Let's assume the latter, hence every subinterval is integrable. Also, the integral from a to x plus the integral from x to b is the integral from a to b. As x approaches b, the integral from x to b is at most q times (b-x), where q bounds the function. Thus that integral goes to 0, and the indefinite integral, from a to x, approaches the definite integral from a to b.
Now assume the indefinite integral exists. Given an ε, choose x and δ such that the integral from a to x is within ε of its limit, and the reiman sums are within ε of the integral from a to x, and the extra points that might be in a net of granularity δ, between x and b, don't contribute more than ε to the Riemann sum. This last constraint is possible because f is bounded. Therefore nets of granularity δ produce Riemann sums that are 3ε away from the indefinite integral. Thus the definite integral exists, and is the same as the indefinite integral.
The above can be modified for the open interval (a,b). The two-sided indefinite integral is the same as the definite integral.
These results hold for complex and vector functions.