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I want to show that $\underline{\int_{a}^{b}} f \ d \alpha \leq \overline{\int_{a}^{b}} f \ d \alpha$

So I want to show that $\sup L(P,f, \alpha) \leq \inf \ U(P, f, \alpha)$. Can I just suppose that $\sup L(P,f, \alpha)> \inf \ U(P, f, \alpha)$ and come up with a contradiction?

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    @Ryan, maybe it seems to be (to you) but it is not. In fact it is a very healthy practice, once you finished concocting a proof, to check whether it is possible to rewrite the parts (if any) where you used contradiction, without this twist. Often, it is quite possible to do so and the resulting proof is much simpler than the original one. The case at hand is a good example of this phenomenon.2011-08-27

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I would first prove this lemma: If $A$ and $B$ are nonempty sets of real numbers such that $a\leq b$ for all $a\in A$ and $b\in B$, then $\sup A\leq \inf B$. You can prove the lemma using only the facts that $\sup A$ is less than or equal to every upper bound for $A$, and $\inf B$ is greater than or equal to every lower bound for $B$.

To apply the lemma to your problem, you can proceed as Didier indicated by showing that if $P$ and $Q$ are arbitrary partitions of $[a,b]$, then $L(P,f,\alpha)\leq U(Q,f,\alpha)$. This would be clear if you had $P=Q$, but typically you don't. However, notice what happens when you refine partitions: Adding more points to a partition can only make the lower sum larger and the upper sum smaller. So what happens if you take a partition that includes both $P$ and $Q$?