Introducing Riemann-Stieltjes Integral

Motivation

Riemann-Stieltjes integral is a generalisation of Riemann integral, the one every college student studies in their calculus class, and is a little bit more difficult to understand. Nevertheless it has advantages of its own, as we will show below. Before seeing the definition and properties of this integral, we first raise some questions that will can motivate our study.

When talking about $\int_a^b fdg$, one may simply think about $\int_a^b fg’dx$. But is it even necessary that $g$ is differentiable? What would happen if $g$ is simply continuous, or even not continuous? Further, given that $g$ is differentiable, can we prove that

$$\int_a^b f(x)dg(x)=\int_a^bf(x)g'(x)dx$$

in a general way(without assuming $f$ is differentiable)? Although integration can be connected to differentiation, it should not be mandatory to lock ourselves into $C^1$ functions, $C^2$ functions or smooth functions all the time.

Another motivation comes from probability theory. Oftentimes one need to consider discrete case ($\sum$) and continuous case ($\int$) separately. One may say that integral is the limit of summation, but it would be weird to write $\int$ as $\lim\sum$ every time. However, if we have a way to write a sum, for example the expected value of a discrete variable ($E(X)$), as an integral, things would be easier. Of course, we don’t want to write such a sum as another sum by adding up the integral on several disjoint segments. That would be weirder.

If you have learned measure theory, you will know that Lebesgue integral does not perfectly cover Riemann integral. For example, $\int_{0}^{\infty}\frac{\sin{x}}{x}dx$ is not integrable in the sense of Lebesgue but Riemann. We cannot treat Lebesgue integral as a perfect generalization of Riemann integral. In this blog post however, we will be studying a faithful generalization of Riemann integral, adding the name of Stieltjes.

We are trying our best to prevent ourselves from using $\sup$, $\inf$, and differentiation theory. But $\varepsilon-\delta$ language is heavily used here, so make sure that you are good at it.

Riemann-Stieltjes Integral

By a partition $P$ on $[a,b]$ we mean a sequence of numbers $(x_n)$ such that

$$a=x_0 \leq x_1 \leq \cdots \leq x_n=b$$

and we associate its size by

$$\sigma(P)=\max_{k}(x_{k+1}-x_k).$$

Let $f$, $g$ be bounded real function on $[a,b]$ (again, no continuity or differentiability required). Given a partition $P$ and numbers $c_k$ with $x_k \leq c_k \leq x_{k+1}$, we define the Riemann-Stieltjes sum (RS-sum) by

$$S(P,f,g)=\sum_{k=0}^{n-1}f(c_k)[g(x_{k+1})-g(x_k)].$$

We say that the limit

$$\lim_{\sigma(P) \to 0}S(P,f,g)$$

exists if there exists some $L \in \mathbb{R}$ such that give $\varepsilon>0$, there exists $\delta>0$ such that whenever $\sigma(P)<\delta$, we have

$$|S(P,f,g)-L|<\varepsilon.$$

In this case, we say $f$ is $RS(g)$-integrable, and the limit is denoted by

$$\int_a^bfdg.$$

This is the so-called Riemann-Stieltjes integral. When $g(x)=x$, we get Riemann integral naturally.

Remarks: Further generalization still available

This integral method can be generalized to Banach space. Let $f$, $g$ be bounded maps of $[a,b]$ into Banach spaces $E$, $F$ respectively. Assume we have a product $E \times F \to G$ denoted by $(u,v) \mapsto uv$ with $\lVert uv \rVert \leq \lVert u \rVert \lVert v \rVert$. Then by replacing the absolute value by norm, still we get the Riemann-Stieltjes integral, although in this case we have

$$\int_a^b fdg \in G$$

and $G$ is not necessary to be $\mathbb{R}$. This is different from Bochner integral, since no measure theory evolved here.

Linearity with respect to $f$ and $g$

First, we shall show that RS(g)-integrable functions form a vector space. To do this, it suffices to show that

$$f \mapsto S(P,f,g)$$

and

$$g \mapsto S(P,f,g)$$

are linear. This follows directly from the definition of RS-sum. Let’s see the result.

Suppose we have

$$\int_a^b fdg=I, \quad \int_a^b hdg=J, \int_a^b fdu=K$$

Then we have the following identities for $\alpha \in I$.

1. $\int_a^b \alpha fdg=\alpha I$.
2. $\int_a^b (f+h)dg=I+J$.
3. $\int_a^bfd(g+u)=I+K$.
4. $\int_a^b fd(\alpha g)=\alpha I$.

