Mathematics III - Foundations of Analysis - Summer 24

The Riemann-Stieltjes integral is a generalization of the well-known (to us) Riemann integral.  I'll give some motivations for this generalization.

• It is extremely useful in probability theory. When, for example, you would like to write the expectation of a random variable, sometimes, people spend time explaining the differences between discrete and continuous variables and give different formulas. Sometimes, they completely ignore random variables that are neither discrete nor continuous in order not to give even more formulas. With the Riemann-Stieltjes integrals, it all boils down to one formula. This makes so many other things in probability theory clearer and more lucid.
• The Ito integral is really a generalization of the Riemann-Stieltjes integral. It is useful for solving stochastic differential equations, partial differential equations and many other things. It is very useful in finanicial mathemtaics and life insurence.
• It makes many arguments in physics where calculus with respect to Dirac functions is done more rigorous.

A quick recap of the Riemann integral.  We defined Riemann integral by the lower and upper sums. A step function on $[a,b]$ is a function $\Phi$ such that there is a partition $P$ such that $\Phi$ is a constant $c_k$ on $(x_{k-1},x_k)$ for all $k=1,...,n.$.  Define the integral of $\Phi$ over $[a,b]$ as

$\int_a^b\Phi(x)dx=\sum_{k=1}^nc_k\Delta x_k$.

For a general function $f$ we say $f$ is Riemann-integrable on $[a,b]$ if and only if for every $\varepsilon>0$ there exist step functions $\Phi$ and $\Psi$ with $\Phi\le f\le\Psi$ such that

$\int_a^b\Psi(x)dx-\int_a^b\Phi(x)dx\le\varepsilon$.

Here we do the similar thing.  Let $\alpha$ be a monotonically increasing function on $[a,b]$ and let the real function $f$ be bounded on $[a,b]$.  We first show that the relation between the lower and upper Riemann-Stieltjes integrals:

$\boxed{\underline{\int_a^b} f d\alpha\le\bar{\int_a^b}f d\alpha}$.

Th proof is based on the inequality of lower sum $L(P,f, \alpha)$ and the upper sum $U(P,f,\alpha)$:

$L(P,f,\alpha)\le L(P^*,f,\alpha)$ and $U(P^*,f,\alpha)\le U(P,f,\alpha),$

where $P^*$ is the refinement of the partition $P$ of $[a,b]$.  Then we can prove the following theorem:

Theorem. (Cauchy criterion for integrability) $f\in{\mathcal R}(\alpha)$ on $[a,b]$ if and only if for every $\varepsilon>0$ there exists a partition $P$ such that

$U(P,f,\alpha)-L(P,f\alpha)\le\varepsilon$.

From this theorem we see that if the previous inequality holds for some $P$ and some $\varepsilon$ it holds for every refinment of $P$ with the same $\varepsilon$. 

If If this inequality holds for $P=\{x_0,..,x_n\}$ and if $t_i,s_i$ are arbitrary points in $[x_{i-1},x_i]$ then $f(s_i), f(t_i)\in[m_i,M_i]$ $\Rightarrow$ $|f(s_i)-f(t_i)|\le M_i-m_i$  $\Rightarrow$ 

$\boxed{\sum_{i=1}^n|f(t_i)-f(s_i)|\Delta\alpha_i \le U(P,f,\alpha)-L(P,f,\alpha)$

Furthermore, we have if $f\in{\mathcal R}(\alpha)$

$L(P,f,\alpha)\le\sum f(t_i)\Delta\alpha_i\le U(P,f,\alpha)$ and $L(P,f,\alpha)\le\int fd\alpha\le U(P,f,\alpha)$

Then

$\boxed{\big|\sum_{i=1}^nf(t_i)\Delta\alpha_i-\int_a^bfd\alpha\big|$.

The sum in the last inequality is called Riemann-Stieltjes sum (like Riemann sum).

As in Riemann integrability a continuous function on $[a,b]$ is Riemann-Stieltjes intebrable, using the fact that the function is in fact uniformly continuous together with the inequality in the Cauchy criterion for integrability:

$\boxed{\text{If } f \text{ is continuous on } [a,b], \text{ then } f\in{\mathcal R}(\alpha).}$

The assumption on $f$ being continuous on $[a,b]$ can be weakened to bounded on $[a,b]$ and discontinuity points are finitely many but $\alpha$ should be continuous at discontinuity points of $f$.

The proof is based on the fact that we can remove the finitely many disjoint intervals (of $[a,b]$) which cover the set of discontinuous points from the interval $[a,b]$ and the remaining set is still compact.  Thus $f$ is uniformaly continuous function.  Then we can use the similar argument in the proof for continuous function after a suitable choice of a partition.

Remark. 

1. A geometric interpretation of Riemann integral $f\ge0$ on $[a,b]$ is well-know to all of us:  area of the region bounded by the curve $y=f(x)$ and the $x$-axis and lines $x=a$ and $x=b$ respectively.    However what geometric interpretation a Riemann-Stieltjes integral can be?  You may be interested in finding an answer to it.  If you are not able to do so, you can consult the article by Bullock.
2. If $f$ and $\alpha$ have a common discontinuous point then the theorem need not hold.  Here is an example.  Take the step function: $f(x)=0$ for $x$ and $f(x)=1$ for $x\ge0$.  Now we take $\alpha=f$.  Clearly $f$ and $\alpha$ have a common discontinuity point at $0$.  We can prove that $\int_{-1}^1f(x)df$ does not exist.  This follows from the Cauchy criterion for integrability since $U(P,f,f)-L(P,f,f)=1$ for all partitions $P$ of $[-1,1]$.  However in practice we would very much like to have such an integral defined. So Riemann-Stieltjes integral is not the "best integral".  How to make the integral be defined? Let's look at  the example again.  If we replace $f$ by its left-continuous version $f_-(x)=0$ for $x\le0$ and $1$ if $x>0$  (note that $f(0)=1$ but $f_-(0)=0$.  We can now see that $\int_{-1}^1f_-df$ exists. The trick is to consider partitions which include a point at $0$.  What we have done is to change the location of the value of the function at a jump, and suddenly we the integral is defined. How can we avoid this artificial fix?  The answer is we should introduce a new types of integral using measure theory.  I hope this gives a motivation to study and understand carefully the measure theory.