# Proof that a function is Riemann integrable if for any $\epsilon > 0$ there exists a partition P such that: $U(P, f) − L(P, f) < \epsilon$

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?

• $\int_{a}^{b}f$ 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. – Alexandros Apr 3 at 21:02

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)