I am trying to prove the following:
Suppose $ f:[a,b]\rightarrow\mathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions $\{P_n\}$ with $\{\mu(P_n)\}\rightarrow0$, the sequence $\{S(P_n,f)\}$ is convergent
,where $\mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.
My attempt at a solution:
Suppose for each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$ $ \{S(P_n,f)\}$ converges.
Let $\epsilon>0$ be given. Then there is an $A\in\mathbb{R}$ and $N\in\mathbb{N}$ such that when $n>N$, there exists $\delta$ such that $\mu(P_n)<\delta\implies |S(P_n,f)-A|<\epsilon$
Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.
Now suppose $f$ is integrable. Then given $\epsilon>0$, there exists $A\in\mathbb{R}$ such that there exists $\delta$ for which $\mu(P)<\delta\implies |S(P,f)-A|<\epsilon, \forall P$.
Then for each sequence each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$ $\mu(P_n)<\delta$.
Then, $|S(P_n,f)-A|<\epsilon$ which means that $\{S(P_n,f)\}$ converges to A. Also by the theorem below, $A=\int f dt$
Best Answer
Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as $\{\mu(P_n)\}\to 0$ and what marked partition mean.