$f \in \mathscr{R}$ on $[a,b]$ if and only if $\forall \epsilon \gt 0, \exists$ a partition P such that $U(P,f,\alpha) – L(P,f,\alpha) \lt \epsilon$

**Question:** **(1)** $(\Leftarrow )$ though $L(P,f,\alpha) \leq \sup L(P,f,\alpha) \leq \inf U(P,f,\alpha) \leq U(P,f,\alpha)$ implies $0 \leq \inf U(P,f,\alpha) – \sup L(P,f,\alpha) \leq \epsilon$ makes complete sense. But I canâ€™t formally prove it.

**(2)** $(\Rightarrow)$ The existence of partition $P_{1}$ and $P_{2}$ depends on the definition of sup and inf.

**Proof**: Assume towards contradiction i.e. for all partition P of $[a,b]$, $ L(P,f,\alpha) + \epsilon /2 \leq \sup L(P,f,\alpha)$. Since $\sup L(P,f,\alpha) \in \Bbb{R}$ is an upper bound of the left side of inequality, we have $ \sup (L(P,f,\alpha) + \epsilon /2) = \sup L(P,f,\alpha) + \epsilon /2 \leq \sup L(P,f,\alpha)$. Which implies $\epsilon /2 \leq 0$. Contradicts our initial assumption. Is this proof correct?

I have actually attempted a different approach to this forward implication.

**My attempt:** Given $\epsilon \gt 0$. $$L(P,f,\alpha) – \epsilon \lt \sup L(P,f,\alpha) = \inf U(P,f,\alpha) \leq U(P,f,\alpha)$$ Thus, $U(P,f,\alpha) – L(P,f,\alpha) \lt \epsilon $. The potential problem with this proof is that, it is true for all partition P. Which I think may not be correct. Anyway, where do you think this proof is going wrong?

## Best Answer

The desired inequality can be derived from the following observations:

Given $\varepsilon>0$, there exists a partition $P_1$ such that $L(f,P_1,\alpha)+\varepsilon>\sup L(f,P,\alpha)$.

Given $\varepsilon>0$, there exists a partition $P_2$ such that $U(f,P_2,\alpha)-\varepsilon<\inf U(f,P,\alpha)$.

There exists a partition $P$ which refines $P_1$ and $P_2$.