 Solution

For distinct indices put arcs in both directions whenever $S_i\cap S_j\ne \varnothing$. If $j$ is not an outgoing neighbour of $i$, every variable determining $A_j$ lies outside $S_i$. The entire family of such events is measurable with respect to variables outside $S_i$, whose sigma-algebra is independent of the variables inside $S_i$. Thus $A_i$ is independent of their whole generated sigma-algebra, establishing the required dependency graph of events. This justifies joint independence rather than just checking each pair.