Proof. We shall show 2 for example. Other three identities follows in the same way.

Notice that the existence of the limit of RS-sum depends only on the size of $P$. For $\varepsilon>0$, there exists some $\delta_1,\delta_2>0$ such that

$$|S(P,f,g)-I|<\frac{\varepsilon}{2},\quad |S(P,h,g)-J| < \frac{\varepsilon}{2}$$

when $\sigma(P)<\delta_1$ and $\sigma(P)<\delta_2$ respectively. By picking $\delta=\min(\delta_1,\delta_2)$, we see for $\sigma(P)<\delta$, we have

$$\begin{aligned} |S(P,f+h,g)-(I+J)|&=|(S(P,f,g)-I)+(S(P,h,g)-J)| \\ &\leq |S(P,f,g)-I| + |S(P,h,g)-J| \\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}$$

Integration by parts but no differentiation

$f \in RS(g)$ if and only if $g \in RS(f)$. In this case, we also have integration by parts:

$$\int_a^b fdg + \int_a^b gdf=f(b)g(b)-f(a)g(a)$$

You may not believe it, but differentiation does not play any role here, as promised at the beginning.

Proof. Using the summation by parts (by Abel), we have

$$\begin{aligned} S(P,f,g)&=\sum_{k=0}^{n-1}f(c_k)[g(x_{k+1})-g(x_k)] \\ &=-\sum_{k=1}^{n-1}g(x_k)[f(c_k)-f(c_{k-1})]+f(c_{n-1})g(b)-f(c_0)g(a). \\ \end{aligned}$$

By writing

$$S(P,f,g)=S(P,f,g)+f(a)g(a)-f(a)g(a)+f(b)g(b)-f(b)g(b)$$

we have

$$S(P,f,g)=f(b)g(b)-f(a)g(a)-S(Q,g,f)$$

where

$$S(Q,g,f)=\sum_{k=1}^{n-1}g(x_k)[f(c_k)-f(c_{k-1})]+[f(b)-f(c_{n-1})]g(b)+[f(c_0)-f(a)]g(a).$$

Consider the partition $Q$ by

$$y_k=\begin{cases} a &\quad k=0 \\ c_{k-1}&\quad k<n \\ b &\quad k=n \end{cases}$$

we have $x_0,x_1,\cdots,x_{n-1},x_k$ to be intermediate points, and

$$S(Q,g,f)=\sum_{k=0}^{n-1}g(x_k)[f(y_{k+1})-f(y_k)].$$

Since $0 < \sigma(Q) \leq 2\sigma(P) \leq 4\sigma(Q)$, when $\sigma(P) \to 0$, we also have $\sigma(Q) \to 0$ and vice versa. Suppose now $\int_a^b gdf$ exists, we have

$$\lim_{\sigma(P) \to 0}S(P,f,g)=f(b)g(b)-f(a)g(a)-\int_a^bgdf=\int_a^bfdg.$$

And integration by parts follows.

Suppose $\int_a^bfdg$ exists, then

$$\lim_{\sigma(Q) \to 0}S(Q,g,f)=f(b)g(b)-f(a)g(a)-\int_a^b fdg=\int_a^b gdf.$$

This proves the proposition. $\square$

The flexibility of Riemann-Stieltjes integral

As said before, we want to represent both continuous and discrete case using integral. For measure theory, we have Lebesgue measure and counting measure. But in some cases, this can be done using Riemann-Stieltjes integral as well. Ordinary Riemann integral and finite or infinite series are all special cases of Riemann-Stieltjes integral.

From integral to series (discrete case)

To do this, we need the unit step function by

$$I(x)=\begin{cases} 0 \quad x \leq 0, \\ 1 \quad x > 0 .\end{cases}$$

If $a<s<b$, $f$ is bounded on $[a,b]$ and continuous at $s$, by putting $g(x)=I(x-s)$, we have

$$\int_a^b fdg=f(s)$$

Proof. A simple verification shows that $\int_a^b fdg=\int_s^b fdg$ (by unwinding the RS-sum, one see immediately that $g(x_k)=0$ for all $x_k\leq s$, therefore the partition before $s$ has no tribute to the value of the integral). Now consider the partition $P$ by

$$s=x_0<x_1<\cdots<x_n=b.$$

We see

$$S(P,f,g)=\sum_{k=0}^{n-1}f(c_k)(g(x_{k+1})-g(x_k))=f(c_0)(g(x_1)-g(x_0))=f(c_0).$$

As $x_1 \to s$, we have $c_0 \to s$, since $f$ is continuous at $s$, we have $f(c_0) \to f(s)$ as desired. $\square$

By the linearity of RS integral, it is easy to generalize this to the case of finite linear combination. Namely, for $g(x)=\sum_{k=1}^{n}c_nI(x-s_n)$, we have

$$\int_a^b fdg=\sum_{k=1}^{n}c_nf(s_n).$$

But now we are discussing the infinite case.

