Prove that a bounded $f$ is integrable if $I_0 := \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$

Solution 1:

Beccause the upper and lower integrals are the infimum and supremum, respectively, of upper and lower sums, it follows that for any $N$ and $M$

$$L(f,P_N) \leqslant \underline{\int}_a^b f \leqslant \overline{\int_a}^bf\leqslant U(f,P_M)$$

Since $I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$, for any $\epsilon > 0$ there exists $N$ and $M$ such that

$$I_0 - \epsilon < L(f,P_N) \leqslant U(f, P_M ) < I_0 +\epsilon$$

Thus,

$$I_0 - \epsilon < \underline{\int}_a^b f \leqslant \overline{\int_a}^bf< I_0 +\epsilon$$

Since $\epsilon$ can be arbitrarily close to $0$, it follows that

$$I_0= \underline{\int}_a^b f = \overline{\int_a}^bf$$

This proves both that $f$ is Riemann integrable and that $I_0$ is the value of the integral.