In @UmbertoP's response to the question, "Partition of real numbers into dense subsets of positive measure," the answer is understandable to a advanced undergraduate; however, I have inadequate knowledge of measure and set theory.
I tried to create a more explicit example here but it's too complicated.
Question: Is there a more explicit version of Umberto P's answer that's understandable to an average undergraduate?
Maybe you'll like this one a bit better. For simplicity, I'll do my construction for $I = [0,1)$ instead of $\mathbb R$: to get a partition of $\mathbb R$, just repeat my $A$ and $B$ periodically.
At stage $n$ (for positive integers $n$) I'll partition $I$ into two sets $A_n$ and $B_n$, each a union of finitely many half-intervals. Start with $A_1 = [0,1/2)$ and $B_1 = [1/2, 1)$.
Given $A_n$ and $B_n$: for each interval $[a,b)$ in one of these sets of length $s = b-a$, remove an interval of length $2^{-n} s$ from the centre of the interval and give it to the other set. Thus from $[0, 1/2)$ in $A_1$, we remove $[1/8, 3/8)$ and put it in $B_2$, while $A_2$ keeps $[0,1/8)$ and $[3/8, 1/2)$, and from $[1/2, 1)$ in $B_1$, we remove $[5/8, 7/8)$ and put it in $A_2$, resulting in $A_2 = [0,1/8) \cup [3/8,1/2) \cup [5/8, 7/8)$ and $B_2 = [1/8, 3/8) \cup [1/2, 5/8) \cup [7/8, 1)$.
Note that in going from stage $n$ to stage $n+1$, the measure of the points transferred is $2^{-n}$. Since $\sum_n 2^{-n}$ is finite, almost every point is transferred only finitely many times. Since sets of measure $0$ are negligible, I'll define $A$ to consist of the points that are eventually in $A_n$, and $B$ as its complement the points that are in $B_n$ for infinitely many $n$, i.e. $$ A = \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k, \ B = [0,1) \backslash A = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty B_k$$
