Riemann-Stieltjes Integral

Definition (Riemann Integrals)

Let $[a,b]$ be a given integral. By a partition $P$ of $[a,b]$ we mean a finite set of points $x_0,\dots ,x_n$ where

$$a=x_0\le x_1\le \cdots \le x_{n-1}\le x_n=b$$

We write

$$\Delta x_i=x_i-x_{i-1}(i=1,\dots ,n)$$

Now suppose $f$ is a bounded real function defined on $[a,b]$. Corresponding to each partition $P$ of $[a,b]$ we put

$$\begin{array}{rl}{M}_i& =\sup f(x)(x_{i-1}\le x\le x_i)\\ m_i& =\inf f(x)(x_{i-1}\le x\le x_i)\\ U(P,f)& =\sum _{i=1}^n{M}_i\Delta x_i\\ L(P,f)& =\sum _{i=1}^n{m}_i\Delta x_i\end{array}$$

$M_i$ and $m_i$ are the Supremum and Infimum of $f$ where the domain is restricted to partition $i$. Then, $U,L$ are just the sums of the areas where $M_i,m_i$ are the “height” of the rectangle.

and finally,

$$\begin{array}{rl}\overline{{\int }_a^b}fdx& =\inf U(P,f)\\ \underline{{\int }_a^b}fdx& =\sup L(P,f)\end{array}$$

where the Infimum and Supremum are taken over all partitions $P$ of $[a,b]$. The LHS are called the upper and lower Riemann integrals of $f$ over $[a,b]$ respectively.

Definition (Riemann-Integrable)

If the upper and lower integrals are equal, then $f$ is Riemann-Integrable on $[a,b]$. we write $f\in \mathcal{R}$ (that is, $\mathcal{R}$ denotes the set of Riemann-Integrable functions), and we denote the common value by

$${\int }_a^{b}fdx\text{or}{\int }_a^{b}f(x)dx$$

This is the Riemann Integral of $f$ over $[a,b]$. Since $f$ is bounded, there exist two numbers, $m$ and $M$, such that

$$m\le f(x)\le M(a\le x\le b)$$

Hence, for every $P$,

$$m(b-a)\le L(p,f)\le U(P,f)\le M(b-a)$$

the numbers $L(P,f)$ and $U(P,f)$ form a bounded set. Thus, the upper and lower integrals are defined for every bounded function $f$.

This is a bit confusing to understand, but what this it is comparing is 4 different rectangles.

1. The first rectangle picks the “smallest height” from the whole $[a,b]$ and is applied to every partition.
2. The $\inf$ at the $i$th partition is not necessarily the $\inf$ over $[a,b]$, so it could be a larger height. Thus it is $\le$.
3. This is quite obvious.
4. The same logic follows as (1) but with the largest height applied to every partition.

Definition (Riemann-Stieltjes Integral)

Let $\alpha$ be a monotonically increasing function on $[a,b]$ (since $\alpha (a),\alpha (b)$ are finite, it follows that $\alpha$ is bounded on $[a,b]$) . Corresponding to each partition $P$ of $[a,b]$, we write

$$\Delta {\alpha }_i=\alpha (x_i)-\alpha (x_{i-1})$$

It is clear that $\Delta {\alpha }_i\ge 0$. For any real function $f$ which is bounded on $[a,b]$ we put

$$\begin{array}{rl}U(P,f\alpha )& =\sum _{i=1}^n{M}_i\Delta {\alpha }_i\\ L(P,f,\alpha )& =\sum _{i=1}^n{m}_i\Delta {\alpha }_i\end{array}$$

where $M_i,m_i$ have the same meaning as Definition (Riemann Integrals) and we define

$$\begin{array}{rl}\overline{{\int }_a^b}fd\alpha & =\inf U(P,f,\alpha )\\ \underline{{\int }_a^b}fd\alpha & =\sup L(P,f,\alpha )\end{array}$$

and the $\inf ,\sup$ is being taken over all partitions. If equal, then

$${\int }_a^{b}fd\alpha \text{or}{\int }_a^{b}f(x)d\alpha (x)$$

This is the Riemann-Stieltjes Integral of $f$ with respect to $\alpha$ over $[a,b]$.

