# The Unapologetic Mathematician

## Increasing Integrators and Order

For what we’re about to do, I’m going to need a couple results about increasing integrators, and how Riemann-Stieltjes integrals with respect to them play nicely with order properties of the real numbers.

When we consider an increasing integrator we have a certain positivity result: if the integrand is nonnegative and the integral exists, then it is nonnegative as well. That is, for $\alpha$ increasing and $f(x)\geq0$ on $\left[a,b\right]$ we have $\int_{\left[a,b\right]}f\,d\alpha\geq0$ as long as it exists. This should be clear, since every Riemann-Stieltjes sum takes the form

$\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\geq0$

where the inequality follows because each value $f(t_i)$ and each difference $\alpha(x_i)-\alpha(x_{i-1})$ is nonnegative. Thus the limit of the sums must be nonnegative as well. From this and the linearity of the integral we see that if $\alpha$ is increasing and $f(x)\geq g(x)$ on $\left[a,b\right]$, then we have the inequality

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha\geq\int\limits_{\left[a,b\right]}gd\alpha$

as long as both integrals exist.

Now, when we talked about absolute values — the metric for the real numbers — we saw that the absolute value of a sum was always less than the sum of the absolute values. That is, $|x+y|\leq|x|+|y|$. And since an integral is just a limit of sums, it stands to reason that a similar result would hold here. Specifically, if $\alpha$ is increasing and $f$ is integrable with respect to $\alpha$ on $\left[a,b\right]$, then so is the function $|f|$, and further we have the inequality

$\displaystyle\left|\int\limits_{\left[a,b\right]}fd\alpha\right|\leq\int\limits_{\left[a,b\right]}|f|d\alpha$

Indeed, given a partition of $\left[a,b\right]$ the difference $M_i(f)-m_i(f)$ between the supremum and infimum of $f$ on the $i$th subinterval is the supremum of $f(x)-f(y)$, where $x$ and $y$ range across $\left[x_{i-1},x_i\right]$. Then, adapting the above inequality we see that

$||f(x)|-|f(y)||\leq|f(x)-f(y)|$

and so we conclude that

$M_i(|f|)-m_i(|f|)\leq M_i(f)-m_i(f)$

Then we can multiply by $\alpha(x_i)-\alpha(x_{i-1})$ and sum over a partition to find

$U_{\alpha,x}(|f|)-L_{\alpha,x}(|f|)\leq U_{\alpha,x}(f)-L_{\alpha,x}(f)$

Riemann’s condition then tells us that $|f|$ is integrable, and the inequality follows by the previous result.

We might hope to extend these results to integrators of bounded variation, but it won’t work right. This is because we go from increasing functions to functions of bounded variation by subtracting, and this operation will break the order properties.

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March 18, 2008 - Posted by | Analysis, Calculus

## 1 Comment »

1. […] that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if and are both integrable over a Jordan-measurable set , and if at each point , then […]

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