Language of all words accepted by a TM by at most $t$ steps is regular
The language $L$ only depends on the first $t$ letters of the input (since the Turing machine never gets to any of the other letters). This implies that it is regular, since
$$
L = (L \cap \Sigma^{&lt;t}) \cup (L \cap \Sigma^t) \Sigma^*,
$$
where $\Sigma^{&lt;t} = \bigcup_{i=0}^{t-1} \Sigma^i$. The right-hand side is regular since $L \cap \Sigma^{&lt;t}$ and $L \cap \Sigma^t$ are both finite, and so, regular.