Final answer:
To prove the integral of a bounded, real-valued function with a finite number of discontinuities on an interval [a,b] exists, we show that the function is Riemann integrable because the set of discontinuities has a measure zero, allowing for the existence of the Riemann integral.
Step-by-step explanation:
To prove that if a real-valued function f on the interval [a,b] is bounded and is continuous except at a finite number of points, then the integral f(x)dx exists, we can invoke the concept of Riemann integrability.
A bounded function like f on a closed interval [a,b] is Riemann integrable if the set of discontinuities has measure zero. For a set to have measure zero, the sum of the lengths of intervals covering the set must be arbitrarily small. In this case, if the function f is only discontinuous at a finite number of points, we can cover each discontinuity with an interval of very small length, making the total sum of these lengths also very small, effectively zero.
Since f is bounded, for any given partition of [a,b], the upper and lower sums of f are finite. When we refine the partition to make intervals around the discontinuities smaller, the difference between the upper and lower sums due to these points of discontinuity can be made arbitrarily small, satisfying the criterion for the existence of the integral of f. Therefore, the Riemann integral of the function exists between a and b.
The continuity of f at all but a finite number of points ensures that the areas under the curve at these discontinuities do not significantly impact the total area, which allows us to conclude that the integral of the function from a to b exists and is well-defined.