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$R$ is a refinement of $Q$

then

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

Rahul Silva
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    Assuming these are upper and lower sums, yes, that's correct. This is one of the key properties in the "Darboux development" of Riemann integration. – Ian Aug 13 '21 at 02:17
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    You can see that by the monotony of the upper and lower bounds. The upper bound is non increasing and the upper bound is non decreasing under refinement. – Fractal Admirer Aug 13 '21 at 02:23

1 Answers1

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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.

papi
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