Final answer:
A Riemann integrable function on an interval is also integrable on any subinterval of that interval, owing to the fact that the Riemann sums converge on the subinterval just as they do on the original interval.
Step-by-step explanation:
If function f is Riemann integrable on the interval [a, b], then it can be shown that for any subinterval [c, d] that is contained within [a, b], the function f restricted to [c, d] is also Riemann integrable on [c, d].
This is due to the definition of Riemann integrability, which requires that for any partition of the interval and any choice of sample points within the subintervals of the partition, the Riemann sums converge to the same limit as the mesh of the partition goes to zero. Since [c, d] is a subinterval of [a, b], any partition of [c, d] is also a partition of [a, b] (potentially with additional points not in [c, d]).
Therefore, if f is integrable on the larger interval, it must be integrable on the smaller interval as well because the definition of integrability is satisfied via the same limiting process of Riemann sums on [c, d] as it is on [a, b]. In essence, the necessary condition for the existence of the Riemann integral on an interval involves the behavior of the function and the formation of Riemann sums within that interval, which by extension applies to any subinterval. Hence, the restriction of f to [c, d] will be Riemann integrable as well.