 Solution

Under the bounded-support convention, false in general, true if $f(0)=0$. For the first claim, take $g=0$ and $f\equiv 1$: the composition is one on all of $\mathbb{R}$, and cannot have bounded support. For the second, refine the finitely many defining intervals into disjoint intervals on which $g$ is constant. The composition is constant on each of them and vanishes outside their bounded union because $f(0)=0$. Hence it is another step function.

Under the unrestricted piecewise-constant convention, true in both cases: composition with any function preserves constancy on the same partition, including its unbounded cells. This also applies to the broader measurable-partition convention. Continuity of $f$ is not needed for this particular assertion.