If $A ⊆ ℝ^n$ , we say that A is a set of zero content if for every $ε > 0$ there are compact rectangles $R_1,…,R_m$ such that $A ⊆ R_1∪…∪R_m$ and $vol(R_1)+…+vol(R_m) < ε$ , with $vol([a_1,b_1]×…×[a_n,b_n]) = (b_1-a_1)⋅…⋅(b_n-a_n)$.
Is it true that if $A$ is not of zero content, then there is a compact rectangle $R ⊆ ℝ^n$ such that $R ⊆ A$ and $vol(R) > 0$?
As a counterexample, take $n=1$ and $A = [0,1]\setminus\mathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.
However, $A$ contains no non-degenerate compact interval, since the rationals are dense.
Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.