Why is the halting problem decidable for LBA?
The halting problem is solvable for any Turing machine which uses a known bounded amount of space, by a generalization of the argument given by Yonatan N. If the amount of space is $S$, the alphabet size is $A$, and the number of states is $Q$, then the number of possible configurations is $QSA^S$. If the machine halts then it must halt within $QSA^S$ steps, since otherwise, by the pigeonhole principle, it has a repeated configuration and so is stuck in an infinite loop. Therefore to determine whether the machine halts, we just run it for $QSA^S$ steps and see whether it halts within that time frame.