“…, d π } (or i, j ∈ N if d π = ∞). Moreover, by [3,Lemma 3.10],h π i,j ∈ S(G, G); thus, as, for any r ∈ G, we have thath π i,j (sr, tr) =h π i,j (s, t) for almost all (s, t), h π i,j = N (u π i,j ) for some element u π i,j ∈ M cb A(G), vanishing on E [1]. By Claim 2, Shπ i,j (T K ) = P (K)Shπ i,j…”