If your prior in \(H_1\) is vacuous—if \(P(H_1) = [0,1]\)—then the above equation shows that your posterior is vacuous as well. That is, if \(p(H_1) = 0\) then \(p(H_1\mid E) = 0\) and likewise for \(p(H_1) = 1 = p(H_1\mid E)\), and since the right hand side of the above equation is a continuous function of \(p(H_1)\), for every \(r\in [0,1]\) there is some \(p(H_1)\) such that \(p(H_1\mid E) = r\). So \(P(H_1\mid E)=[0,1]\).