# The Unapologetic Mathematician

## Integrators of Bounded Variation

Today I want to start considering Riemann-Stieltjes integrals where the integrator $\alpha$ is a function of bounded variation.

From our study of functions of bounded variation, we know that $\alpha$ can be written as the difference between two increasing functions $\alpha=\alpha_1-\alpha_2$. Then the linearity of the integral in the integrator tells us that

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[a,b\right]}fd\alpha_1-\int\limits_{\left[a,b\right]}fd\alpha_2$

Or does it? Remember that we have to understand this equation as saying that if two of the integrals exist then the third one does and the equality holds. But we also know that we’ve got a lot of choice in how to carve up $\alpha$, and it’s easily possible to do it in such a way that the integrals on the right don’t exist.

Luckily, we can show that there’s always at least one splitting of $\alpha$ into two parts like we want, and further so that all the integrals above exist and the equation holds. And it turns out that the first function is just the variation $V$! That is, I assert that if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on the interval $\left[a,b\right]$ and there exists some $M$ with $|f(x)|\leq M$ on $\left[a,b\right]$, then $f$ is Riemann-Stieltjes integrable with respect to $V$ on the same interval. And because $V$ is increasing, we can prove it by showing that $f$ satisfies Riemann’s condition with respect to $V$ on $\left[a,b\right]$.

So, given $\epsilon>0$ we need to find a partition $x_\epsilon$ so that for any finer partition $x$ we have $0\leq U_{V,x}(f)-L_{V,x}(f)<\epsilon$.

We start by picking a partition so that for any finer partition $x$ and any choices $t_i$ and $t_i'$ of tags for $x$ we have

$\displaystyle\left|\sum\limits_{i=1}^n\left(f(t_i)-f(t_i')\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\right|<\epsilon/4$

Such a partition exists because the sum is the difference between two Riemann-Stieltjes sums for $\int_{\left[a,b\right]}f\,d\alpha$, and we’re assuming that these sums converge to the value of the integral.

Now let’s refine it a bit. Any refinement will still satisfy the same condition we already have, but let’s make it also satisfy

$\displaystyle V(b)<\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|+\frac{\epsilon}{4M}$

This we can do because the total variation is the supremum of such sums over all partitions, and so we can find a fine enough partition to come within $\frac{\epsilon}{4M}$ of it. This new partition $x_\epsilon$ will be the one we need to establish Riemann’s condition. Now we must actually prove that it works.

First, notice that $\left(V(x_i)-V(x_{i-1})\right)-\left|\alpha(x_i)-\alpha(x_{i-1})\right|\geq0$, since the variation only grows as fast as the function itself changes. Also, the supremum $M_i(f)$ and infimum $m_i(f)$ in the $i$th subinterval are each less than $M$ in absolute value, so $M_i(f)-m_i(f)\leq2M$. Together, these show us that

$\displaystyle\sum\limits_{i=1}^n\left(M_k(f)-m_k(f)\right)\left(\left(V(x_i)-V(x_{i-1})\right)-\left|\alpha(x_i)-\alpha(x_{i-1})\right|\right)$
$\displaystyle\leq2M\sum\limits_{i=1}^n\left(\left(V(x_i)-V(x_{i-1})\right)-\left|\alpha(x_i)-\alpha(x_{i-1})\right|\right)$
$\displaystyle=2M\left(V(b)-\sum\limits_{i=1}^n\left|\alpha(x_i)-\alpha(x_{i-1})\right|\right)<\frac{\epsilon}{2}$

by the second property of $x_\epsilon$ we established when we refined our partition.

Now we set $h=\frac{\epsilon}{4V(b)}$. We want to pick two sets of tags so that $f(t_i)-f(t_i')>M_i(f)-m_i(f)-h$ for subintervals where $\alpha(x_i)\geq\alpha(x_{i-1})$, and so that $f(t_i')-f(t_i)>M_i(f)-m_i(f)-h$ for subintervals where $\alpha(x_i)<\alpha(x_{i-1})$. Then we find the inequality

$\displaystyle\sum\limits_{i=1}^n\left(M_i(f)-m_i(f)\right)\left|\alpha(x_i)-\alpha(x_{i-1})\right|$
$\displaystyle<\sum\limits_{i=1}^n\left(f(t_i)-f(t_i')\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)+h\sum\limits_{i=1}^n\left|\alpha(x_i)-\alpha(x_{i-1})\right|$
$\displaystyle<\frac{\epsilon}{4}+hV(b)=\frac{\epsilon}{4}+\frac{\epsilon}{4}=\frac{\epsilon}{2}$

by the first property of the partition we chose. Notice that in some of the subintervals I took the absolute value off of the change in $\alpha$ by switching which sample of $f$ I subtracted from which, introducing a new negative sign.

So, adding these two big inequalities together, we find that

$\displaystyle\sum\limits_{i=1}^n\left(M_i(f)-m_i(f)\right)\left(V(x_i)-V(x_{i-1})\right)|<\epsilon$

which shows that the upper and lower sums for the partition $x$ differ by less than $\epsilon$. Therefore $f$ indeed satisfies Riemann’s condition, and it thus integrable with respect to $V$ on $\left[a,b\right]$.

What this means is that integrals with respect to integrators of bounded variation can always be reduced to those with respect to increasing integrators, and thus to situations where Riemann’s condition can be brought to bear.

March 17, 2008 - Posted by | Analysis, Calculus

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