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I am working my way through the proof of the following: Let $f$ be bounded on [a,b]. Then $f$ is Riemann integrable if and only if for every $\epsilon$ there is a partition on $[a,b]$ such that: $0 \leq U(f,P) - L(f,P) \leq \epsilon$. I have a question regarding the beginning of the proof:

Since $$\int_{a}^{b}f = \inf\{U(f,P)\}=\sup\{L(f,P)\}$$ there should exist two partitions $P_1$ and $P_2$ such that: $$0\leq U(f,P_1)-\int_{a}^{b}f < \frac{\epsilon}{2}$$ and $$0 \leq \int_{a}^{b}f-L(f,P_2) < \frac{\epsilon}{2}$$ Why do the partitions give those inequalities?

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    $\begingroup$ $\int_{a}^{b}f$ is defined to be the infimum of $U(f,P)$over all possible partitions. By the definition of infimum, you can find one partition, say P1, such that $U(f,P_1)$ is "as close to the infimum as you'd like", ie $U(f,P_1)-\int_{a}^{b}f < \frac{\epsilon}{2}$. Similarly, for supremum. $\endgroup$ – Alexandros Apr 3 at 21:02
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Since $\int_{a}^{b}f = inf\{U(f,P)\}$, we can keep finding partitions that such that $U(f,P)$ are closer and closer to $\int_{a}^{b}f$ approaching from above. So choose a partition, call it $P_1$, such that it is closer than $\frac{\epsilon}{2}$ to $\int_{a}^{b}f$. That means $$ U(f,P_1) < \int_{a}^{b}f +\frac{\epsilon}{2}$$ $$ \iff U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2}$$

Finally note that $U(f,P_1) \geq inf\{U(f,P)\}$, so $$ 0 = \int_{a}^{b}f - \int_{a}^{b}f = inf\{U(f,P)\} - \int_{a}^{b}f \leq U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2}$$ $$ \iff 0 \leq U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2} $$

The argument is almost identical for $L(f,P)$ (except flipping some inequalities)

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  • $\begingroup$ I see now, thank you very much. $\endgroup$ – Keighleyite Apr 3 at 21:30

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