Q. Let $f:[a,b]\to R$ be a bounded function. Suppose there exists a sequence of partitions $\{P_k\}$ $$\lim_{k \to \infty}(U(P_k,f)-L(P_k,f))=0.$$ Show that $f$ is Riemann integrable and that $$\int_a^b f = \lim_{k \to \infty}U(P_k,f)=\lim_{k\to \infty}L(P_k,f).$$
Let $P_k=(a=x_0,x_1,...,x_{n-1},x_k=b)$ such that $x_i = \frac ik$ for $k \in N, i\le k.$
Let $\int_a^bf=I$. Then, we know that $L(P_k,f)\le I \le U(P_k,f).$ Thus, the sequence of $L(P_k,f)$ and $U(P_k,f)$ are bounded. I was trying to prove the question by using the monotone bounded sequence theorem. But, I realize that the sequence is not monotone increasing and decreasing, respectively in this partition. What should be an alternative approach?
Start from $$L(f,P) \le \underline\int_a^b f \le \overline\int_a^b f \le U(f,P)$$ valid for every partition $P$ of $[a,b]$ .
Then $$ 0 \le \overline\int_a^b f - \underline\int_a^b f \le U(f,P) - L(f,P) $$Edit
In the same way you can get also $$0 \le U(f,P) - \overline\int_a^b f \le U(f,P) - L(f,P)$$ and $$0 \le \underline\int_a^b f - L(f,P) \le U(f,P) - L(f,P)$$