Exotic Lebesgue Measurable Function Measurable functions whose graphs are dense in the plane are well known. Examples include, the Conway 13 function, as given in the answer in this link: When is the graph of a function a dense set?
However, is it possible to impose the extra condition that the (one dimensional) measure of the set formed by taking the projection (on the horizontal axis) of all points on the graph (of our measurable function) that are contained in any given bounded rectangle (with sides parallel to the axes) is strictly positive?   
EDIT: Made the question less notational.      
 A: Given $x \in [0,1)$, let $K(x) \in \{1,2,\ldots,+\infty\}$ be the largest integer $k$ such that the binary expansion of $x$ is of the form
$$
x = 0.b_1b_2\ldots b_k 1\underbrace{0\cdots 0}_{\ge k\; \text{times}}1c_1c_2\ldots
$$
By Borel-Cantelli, one has $K(x) < \infty$ for some set $A$ of full Lebesgue measure. Let now $A_{k,l}$ be the set of $x \in A$ with $K(x) = k$ and such that, in the above expansion, one has exactly $k+l$ zeroes appearing between $b_k$ and $c_1$. Obviously the sets $A_{k,l}$ are all disjoint. They also have the property that one has $p_{k,l} > 0$ such that, for each integer $m < 2^k$, the Lebesgue measure of the intersection of $A_{k,l}$ with $[m2^{-k},(m+1)2^{-k}]$ (which corresponds to fixing the $b$'s in the above expansion) is given by $p_{k,l}$. In particular, setting $B_l = \bigcup_{k \ge 1} A_{k,l}$, each set $B_l$ gives positive Lebesgue measure to each dyadic interval. 
Now we are done: just take any enumeration $q_l$ of the rationals in $[0,1]$ and define $f(x)$ to be equal to $q_l$ for $x\in B_l$ and for example $0$ if $x \not \in \bigcup B_l$. 
A: Of course, there are no problems. Enumerate all rational intervals $\Delta_1,\Delta_2,\dots$ on the real line. Then enumerate all pairs $(p,q)$ of positive integers. On $i$-th step choose $i$-th pair $(p,q)$ and choose a nowhere dense Cantor-type set $K_i$ of positive measure inside $\Delta_p$ which maps into $\Delta_q$ (say, linearly). Sets $K_i$ may be chosen mutually disjoint. Then our function $f$ is inductively defined on their union $U$, put $f=0$ outside $U$.
