it may have isolated points; you want to know how to get rid of the isolated points.
Let $\mathcal I$ be the set of all rational intervals $I$ such that $I\cap F$ is countable; let $A=\bigcup_{I\in\mathcal I}(I\cap F),$ and let $E=F\setminus A.$ I claim that $E$ has all four properties.
$E$ is closed because $E=F\setminus\bigcup\mathcal I$ and $\bigcup I$ is open.
$E$ is nowhere dense because $E\subseteq F.$
$E$ has positive Lebesgue measure because $F\setminus E=\bigcup_{I\in\mathcal I}(I\cap F)$ is countable.
Suppose $x$ is an isolated point of $E.$ Choose a rational interval $I$ containing $x$ so that $I\cap E=\emptyset.$ Then $I\cap F\subseteq F\setminus E,$ so $I\cap F$ is countable, so $I\in\mathcal I,$ so $x\notin F\setminus\bigcup\mathcal I= E,$ a contradicton. Therefore $E$ has no isolated points.
A: Take the set $F$ described in the other answer, and then consider the set $$F'=\overline{F^{(1)}}$$
where $F^{(1)}$ is the set of points of $F$ with Lebesgue density $1$.
Now condition 1 is trivial. Condition 2 is satisfied, the new set being smaller. Condition 4 is satisfied because $F^{(1)}$ does not contain isolated points, and taking the closure does not add any. By Lebesgue density theorem $F'$ has the same measure of $F$, thus also condition 3 is satisfied.
A: Thm: If $F\subset [0,1]$ is compact, then $F$ is the disjoint union of a perfect set and a set that is at most countable.
Taking this as known, let $U$ be any open subset of $[0,1]$ containing $\mathbb Q \cap [0,1]$ such that $m(U) < 1.$ Let $F = [0,1]\setminus U.$ Then $F$ is compact, nowhere dense, and has positive measure. Apply the theorem to write $F=F_1 \cup F_2,$ where $F_1$ is perfect and $F_2$ is countable. Then $F_1$ satisfies the requirements 1-4. 
