prove $L(f)\leq U(f)$

Question

How exactly would I go about proving the following statement?

Given $f:[a,b]\to\mathbb{R}$ show that $$L(f)\leq U(f)$$ where $$L(f)=\sup_{P\in\mathscr{P}}L(f,P) \text{ and } U(f)=\inf_{P\in\mathscr{P}}U(f,P)$$ where $P$ is any partition of $[a,b]$ and $\mathscr{P}$ is the set of all partitions of $[a,b]$.

Intuitively this makes sense because for any partition $P$ made up of the intervals $I_{1},...,I_{n}$ we know that $$m_{k}=\inf_{x\in I_{k}}f(x)\leq M_{k}=\sup_{x\in I_{k}}f(x)$$

Would I be along the right lines to consider $P$ being the partition used for $L(f)$ and $Q$ being the partition used for $U(f)$ and then let $R$ be a partition defined as $R=P\cup Q$ so that $R$ is a refinement of both $P$ and $Q$?

Thanks!

Answer 1

Say $\displaystyle L(f)>U(f)=\inf_{P\in\scr P}U(f,P)$

$L(f)$ is not a lower bound for $\{U(f,P):P\in\mathscr P\}\therefore\exists P_1\in\mathscr P$ such that $U(f)\le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for $\{L(f,P):p\in\mathscr P\}\therefore \exists P_2\in\mathscr P$ such that $U(f,P_1)<L(f,P_2)\le L(f)$.

This is a contradiction, since $U(f,P_1)\ge L(f,P_2)\ \forall P_1,P_2\in\mathscr P$.

Answer 2

If $P,Q \in \mathscr{P}$, then we have

$L(F,P) \le U(f,Q)$.

This gives

$L(f) \le U(f,Q)$ for all $Q \in \mathscr{P}$,

hence $L(f) \le U(f).$