I'd like your help with understanding how come the following language is decidable (in $R$).
Let $M_0,M_1,M_2,...$ numbering of all Turing machines over $\sum$, why does $B=\{ \langle M_i: \exists j < i . L(M_j)=\sum^* \} \in R$?
I understand that we have only have finite number of machines to go through for checking but the check itself for $L(M_j)=\sum^*$ is not decidable, and we might be stuck in infinite loop, so I thought that the language is in $co-RE \setminus R$, since we can check in parallel all the turing machines until $i$, and if $\exists j$ such that $L(M_j)\neq\sum^*$ stop and accept.