Dear all,
When I said that the A takes only S_j and not S_j^i that is not true. The algo/rule takes the labelled training set S = set of (x_k, y_k)'s k =1...m. So the argument of A is indeed A(S_j^i) as written in the notes and white board. The distribution D_i is peaked on y = f_i(x) so indeed this selects the i-th labelling of the j-th sequence S_j --> S_j^i.
Then there was a question at the very end when we pair functions. The construction is such that in the pairing the two pairs f_i and f_iprime have the same labels on the seen vectors: so there we have that A(S_j^i) = A(S_j^iprime) are the same within a pair i and iprime.
So all works fine exactly as written. Indeed all these notations are necessary to be precise enough even though it makes things difficult to follow.
Best, N.M
