# upper and lower sums in Riemann integral

I want to prove this:

If $P^{*}$ is a finer partition than $P$, then show that $L(f,P, \alpha) \leq L(f,P^{*}, \alpha)$ and $U(f,P^{*}, \alpha) \leq U(f,P, \alpha)$.

If you have a set $S = \{1.2.3 \}$ then adding an element can change the infimum. If $S' = \{\frac{1}{2},1.2.3 \}$ then $\inf S' \leq \inf S$. I don't get how the above holds then.

-
smaller the interval bigger the infimum and smaller the supremum. Now apply this fact to your example. –  timhortons Aug 25 '11 at 4:24
What's $\alpha$? (Edit: Probably a Stieltjes integral) –  Dylan Moreland Aug 25 '11 at 5:34
When we talk about partitions we are usually fixing an inverval $[a, b]$, and our partitions then start at $a$ and end at $b$, so usually it isn't correct to introduce points outside of that interval and call the result a refinement. If this were allowed, then the statement of your problem wouldn't be true. –  Dylan Moreland Aug 25 '11 at 5:59

For this, I suppose that $f:[a,b]\to \mathbb{R}$, and $$P=\{a=x_0\lt x_1\lt\ldots\lt x_m=b\}$$ is a partition of $[a,b]$. Take care of the considerations about partitions given by Dyland's comment. Let $$S_m=\{1,2,\ldots, m\}.$$ Note that any partition $P^*$ finer than $P$ can be obtained from $P$ by adjoining it some points, we say $n$. So, we proceed by induction on $n$.

Suppose that $P^*$ is obtained by adding a point $x$ to $P$. Then $x\in (x_{r-1},x_r)$, for some $r\in S_m$. Let $$m_i=\inf f([x_{i-1},x_i]) \text{ for } i\in S_m$$ $$m_r'=\inf f([x_{r-1},x]) \text{ and } m_r''=\inf f([x,x_r])$$ Then \begin{align*} L(f,P^*)-L(f,P) &= m_r'(x-x_{r-1}) + m_r''(x_r-x)-m_r(x_r-x_{r-1})\\ &= (m_r'-m_r)(x-x_{r-1}) + (m_r''-m_r)(x_r-x)\\ &\geq 0. \end{align*} You can verify that the factors that involve infs are nonnegative, and this proves the claim. I'll leave the inductive step to you.

For upper sums is similar. For the Riemann-Stieltjes integral the lower and upper sums are studied usually for increasing integrators $\alpha$, and then the proof is the same thing.

-
@integratethis: Is not this a correct answer to your question? –  leo Aug 28 '11 at 18:53

Remember that you don't take infimum on the set of division points; you take the infimum value of the function in each of the intervals in the division. Now finer division induces smaller intervals, hence you take the infimum on smaller sets and it can only rise.

Think of it this way: the finer the partition, the more intervals there are, and so each interval is smaller, and so the potential error is smaller (you need more extremal points to cause large errors than you needed before).

-