# The Unapologetic Mathematician

## Upper and Lower Integrals

Way back when, we talked about Darboux sums, where we used a particular recipe to pick the tags. Specifically, we defined the upper sum by picking a local maximum of $f$ in each subinterval as our tag, and the lower sum similarly. Today, let’s consider how this works with Riemann-Stieltjes sums, and specifically with an increasing integrator $\alpha$.

So given a partition $x$, we define the upper and lower Riemann-Stieltjes sums as follows:

$\displaystyle U_{\alpha,x}(f)=\sum\limits_{i=1}^n\max\limits_{x_{i-1}\leq t\leq x_i}\{f(t)\}\left(\alpha(x_i)-\alpha(x_{i-1})\right)$
$\displaystyle L_{\alpha,x}(f)=\sum\limits_{i=1}^n\min\limits_{x_{i-1}\leq t\leq x_i}\{f(t)\}\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

Now since we’ve chosen $\alpha$ to be increasing we can see that $\alpha(x_i)-\alpha(x_{i-1})\geq0$. Therefore we can find the inequalities

$\min\limits_{x_{i-1}\leq t\leq x_i}\{f(t)\}\left(\alpha(x_i)-\alpha(x_{i-1})\right)\leq f(t_i)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\leq\max\limits_{x_{i-1}\leq t\leq x_i}\{f(t)\}\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

for any possible tag $t_i\in\left[x_{i-1},x_i\right]$. And so any Riemann-Stieltjes sum for any collection of tags in the partition $x$ lies between the lower and upper sums: $L_{\alpha,x}(f)\leq f_{\alpha,x}\leq U_{\alpha,x}(f)$. Notice that we need $\alpha$ to be increasing here — if not, we can construct some pathological function $f$ that makes any combination of these inequalities fail.

Now the next step in Darboux integration was noting that any refinement of a partition drops the upper sum and raises the lower sum. Just like then, we can simply consider the process of adding a single new partition point, since any further refinement is just a sequence of new partition points. Then since any two partitions have a common refinement, we will see that the upper sum for any partition is greater than the lower sum for any other partition.

As before, adding a new point $c$ between $x_{i-1}$ and $x_i$ replaces the $i$th term in the sum with two terms:

$\max\limits_{x_{i-1}\leq t\leq c}f(t)\left(\alpha(c)-\alpha(x_{i-1})\right)+\max\limits_{c\leq t\leq x_i}f(t)\left(\alpha(x_i)-\alpha(c)\right)$

Each of the two maxima is at most the one maximum we had before, so we find

$\max\limits_{x_{i-1}\leq t\leq c}f(t)\left(\alpha(c)-\alpha(x_{i-1})\right)+\max\limits_{c\leq t\leq x_i}f(t)\left(\alpha(x_i)-\alpha(c)\right)\leq\max\limits_{x_{i-1}\leq t\leq c}f(t)\left(\alpha(x_i)-\alpha(c)+\alpha(c)-\alpha(x_{i-1})\right)$

which establishes this inequality. Notice that again we’ve had to multiply by differences between values of $\alpha$, and so as above this inequality hinges on the fact that our integrator is monotonically increasing.

Now that we know upper sums are always greater than lower sums (for increasing integrators!) we know that if they meet at all, it will be at the bottom of all the upper sums and the top of all the lower sums. Thus we define the “upper Stieltjes integral” $\overline{I}_{\alpha,\left[a,b\right]}(f)$ as the greatest lower bound of all the upper sums. Notice that if any lower sum exists then it’s a lower bound for the set of upper sums, and so Dedekind completeness tells us that this upper integral exists. Similarly, we define the lower integral $\underline{I}_{\alpha,\left[a,b\right]}(f)$ as the least upper bound of all the lower sums, with similar comments on its existence.

Since the upper sums are greater than the lower sums, we can see that the upper integral will be greater than the lower integral. Indeed, if $\epsilon>0$ is given then there is some partition $x$ so that $U_{\alpha,x}(f)\leq\overline{I}_{\alpha,\left[a,b\right]}(f)+\epsilon$, since the upper integral is a greatest lower bound. Then $U_{\alpha,x}(f)$ is an upper bound for the lower sums, and so $\underline{I}_{\alpha,\left[a,b\right]}(f)\leq\overline{I}_{\alpha,\left[a,b\right]}(f)+\epsilon$. Since $\epsilon$ was arbitrary, the lower integral is less than or equal to the upper integral.

Upper and lower integrals are in some ways as nice as Riemann-Stieltjes integrals. For instance, they’re both linear over the region of integration:

$\overline{I}_{\alpha,\left[a,b\right]+\left[b,c\right]}(f)=\overline{I}_{\alpha,\left[a,b\right]}(f)+\overline{I}_{\alpha,\left[b,c\right]}(f)$
$\underline{I}_{\alpha,\left[a,b\right]+\left[b,c\right]}(f)=\underline{I}_{\alpha,\left[a,b\right]}(f)+\underline{I}_{\alpha,\left[b,c\right]}(f)$

However, the upper integral is only convex over its integrand, while the lower integral is concave:

$\overline{I}_{\alpha,\left[a,b\right]}(f+g)\leq\overline{I}_{\alpha,\left[a,b\right]}(f)+\overline{I}_{\alpha,\left[a,b\right]}(g)$
$\underline{I}_{\alpha,\left[a,b\right]}(f+g)\geq\underline{I}_{\alpha,\left[a,b\right]}(f)+\underline{I}_{\alpha,\left[a,b\right]}(g)$

March 10, 2008 - Posted by | Analysis, Calculus

1. […] where attained its minimum. We even extended these to the Riemann-Stieltjes case and built up upper and lower integrals. And we can do the same thing […]

Pingback by Upper and Lower Integrals and Riemann’s Condition « The Unapologetic Mathematician | December 2, 2009 | Reply

2. Thank you for writing this post. I just had a few questions however (sorry, I do realize that this was written a very long time ago).

First, you state that the lower Riemann-Stieltjes sum is less than or equal to $f_{\alpha, x}$; by that do u mean the sum of the products of $f(t)$ and $\alpha(x_{i}) – \alpha{x_{i-1}}$, where $t$ is an any point in the ith interval $\left[ x_{i-1}, x_{i} \right]$?

Second, you state that the upper integral is a glb of the upper sums and yet state that U(f) is less than or equal to the upper integral plus epsilon for some partition x; do you mean strictly less than epsilon since the condition for Riemann-Stieltjes integrability is the equality of the upper and lower integrals?

Comment by jrupac | December 2, 2009 | Reply

3. For your first question, yes, I used that notation in the aforelinked post where I defined general Riemann-Stieltjes sums.

For the second, it really doesn’t matter since the result is that the difference is less than or equal to $\epsilon$ for arbitrarily small $\epsilon$.

Comment by John Armstrong | December 2, 2009 | Reply

4. […] this end, we have five assertions relating the upper and lower single and double integrals involving a function which is defined and bounded on the rectangle above. […]

Pingback by Iterated Integrals I « The Unapologetic Mathematician | December 16, 2009 | Reply

5. […] And we easily see that , where is the supremum of on the rectangle , so this is a bounded function as well. Thus the upper integral […]

Pingback by Iterated Integrals II « The Unapologetic Mathematician | December 17, 2009 | Reply

6. […] then for every . In fact, I used this term once before, when talking about upper and lower Darboux integrals. […]

Pingback by How They Get You | Drmathochist's Blog | August 28, 2010 | Reply