This is a generalization of the Riemann Integral. Think of $\alpha$ as a weighing the standard area. The Riemann Integral can be computed when $\alpha (x)=x$.

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From here on out, assume that $f$ is real and bounded, and that $\alpha$ is monotonically increasing.

Definition (Refinement)

We say that the partition $P^{\ast }$ is a refinement of $P$ if $P^{\ast }\supset P$. Given two partitions $P_1,P_2$, we say that $P^{\ast }$ is a common refinement if $P^{\ast }=P_1\cup P_2$.

Theorem (Upper Limits of Refinements)

If $P^{\ast }$ is a refinement of $P$ then

$$L(P,f,\alpha )\le L(P^{\ast },f,\alpha )\text{and}U(P^{\ast },f,\alpha )\le U(P,f,\alpha )$$

Proof: It suffices to prove when $P^{\ast }$ has one more point than $P$. Let this point be $c$ such that $\{c\}=P^{\ast }\setminus P$. We can assume that $x_{i-1}<c<x_i$ for some $1\le i\le n$. Then

$$\begin{array}{rl}L(P^{\ast },f,\alpha )-L(P,f,\alpha )& =\underset{[x_{i-1},c]}{\inf }f\cdot ([\alpha (c)-\alpha (x_{i-1})])\\ & +\underset{[c,x_i]}{\inf }f\cdot (\alpha (x_i)-\alpha (c))\\ & -\underset{[x_{i-1},x_i]}{\inf }f\cdot (\alpha (x_i)-\alpha (x_{i-1}))\end{array}$$

Recall that these are sums.

Now since

$$\underset{[x_{i-1},c]}{\inf }f+\underset{[c,x_i]}{\inf }\ge \underset{[x_{i-1},x_{i]}}{\inf }f$$

we get that $L(P^{\ast },f,\alpha )-L(P,f,\alpha )\ge 0$.

Proposition (Separate Partition Limits)

For any two partitions $P,P^{\prime }$ we have $L(P,f,\alpha )\le U(P^{\prime },f,\alpha )$.

Proof: Take $P^{\ast }=P\cup P^{\prime }$, the common refinement. Then

$$L(P,f,\alpha )\le \stackrel{(a)}{L}(P^{\ast },f,\alpha )\le \stackrel{(b)}{U}(P^{\ast },f,\alpha )\le \stackrel{(a)}{U}(P^{\prime },f,\alpha )$$

which is true from

• $(a):$ Theorem (Upper Limits of Refinements)
• $(b):$ For any ${\xi }_i$, we see $m_i\le f({\xi }_i)\le M_i$.

Proposition (Riemann LB < UB)

$$\underline{\int }\le \overline{\int }$$

Proof: Since this is essentially comparing the Infimum and Supremum, this follows from Proposition (Separate Partition Limits).

Lemma (Riemann-Stieltjes Integrability)

We say $f$ is Riemann-Stieltjes Integral iff $\forall \epsilon >0,\exists$ partition $P$ such that $U(P,f,\alpha )-L(P,f,\alpha )<\epsilon$.

Proof: $(⟹)$. If Riemann-Stieltjes integrable, then $\overline{\int }=\underline{\int }$. Then for $\epsilon /2,\exists P$ such that

$$U(P,f,\alpha )<\overline{\int }+\epsilon /2$$

Similarly, $\exists P^{\prime }$ such that

$$L(P^{\prime },f,\alpha )\ge \underline{\int }-\epsilon /2$$

So, we have that

$$U(P,f,\alpha )-L(P^{\prime },f,\alpha )<\epsilon$$

Then take $P\cup P^{\prime }$ and apply Theorem (Upper Limits of Refinements).

