# Theorem 6.6 of Baby Rudin

alternative-proofproof-writingreal-analysissolution-verificationsupremum-and-infimum

$$f \in \mathscr{R}$$ 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?

1. Given $$\varepsilon>0$$, there exists a partition $$P_1$$ such that $$L(f,P_1,\alpha)+\varepsilon>\sup L(f,P,\alpha)$$.
2. Given $$\varepsilon>0$$, there exists a partition $$P_2$$ such that $$U(f,P_2,\alpha)-\varepsilon<\inf U(f,P,\alpha)$$.
3. There exists a partition $$P$$ which refines $$P_1$$ and $$P_2$$.