Theorem – Upper/Lower sums inequality

Contents
1. Theorem
2. Proof

Theorem

Let $P$ and $Q$ be two partitions of an interval $[a,b]$ with $a\neq b$ . Let $P$ be a refinement of $Q$ . If the $L(f,Q)$ , $L(f,P)$ , $U(f,Q)$ and $U(f,P)$ are the upper and lower sums of a bounded function $f$ , then

$L(f, Q) \leq L(f, P) \leq U(f, P) \leq U(f, Q)$

Proof

The proof makes use of the following fact. If the maximum value of $f(x)$ on an interval $[a,b]$ is $f(i)$ , and we separate the interval into two different intervals, $[a, s]$ and $[s, b]$ , then either the maximum of $[a, s]$ is equal to the maximum of $[s, b]$ , which equals $f(i)$ , or the maximum of one of the two intervals is strictly smaller than $f(i)$ . A similar fact is also true for the minimum.

Now, recall the definitions of $L(f,Q)$ , $L(f,P)$ , $U(f,Q)$ and $U(f,P)$ . The inequality $L(f, P) \leq U(f, P)$ follows from the definitions of the lower and upper sums.

To prove the inequality $L(f, P) \leq L(f, Q)$ , first note that there must be at least one sub-interval (as the interval is closed, so we can simply take $x_1 = a$ and $x_n = b$ ). Then, let us consider any sub-interval $[x_{i-1} x_i]$ for some $2\leq i \leq |Q|$ . Then, let us split this interval into two disjoint intervals, $[x_{i-1}, x_i] = [x_{i-1}, x*] \cup [x*, x_i]$ .

Then, let us denote $\Delta x_i = |x_i - x_{i-1}|$ , $\Delta y = |x_{i-1} - x*|$ and $\Delta z = |x_i - x*|$ . This gives $\Delta x_i = \Delta y + \Delta z$ . Because the summand of the lower sum (for the interval $[x_{i-1}, x_i]$ is the product of $c_i = inf\{f(x)~|~x\in[x_{i-1}, x_i]\}$ and $\Delta x_i$ , we obtain (for the interval $[x_{i-1}, x_i]$ ), $c_i \Delta x_i$ .

Similarly, for the two separate intervals $[x_{i-1}, x*], [x*, x_i]$ , we have

$sup\{f(x)~|~x\in [x_{i-1}, x*] \} \times \Delta y$

And,

$sup\{f(x)~|~x\in [x*, x_i] \} \times \Delta z$

We now have three situations. Firstly, if $sup\{f(x)~|~x\in [x*, x_i] \} = sup\{f(x)~|~x\in [x_{i-1}, x*] \} = sup\{f(x)~|~x\in [x_{i-1}, x_i] \}$

then, $\Delta x_i = \Delta y + \Delta z$ implies that

$(sup\{f(x)~|~x\in [x*, x_i] \}\times \Delta y) + (sup\{f(x)~|~x\in [x_{i-1}, x*] \}\times \Delta z) \\\\ = (sup\{f(x)~|~x\in [x_{i-1}, x_i] \}\times \Delta y) + (sup\{f(x)~|~x\in [x_{i-1}, x_i] \} \times \Delta z) \\\\ = sup\{f(x)~|~x\in [x_{i-1}, x_i] \} \times ( \Delta y + \Delta z) \\\\ = sup\{f(x)~|~x\in [x_{i-1}, x_i] \}\times \Delta x_i$ .

So this means that the lower sum will be unaltered. However, in the case where $sup\{f(x)~|~x\in [x*, x_i] \} \neq sup\{f(x)~|~x\in [x_{i-1}, x*] \}$ , we have that one of these values is lower than the other, which means that the value of the lower sum is lower when using the refinement of $Q$ . This proves the first inequality.

The third inequality is proven in a similar way.