$(⟸)$. The assumption implies that $\overline{\int }-\underline{\int }<\epsilon$ for all $\epsilon >0$. Then clearly they are equal and thus integrable by Definition (Riemann-Stieltjes Integral).

Theorem (Continuity Implies Riemann-Stieltjes Integrable)

If $f$ is continuous then $f$ is Riemann-Stieltjes Integrable for any $\alpha$.

• Recall that we let $f$ be real and bounded.
• $\alpha$ is monotonically increasing.

Proof: We know $U(P,f,\alpha )-L(P,f,\alpha )={\sum }_{i=1}^n(M_i-m_i)\Delta {\alpha }_i$. Then over any closed interval, continuous functions are uniformly continuous. So, for any $\eta >0,\exists \delta >0$ such that if $|x-y|<\delta ,|f(x)-f(y)|<\eta$. So if $|x_i-x_{i-1}|<\delta ,$ then $M_i-m_i\le \eta$.

Note that $M,m$ are defined as the Supremum and Infimum of $f$ on partition $P_i$.

We take a $P$ such that $|x_i-x_{i-1}|<\delta$ for any $i$. Then

$$U(P,f,\alpha )-L(P,f,\alpha )\le \sum _{i=1}^n\eta (\alpha (x_i)-\alpha (x_{i-1}))=\eta (\alpha (b)-\alpha (a))$$

This goes to $0$ so $\eta (\alpha (b)-\alpha (a))<\epsilon$.

Proposition (Finite Discontinuities on Riemann-Stieltjes)

If $f$ has finite discontinuities, (i.e. $\exists u_1,\dots ,u_r$ such that $f$ is continuous on $[a,b]\setminus \{u_1,\dots ,u_r\}$) and if $\alpha$ is increasing, and continuous at $u_1,\dots ,u_r$ then $f\in \mathcal{R}(\alpha )$.

Proof: Fix $\epsilon >0$. Let $|f|\le M$ for $M>0$. Then since $\alpha$ is continuous at $u_i$, $\exists {\epsilon }_i$ such that

$$\alpha (u_i+{\epsilon }_i)-\alpha (u_i-{\epsilon }_i)<\frac{\epsilon }{4rM}$$

Since $\alpha$ is continuous, the oscillation must approach $0$. It essentially forces the change in $\alpha$ in neighborhoods at the discontinuities to be arbitrarily small. Define

$$E:=\bigcup _{i=1}^n(u_i-{\epsilon }_i,u_i+{\epsilon }_i)$$

and define

$$F:=[a,b]\setminus E$$

which is closed and hence compact. Then since $f$ is continuous on $[a,b]$ it is uniformly continuous on $f$ by theorem. We cut $f$ into subintervals to guarantee

$$M_i-m_i<\frac{\epsilon }{2(\alpha (b)-\alpha (a))}$$

Now put all subintervals into partition $P$. Then

$$U-L=\underset{\text{intervals covering }E}{\underbrace{\sum _{i=1}^r(M_i-m_i)\Delta {\alpha }_i}}+\underset{\text{intervals covering }F}{\underbrace{\sum _{i=1}^n(M_i-m_i)\Delta {\alpha }_i}}\le \left(\sum _{i=1}^{r}2M\cdot \frac{\epsilon }{4rM}\right)+\left(\frac{\epsilon }{2(\alpha (b)-\alpha (a))}\sum _{i=1}^n\Delta {\alpha }_i\right)$$

which is $<\epsilon /2+\epsilon /2=\epsilon$.

Proposition (Increasing Riemann-Stieltjes)

If $f$ is increasing or decreasing then $f$ is Riemann-Stieltjes Integrable for any increasing continuous $\alpha$.

