I've come across this question and I think I'm on the right track with the idea I just don't really know how to formalize it properly or understand why everything said in the question is required for the proof. The question is as follows:
Let $\mathcal X$ be a finite example domain and let $\mathcal Y = \{0, 1\}$. Let $H ⊆ \mathcal Y^ \mathcal X$ be a finite hypothesis class. Let $\mathcal D$ be a distribution over $\mathcal X \times \mathcal Y$. Let $\mathcal G$ be a different distribution over $\mathcal X × \mathcal Y$, such that $G_{Y |X} = D_{Y |X}$. Suppose that an ERM algorithm $\mathcal A$ receives a sample from $\mathcal G$ instead of a sample from $ \mathcal D$, that is: $S \sim \mathcal G^m$. Prove that if $\mathcal D$ and $\mathcal G$ are realizable by $\mathcal H$, and $supp(\mathcal D_\mathcal X) ⊆ supp(\mathcal G_\mathcal X)$, then for every $\epsilon, \delta \in (0, 1)$, there exists a sample size $m$ (that can depend on $\mathcal D, \mathcal G, \mathcal H, \epsilon, \delta)$ such that $\mathbb P_{S\sim \mathcal G^m}[err(\mathcal A[S], \mathcal D) \geq \epsilon)] \leq \delta$.
My idea was that since the conditional distributions are equal and the realizability assumption holds it must be that the labeling function for each distribution, say f and g, are the same as well. Now since the support of D is contained in the support of G, any example that can be observed in the sample drawn from D can also be observed in a sample drawn from G, hence enough examples drawn from G will suffice to achieve a training set S that contains all of the support of D. However, I can't quite formalize this idea.
Thanks in advance!