# 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 | 6 Comments