Proof: Since $f$ is increasing, $M_i-m_i=f(x_i)-f(x_{i-1})$. If we cut the subintervals such that for some $N$,

$$\Delta {\alpha }_i<\frac{1}{N(f(b)-f(a))}$$

is the length of each subinterval. Then

$$U-L\le \sum _{i=1}^n(f(x_i)-f(x_{i-1}))\frac{1}{N(f(b)-f(a))}=\frac{1}{N}<\epsilon$$

Note that we cancel out $f(b)-f(a)$ because we have a telescoping sum.

Proposition (Composition of Integrable is Integrable)

If $g$ is continuous and $f$ is Riemann-Stieltjes Integrable, and $g\circ f$ is well defined, then $g\circ f$ is integrable.

Properties of the Integral

1. If $f_1,f_2\in \mathcal{R}(\alpha )$ on $[a,b]$ then $f_1+f_2\in \mathcal{R}(\alpha )$ and if $cf\in \mathcal{R}(\alpha )$ for every constant $c$, and

$$\begin{array}{rl}{\int }_a^b(f_1+f_2)d\alpha & ={\int }_a^b{f}_{1}d\alpha +{\int }_a^b{f}_{2}d\alpha \\ {\int }_a^{b}cfd\alpha & =c{\int }_a^{b}fd\alpha \end{array}$$

2. If $f_1(x)\le f_2(x)$ on $[a,b]$ then

$${\int }_a^b{f}_{1}d\alpha \le {\int }_a^b{f}_{2}d\alpha$$

3. If $f\in \mathcal{R}(\alpha )$ on $[a,b]$ and if $a<c<b$, then $f\in \mathcal{R}(\alpha )$ on $[a,c]$ and $[c,b]$, and

$${\int }_a^{c}fd\alpha +{\int }_b^{c}fd\alpha ={\int }_a^{b}fd\alpha$$

4. If $f\in \mathcal{R}(\alpha )$ on $[a,b]$ and if $f(x)$ is bounded by $M$, then

$$\left|{\int }_a^{b}fd\alpha \right|\le M[\alpha (b)-\alpha (a)]$$

5. If $f\in \mathcal{R}({\alpha }_1)$ and $f\in \mathcal{R}({\alpha }_2)$ then $f\in \mathcal{R}({\alpha }_1+{\alpha }_2)$ and

$${\int }_a^{b}fd({\alpha }_1+{\alpha }_2)={\int }_a^{b}fd{\alpha }_1+{\int }_a^{b}fd{\alpha }_2$$

then if $c$ is some positive constant and $f\in \mathcal{R}(\alpha )$ then $f\in c\alpha$ and

$${\int }_a^{b}fd(c\alpha )=c{\int }_a^{b}fd\alpha$$

6. If $f,g\in \mathcal{R}(\alpha )$ on $[a,b]$ then

$$fg\in \mathcal{R}(\alpha )|f|\in \mathcal{R}(\alpha )\left|{\int }_a^{b}fd\alpha \right|\le {\int }_a^b|f|d\alpha$$

Proof is quite long so I will omit. However, this is just Rudin Theorem 6.12 and 6.13 copied verbatim.

Definition (Step Function)

The unit step function $I$ is defined by

$$I(x)=\{\begin{array}{ll}0& (x\le 0)\\ 1& (x>0)\end{array}$$

Theorem (Derivative Accumulator)

Assume $\alpha$ increases monotonically and ${\alpha }^{\prime }\in \mathcal{R}$ on $[a,b]$. Let $f$ be a bounded real function on $[a,b]$. Then $f\in \mathcal{R}(\alpha )$ if and only if $f{\alpha }^{\prime }\in \mathcal{R}$. In that case

$${\int }_a^{b}fd\alpha ={\int }_a^{b}f(x){\alpha }^{\prime }(x)dx$$

Proof:

TODO

Theorem (Lebesgue)

A bounded function $f:[a,b]\to \mathbb{R}$ is Riemann-Integrable iff the set

$$E:=\{x\in [a,b]\mid f\text{ is not continuous on }x\}$$

has Lebesgue measure $0$.

Acknowledgement

Special thanks to Ben for giving me his notes for this content.