Set with no full or empty intervals Let $m(\cdot)$ denote Lebesgue measure.
Does there exist an $S\subset [0,1]$ such that for all $a<b$ we would have $0<m([a,b]\cap S)<b-a$?
I believe that the answer is yes, and I have a rough construction in mind.
On the first step, allocate $1/4$ "mass" to either side of $1/2$ (this gives $S_1$).  Then recursively allocate $1/8$ mass to either side of $1/4$ and of $3/4$ (to get $S_2$).  Continue this process.
At depth $d$ of the recursion, we have that for $a,b$ with binary expansions with at most $d$ digits, $m([a,b]\cap S_d)=(b-a)/2$.  Then in the limit, any interval can be written as the countable disjoint union of intervals with finite binary expansion, so (I think) $m([a,b]\cap S_d)=(b-a)/2$ holds for all $0<a<b<1$.
I can describe the construction a bit more formally:
For every $n$, consider the set of points with $n$ digits in its binary expansion.  There are $2^{n-1}$ such points since every digit is free, except for the last one which must be 1.  Place an interval of length $2^{-n}$ centered at each of those points.  Let $S_n$ be the union of those intervals.
This would give us $S_1=[1/4,3/4],S_2=[1/8,3/8]\cup [5/8,7/8],\cdots$
 A: It is not clear to me how you intend to "take the limit" of your construction and get a final set that works.  Here is an alternate construction that works.
We will repeatedly use the following fact: for any $c>0$ and any interval $(a,b)$, there exists $C\subset(a,b)$ which is closed, has empty interior, and for which $0<m(C)<c$ (proof: take an appropriate fat Cantor set).
Now enumerate the open intervals in $[0,1]$ with rational endpoints as $(a_n,b_n)$.  Also, fix some sequence $(r_n)$ of elements of $(0,1)$ such that $\prod_{n=0}^\infty (1-r_n)>0$.  We choose a sequence of closed empty interior sets $C_n$ as follows.  Having chosen $C_m$ for all $n<m$, let $d_m=m((a_n,b_n)\setminus\bigcup_{m<n} C_m)$ and let $d=\min_{m\leq n} d_m$.  Note that $d_m>0$ for each $m$, since $\bigcup_{m<n} C_m$ is a closed set with empty interior.  Now choose $C_n\subset(a_n,b_n)$ to be a closed set with empty interior such that $0<m(C_n)<r_nd$.
I now claim that the set $S=\bigcup_{n=0}^\infty C_n$ has the desired property.  Indeed, any interval $[a,b]\subseteq[0,1]$ contains some $(a_n,b_n)$, and then $m(S\cap[a,b])\geq m(C_n\cap(a_n,b_n))>0$.  On the other hand, I claim that $S$ cannot have full measure in $(a_n,b_n)$.  First, $\bigcup_{m<n}C_m$ does not have full measure in $(a_n,b_n)$ since it is closed and has empty interior.  Now observe that each new set $C_m$ for $m\geq n$ takes up at most an $r_m$ fraction of the measure of $(a_n,b_n)$ that is not yet covered by any $C_k$.  Since $\prod_{m=n}^\infty (1-r_m)>0$, these sets cannot add up to the full measure of $(a_n,b_n)$.  Thus $m(S\cap(a_n,b_n))<b_n-a_n$ and it follows that $m(S\cap[a,b])<b-a$.
