Does $\lVert P\rVert \leq \lVert Q\rVert$ imply $U(P,f) \leq U(Q,f)$? Let $P = \{a=x_0,x_1, \ldots, x_n=b\}$ be a partition on the interval $[a,b]$. Then $\lVert P\rVert$ denotes the norm of partition $P$, defined to be the length of the greatest subinterval of the form $[x_{i-1},x_i]$.
Let $f:[a,b] \to \mathbb{R}$ be a bounded function. Define $M_i = \operatorname{sup} f([x_{i-1},x_i])$ for all $i = 1, \ldots,n$ and the upper sum of $f$ corresponding to $P$ is $U(P,f)= \sum_{i=1}^{n}{M_i\Delta x_i}$ (where $\Delta x_i = x_i - x_{i-1}$). Here is the real question:

Suppose $P$, $Q$ are two partitions on $[a,b]$ such that $\lVert P\rVert \leq \lVert Q\rVert$. Then is it true that $U(P,f) \leq U(Q,f)$?

This seems to be true intuitively, but I can't find a solid argument to claim so. I tried to look for counter-examples too, but without any success. I would appreciate any help regarding this problem. Thanks and regards.
 A: This isn't necessarily true. Here is a counterexample:
$$
f(x) = \begin{cases}
0 & \text{if } 0 \le x \le 2 \\
x & \text{if } 2 \lt x \le 3 \\
\end{cases}
$$
Consider two partitions: $P = [0, 1, 2, 3]$, $Q = [0, 2, 2.5, 3]$.
Clearly $\|P\| = 1$, $\|Q\| = 2$. Therefore $\|P\| \le \|Q\|$.
However $U(P, f) = 3$, $U(Q, f) = 2.75$. Therefore $U(P, f) > U(Q, f)$.
A: As a strengthening of Ayman's answer, suppose $P$ and $Q$ are partitions such that there is even a single point in $P$ that is not in $Q$. Then for any $m$, $M$ with $m\le M$, there exists an $f$ such that $U(P,f) = M$, $U(Q,f)=m$, i.e. I can make $U(P,f)$ as large as I like while keeping $U(Q,f)$ small.
The technique is simple: find a point $x$ in $P\setminus Q$, and pick $y_k$ in $Q$ such that $y_k<x<y_{k+1}$. Then give $f$ a huge spike on the interval $(y_k, y_{k+1})$, and $Q$ will not pick it up at all, but $U(P,f)$ can be made arbitrarily large.
Hence you really need $P\subseteq Q$ to conclude anything about $U(P,f)$ relating to $U(Q,f)$.
A: Ayaman's counterexample works even if you restrict yourself to equidistant partitions. If $0<\alpha<1$ and
$$f(x) = \begin{cases}1&0\le x <\alpha\\0&\alpha\le x\le1\end{cases}$$
and let $P_n$ be the equidistant partition $\{x_0=0, x_1=\frac1n,\ldots, x_n=1$}, then $\Vert P_n\Vert=\frac1n$ and $U(P_n,f)=\frac1n(\lfloor n \alpha\rfloor+1)$. For example, $\alpha=\frac12$ leads to the sequence 
$$1, 1, \frac 23, \frac34,\frac35,\frac23, \frac47, \frac58,\ldots $$
that converges to $\frac12$, but not monotonously.
A: Here is a different counterexample. Let $f$ be the function $f(x)=0$ for $x \in [0,2] $ except $f(.75)=f(1.25)=1$.
Then for $P=\{0,.75,1.25,2\},\;\;Q=\{0,.5,1.5,2\}$ we have$||P||=.75<1=||Q||$ but 
$U(P,f)=2>1=U(Q,f)$.
