Consider $f$ on $[0,1]$ defined as $f(0)=0$ $$f(x)=2^{-n}\quad \text{if}\quad 2^{-n-1}<x\le2^{-n},$$ for $n=0,1,2,3,...$ I'm looking for two reasons why $\int^{1}_{0}f(x) \,dx$ exists?
One of them could be because $f$ is a step function we can therefore integrate $f$ over $[0,1]$ and can calculate the integral. But what could be other reason for integral to exists?
The fact that you have an "infinite" step function (and this is abuse of the definition of a step function) doesn't imply it's (Riemann) integrable. Consider the classical example $f(x)=1$ whenever $x$ is rational and 0 otherwise. The example is not considered a step function since it's not a sum over a finite number over intervals. If you want to show it's integrable (or not), you need to show that the supremum over lower sums equals the infimum over upper sums.
One idea is to show the function is a bounded and monotonically increasing, which are known to be always Riemann Integrable.