Theorem 7.4.7: Riemann implies Lebesgue Integrable

If f is a bounded function defined on [a, b] such that f is Riemann integrable, then f is Lebesgue integrable and
f(x) dx = [a,b] f(x) dx
Context Context
The proof is simple. Recall that for a given function f we defined I*(f) to be the infimum over all upper sums and I*(f) to be the supremum over all lower sums.

Since every step function is also a simple function, every upper sum is a simple function that is bigger than f, and every lower function is a simple function less than f. Therefore:

I*(f) I*(f)L I*(f)L I*(f)
But if f is Riemann integrable, the first and last quantities agree, so that f must be Lebesgue integrable as well with the same value for the integral.


Interactive Real Analysis, ver. 1.9.5
(c) 1994-2007, Bert G. Wachsmuth
Page last modified: Mar 26, 2007