Suppose $c_n \geq 0$ for all $n \ge 0$ and $\sum_{n \ge 0} c_n$ converges, $(s_n)$ is a sequence of distinct points in $(a,b)$, and

$$g(x)=\sum_{n \ge 0}c_nI(x-s_n).$$

Let $f$ be continuous on $[a,b]$. Then

$$\int_a^b fdg=\sum_{n}c_nf(s_n)$$

Proof. First it’s easy to see that $g(x)$ converges for every $x$, and is monotonic with $g(a)=0$, $g(b)=\sum_n c_n$. For given $\varepsilon>0$, there exists some $N$ such that

$$\sum_{N+1}^{\infty}c_n<\varepsilon.$$

Putting

$$g_1(x)=\sum_{n=1}^{N}c_nI(x-s_n),\quad g_2(x)=\sum_{N+1}^{\infty}c_nI(x-s_n)=g(x)-g_1(x)$$

we have

$$\int_a^b fdg_1=\sum_{n=1}^{N}c_nf(s_n).$$

By putting $M=\sup|f(x)|$, we see

$$\left\vert\int_a^b fdg_2 \right\vert=\left\vert\int_a^b fdg-\int_a^bfdg_1 \right\vert=\left\vert\int_a^b fdg-\sum_{n=1}^{N}c_nf(s_n)\right\vert \leq M\varepsilon$$

The inequality holds since $|g_2(b)-g_2(a)|<\varepsilon$. Since $M$ is finite, when $N \to \infty$, we have the desired result. $\square$

Transformed into ordinary Riemann integral (continuous case)

Finally we will discuss differentiation. The following theorem shows the connection between RS integral and Riemann integral.

Let $f$ be continuous and suppose that $g$ is real differentiable on $[a,b]$ while $g’$ is Riemann integrable as well, then $f \in RS(g)$ and

$$\int_a^b fdg=\int_a^b fg'dx$$

Proof. By mean value theorem, for each $k$, we have

$$g(x_{k+1})-g(x_k)=g'(\zeta_k)(x_{k+1}-x_k).$$

The RS-sum can be written as

$$S(P,f,g)=\sum_{k=0}^{n-1}f(c_k)[g(x_{k+1})-g(x_k)]=\sum_{k=0}^{n-1}f(c_k)g'(\zeta_k)(x_{k+1}-x_k).$$

Since $g’$ is Riemann integrable, we have

$$\sum_{k=0}^{n-1}|g'(c_k)-g'(\zeta_k)|(x_{k+1}-x_k) <\varepsilon$$

given that $|S(P,g’,x)-\int_a^b g’dx|<\varepsilon$. Therefore

$$\left\vert\sum_{k=0}^{n-1}f(c_k)g'(\zeta_k)(x_{k+1}-x_k)-\sum_{k=0}^{n-1}f(c_k)g'(c_k)(x_{k+1}-x_k)\right\vert\leq M\varepsilon$$

where $M=\sup|f(x)|<\infty$ ($f$ is assumed to be bounded.) . Also notice that $fg’$ is integrable since $f$ is continuous. Therefore

$$\begin{aligned} \left\vert S(P,f,g)-\int_a^bfg'dx \right\vert&=\left\vert S(P,f,g)-S(P,fg',x)+S(P,fg',x)-\int_a^bfg'dx \right\vert \\ &\leq \left\vert S(P,f,g)-S(P,fg',x) \right\vert+\left\vert S(P,fg',x)-\int_a^bfg'dx \right\vert \\ &< (M+1)\varepsilon. \end{aligned}$$

Therefore,

$$\int_a^bfdg=\int_a^b fg'dx,$$

which proves the theorem. $\square$

To sum up, given $\varepsilon>0$, there exists some $\delta>0$ such that if $\sigma(P)<\delta$, we have

$$\left|S(P,g',x)-\int_a^b g'dx\right|<\varepsilon/(M+1)$$

and

$$\left\vert S(P,fg',x)-\int_a^bfg'dx \right\vert<\varepsilon/(M+1).$$

After some estimation, we get

$$\left|S(P,f,g)-\int_{a}^{b}fg'dx \right|<(M+1)\frac{\varepsilon}{M+1}=\varepsilon.$$