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Prove the statement that for a bounded function $f:[a,b]\to \mathbb{R}$ and $K=\left\{P:\text{ partition of [a,b] }\right\}$
$f$ is integrable iff $\exists (P_{n}):\mathbb{N}\to K$ such as that $U(f,P_{n})-L(f,P_{n})\to 0$.
Here $U(f,P_{n})$ and $L(f,P_{n})$ are obviously the upper and lower Riemann Sums.

You can use the Riemann Criterion that is, $f$ is integrable iff $\forall \epsilon>0 \exists P_{\epsilon}\in K: U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon\\$

Proof: ($\Rightarrow$) $$ U(f,P_{n})-L(f,P_{n})\to 0\Rightarrow \forall \epsilon>0 \exists N\in \mathbb{N}: n\ge N\Rightarrow \left|U(f,P_{n})-L(f,P_{n})\right|<\epsilon\Rightarrow \\ \forall \epsilon>0 \exists N\in \mathbb{N}: U(f,P_{N})-L(f,P_{N})<\epsilon\Rightarrow \forall \epsilon>0 \exists P_N\in K: U(f,P_{N})-L(f,P_{N})<\epsilon $$ and set $P_{N}=P_{\epsilon}$.

How do I show the ($\Leftarrow$)? More specifically how do I construct a sequence of $P_{\epsilon}$ partitions (that is unique regardless of $\epsilon$) such as that $U(f,P_{n})-L(f,P_{n})\to 0$?

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up vote 1 down vote accepted

Let $n \in \mathbb N$. By assumption, applied with $\epsilon = \frac 1n$, there is a $P_n \in K$ with $U(f, P_n) - L(f, P_n) < \frac 1n$. We do this for every $n \in \mathbb N$ and obtain a sequence $(P_n)$. We now have $U(f, P_n) - L(f, P_n) \to 0$ as $n \to \infty$.

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