Is there a Lebesgue measurable subset $A \subset R$ such that for every interval $(a,b)$ we have $0 < \lambda(A\cap(a,b))< (b-a)$ [duplicate]

Instead of trying to understand your approach (which sounds complicated), let me tell you how I'd do it. I guess you know how to construct a closed nowhere dense set of positive measure inside a given interval, right? Enumerate all the rational intervals in a sequence $I_1,I_2,I_3,\dots$. Now construct an infinite sequence $M_1,N_1,M_2,N_2,M_3,N_3,\dots$ of pairwise disjoint closed nowhere dense sets of positive measure, with $M_k,N_k\subset I_k$. [1] The $F_\sigma$-set $M=M_1\cup M_2\cup M_3\cup\cdots$ does what you want. [2]

[1] Note that, at each step of the construction, you have constructed a finite number of closed nowhere dense sets, whose union is therefore a closed nowhere dense set. Hence the interval $I_k$ you are currently working in will contain a subinterval which is disjoint from that nowhere dense set. Construct the next closed nowhere dense set inside that subinterval.

[2] Any interval $I$ contains some rational interval $I_k$. Since $N_k\subseteq I_k\subseteq I$ and $N_k\cap M=\emptyset$, it follows that $M\cap I\subseteq I\setminus N_k$ and $m(M\cap I)\le m(I\setminus N_k)=m(I)-m(N_k)\lt m(I)$.