No.

Take N = 2^n and fix any constant edge density p in (0,1). Let G be a d-regular bipartite graph on (N,N) with d = pN (such regular graphs exist for all large N). Define (X,Y) to be uniformly distributed over the edges of G, i.e., each edge has probability 1/(Nd) and non-edges have probability 0.

• The marginals are uniform: since G is d-regular on both sides, each left (resp. right) vertex is incident to exactly d edges, hence P(X=x) = d/(Nd) = 1/N and P(Y=y) = 1/N.

• The mutual information is small: H(X)=H(Y)=log N = n, while H(X,Y)=log(number of edges)=log(Nd)=log(pN^2)=2n+log p, so I(X;Y)=H(X)+H(Y)-H(X,Y)=2n-(2n+log p) = -log p, which is a constant independent of n (certainly ≤ sqrt(n) for large n). Since p<1, the variables are dependent.

• Yet the support contains no large product set: the support is exactly the edge set of G. A set Xset×Yset is contained in the support iff the induced bipartite subgraph between Xset and Yset is complete (a biclique). For a random d-regular bipartite graph (and similarly for G(N,N,p)), with probability tending to 1 as N→∞: • there is no biclique K_{αN,βN} for any fixed α,β>0 (a simple union bound shows the expected number of such bicliques is at most C(N,αN)C(N,βN)p^{αβN^2}, which tends to 0), and in fact • the largest complete bipartite subgraph has both sides only O(log N) in size.

Thus, even with mutual information as small as 1 bit (take p=1/2), there need not exist “large” sets Xset, Yset (in any reasonable sense such as linear in N or even N^ε) whose product is contained in the support.

Equivalently, the graph-theoretic variant is false: high edge density alone does not force the presence of a large complete bipartite subgraph; dense random (regular) bipartite graphs typically contain no K_{αN,βN} for any fixed α,β>0, and their largest bicliques are only logarithmic in N.

This shows that small mutual information (even with uniform marginals) does not imply the existence of large product sets contained in the support.