# The Unapologetic Mathematician

## Riemann’s Condition

If we want our Riemann-Stieltjes sums to converge to some value, we’d better have our upper and lower sums converge to that value in particular. On the other hand, since the upper and lower sums sandwich in all the others, their convergence is enough for the rest. And their convergence is entirely captured by their lower and upper bounds, respectively — the upper and lower Stieltjes integrals. So we want to know when $\underline{I}_{\alpha,\left[a,b\right]}(f)=\overline{I}_{\alpha,\left[a,b\right]}(f)$.

We’ll prove this equality in general by showing that the difference has to be arbitrarily small. That is, for any partition $x$ of $\left[a,b\right]$ we have the inequalities

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

by definition. Subtracting the one from the other we find

$\overline{I}_{\alpha,\left[a,b\right]}(f)-\underline{I}_{\alpha,\left[a,b\right]}(f)\leq U_{\alpha,x}(f)-L_{\alpha,x}(f)$

So if given an $\epsilon>0$ we can find a partition $x$ for which the upper and lower sums differ by less than $\epsilon$ then the difference between the upper and lower integrals must be even less. If we can do this for any $\epsilon>0$, we say that the function $f$ satisfies Riemann’s condition with respect to $\alpha$ on $\left[a,b\right]$.

The lead-up to the definition of Riemann’s condition shows us that if $f$ satisfies this condition then the lower and upper integrals are equal. Then just like we saw happen with Darboux sums we can squeeze any Riemann-Stieltjes sum between and upper and a lower sum. So if the upper and lower integrals are both equal to some value, then the limit of the Riemann-Stieltjes sums over tagged partitions must exist and equal that value, and thus $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\left[a,b\right]$.

Now what if the $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\left[a,b\right]$? We would hope that $f$ then satisfies Riemann’s condition with respect to $\alpha$ on $\left[a,b\right]$, and so these three conditions are equivalent. So given $\epsilon>0$ we need to find an actual partition $x$ of $\left[a,b\right]$ so that $0\leq U_{\alpha,x}(f)-L_{\alpha(x)}<\epsilon$.

Since we’re assuming that $f$ is Riemann-Stieltjes integrable, we’ll call the value of the integral $A$. Then we can find a tagged partition $x_\epsilon$ so that for any finer tagged partitions $x=((x_0,...,x_n),(t_1,...,t_n))$ and $x'=((x_0,...,x_n),(t_1',...,t_n'))$ we have

$\displaystyle\left|A-\sum\limits_{i=1}^nf(t_i)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\right|<\frac{\epsilon}{3}$
$\displaystyle\left|A-\sum\limits_{i=1}^nf(t_i')\left(\alpha(x_i)-\alpha(x_{i-1})\right)\right|<\frac{\epsilon}{3}$

Combining these we find that

$\displaystyle\left|\sum\limits_{i=1}^n\left(f(t_i)-f(t_i')\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\right|<\frac{2}{3}\epsilon$

Now as we pick different $t$ and $t'$ we can make the difference in values of $f$ get as large as that between $M_i=\max\limits_{x_{i-1}\leq t\leq c}f(t)$ and $m_i=\min\limits_{x_{i-1}\leq t\leq c}f(t)$. So for any $h>0$ we can choose tags so that $f(t_i)-f(t_i')>M_i-m_i-h$. In particular, we can consider $h=\frac{\epsilon}{3(\alpha(b)-\alpha(a))}$, which is positive because $\alpha$ is increasing.

The difference between the upper and lower sums is

$\displaystyle\sum\limits_{i=1}^n\left(M_i-m_i\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

which is then less than

$\displaystyle\sum\limits_{i=1}^n\left(f(t_i)-f(t_i')\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)+h\sum\limits_{i=1}^n\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

which is then less than $\epsilon$.

Thus we establish the equivalence of Riemann’s condition and Riemann-Stieltjes integrability, as long as the integrator $\alpha$ is increasing.