Let $f:[a,b] \to \mathbb{R}$ be bounded, prove that if $L(f,P_1)<L(f,P_2)$ then $U(f,P_2)<U(f,P_1)$

If $P_1$ and $P_2$ are partitions of $[a,b]$ and $f:[a,b] \rightarrow \mathbb{R}$ is bounded, prove that if $L(f,P_1)<L(f,P_2)$ then $U(f,P_2)<U(f,P_1)$.

I believe that this is essentially an extension of what my book calls the Refinement Lemma, that a function that is bounded and P is a partition of its domain, then $L(f,P) \leq L(f,P^*)$ and $U(f,P^*) \leq U(f,P)$, which is essentially a rewriting of my listed problem.

Though I know how to prove the general Refinement Lemma, I have no idea how to show that $P^*$ is a refinement of $P$, if that is even the case or the correct strategy, and more pertinently to my problem, if $L(f,P_1)<L(f,P_2)$ then $U(f,P_2)<U(f,P_1)$.

I was hoping for someone to confirm whether or not this is the correct strategy, or to correct me if I am wrong, and to help me show that if $L(f,P_1)<L(f,P_2)$, then $P_2$ is a refinement of $P_1$, after which, I believe I am golden, as the following proof is simple, I believe.

• By refinement do you mean that $P_1\subseteq P_2$? If so, I don't think that would apply here; two $n-$partitions certainly need not have the same value for $L(f,P)$ – AnotherJohnDoe Jun 26 '18 at 20:17

This is not true.

Consider $f : [0,1] \to \mathbb{R}$ be given by

$$f(x) = \begin{cases} \frac12, & \text{if x = \frac14} \\ 1, & \text{otherwise} \end{cases}$$

and define $P_1 = \left\{0, \frac34, 1\right\}$, $P_2 = \left\{0, \frac12, 1\right\}$.

We have

$$L(f, P_1) = \left(\frac34 - 0\right)\inf_{x \in \left[0, \frac34\right]} f(x) + \left(1-\frac34\right)\inf_{x \in \left[0, \frac34\right]} f(x) = \frac34 \cdot\frac12 + \frac14 \cdot 1 = \frac58$$

$$L(f, P_2) = \left(\frac12 - 0\right)\inf_{x \in \left[0, \frac12\right]} f(x) + \left(1-\frac12\right)\inf_{x \in \left[0, \frac12\right]} f(x) = \frac12 \cdot\frac12 + \frac12 \cdot 1 = \frac34$$

Therefore $L(f, P_1) < L(f, P_2)$.

However

$$U(f, P_1) = \left(\frac34 - 0\right)\sup_{x \in \left[0, \frac34\right]} f(x) + \left(1-\frac34\right)\sup_{x \in \left[0, \frac34\right]} f(x) = \frac34 \cdot1 + \frac14 \cdot 1 = 1$$

$$U(f, P_2) = \left(\frac12 - 0\right)\sup_{x \in \left[0, \frac12\right]} f(x) + \left(1-\frac12\right)\sup_{x \in \left[0, \frac12\right]} f(x) = \frac12 \cdot1 + \frac12 \cdot 1 = 1$$

so $U(f, P_1) = U(f, P_2)$.

• To be abundantly clear, the statement is false? That was part of the original question but I, due to reasons explained in the OP, seemingly falsely assumed that the statement was true and tried to prove it. – Bad at algebra and proofs Jun 26 '18 at 23:58
• @Badatalgebraandproofs Yes, the statement is false, and $L(f,P_1)<L(f,P_2)$ does not imply $P_1 \subseteq P_2$ (otherwise it would be true). – mechanodroid Jun 27 '18 at 6:45