Is this a valid property for Riemann Sums

Question

$R$ is a refinement of $Q$

then

$$U(R, f) − L(R, f) ≤ U(Q, f) − L(Q, f)$$

Answer 1

If $R$ is a refinement of $Q$ then $U(R,f) \leq U(Q,f)$ and $L(Q,f) \leq L(R,f).$

Proof:

Let $Q = \{a = x_0 ,...,x_{i-1},x_i,...,x_n=b\}$. WLOG, suppose $R = \{a=x_0,...x_{i-1},z,x_i,...,x_n = b\}$.

By definition, $U(Q,f) = \sum_{k=1}^nM_k\Delta x_k$, where $M_k$ and $\Delta x_k$ have the usual definitions.

Also, $U(R,f) = \sum_{k=1}^{i-1}M_k\Delta x_k + M_{z}^1(z - x_{i-1}) +M_{z}^{2}(x_i -z) +\sum_{k=i+1}^nM_k\Delta x_k$, where $M_{z}^1 = sup_{x\in[x_{i-1},z]}f(x)$ and $M_z^2 = sup_{x\in [z,x_i]}f(x)$.

It is easy to see that $M_i\Delta x_i \geq M_z^1(z-x_{i-1})+M_z^2(x_i-z)$, so $U(R,f)\leq U(Q,f)$.

A somewhat similar argument can be made for the lower sums. QED.

Your statement is a direct consequence of these inequalities.