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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, $A=\int f dt$

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What is $\mu(P_n)$? –  leo Feb 1 '12 at 2:13
    
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition. –  ncmathsadist Feb 1 '12 at 2:16
    
Sorry, clarified –  Shafat Arbaz Alam Feb 1 '12 at 2:21
    
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $\epsilon\gt 0$ exist $\delta\gt 0$ s.t. if $P$ is partition of $[a,b]$ with $\mu(P)\lt\delta$, then $$|S(P_n,f)-I|\lt\epsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does. –  leo Feb 1 '12 at 3:10
    
I know that definition. I'm curious about how I use the sequential properties to get to there. Thanks for your help! –  Shafat Arbaz Alam Feb 1 '12 at 3:19

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