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