## Theorem 7.4.7: Riemann implies Lebesgue Integrable

If

The proof is simple. Recall that for a given function *f*is a bounded function defined on*[a, b]*such that*f*is Riemann integrable, then*f*is Lebesgue integrable andf(x) dx =_{[a,b]}f(x) dx

*f*we defined

*I*to be the infimum over all upper sums and

^{*}(f)*I*to be the supremum over all lower sums.

_{*}(f)
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:

But ifI_{*}(f) I_{*}(f)_{L}I^{*}(f)_{L}I^{*}(f)

*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.