# 8.9: Riemann Integration. Stieltjes Integrals

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I. In this section, $$\mathcal{C}$$ is the family of all intervals in $$E^{n},$$ and $$m$$ is an additive finite premeasure on $$\mathcal{C}$$ (or $$\mathcal{C}_{s}$$), such as the volume function $$v$$ (Chapter 7, §§1-2).

By a $$\mathcal{C}$$-partition of $$A \in \mathcal{C}$$ (or $$A \in \mathcal{C}_{s}$$), we mean a finite family

$\mathcal{P}=\left\{A_{i}\right\} \subset \mathcal{C}$

such that

$A=\bigcup_{i} A_{i} \text { (disjoint).}$

As we noted in §5, the Riemann integral,

$R \int_{A} f=R \int_{A} f dm,$

of $$f : E^{n} \rightarrow E^{1}$$ can be defined as its Lebesgue counterpart,

$\int_{A} f,$

with elementary maps replaced by simple step functions ("$$\mathcal{C}$$-simple" maps.) Equivalently, one can use the following construction, due to J. G. Darboux.

## Definitions

(a) Given $$f : E^{n} \rightarrow E^{*}$$ and a $$\mathcal{C}$$-partition

$\mathcal{P}=\left\{A_{1}, \ldots, A_{q}\right\}$

of $$A,$$ we define the lower and upper Darboux sums, $$\underline{S}$$ and $$\overline{S},$$ of $$f$$ over $$\mathcal{P}$$ (with respect to $$m$$) by

$\underline{S}(f, \mathcal{P})=\sum_{i=1}^{q} m A_{i} \cdot \inf f\left[A_{i}\right] \text { and } \overline{S}(f, \mathcal{P})=\sum_{i=1}^{q} m A_{i} \cdot \sup f\left[A_{i}\right].$

(b) The lower and upper Riemann integrals ("R-integrals") of $$f$$ on $$A$$ (with respect to $$m)$$ are

$\left. \begin{array}{l}{R \underline{\int}_{A} f=R \underline{\int}_{A} f dm=\sup_{\mathcal{P}} \underline{S}(f, \mathcal{P}) \text { and }} \\ {R \overline{\int}_{A} f=R \overline{\int}_{A} f dm=\inf_{\mathcal{P}} \overline{S}(f, \mathcal{P}),}\end{array} \right\}$

where the "inf" and "sup" are taken over all $$\mathcal{C}$$-partitions $$\mathcal{P}$$ of $$A$$.

(c) We say that $$f$$ is Riemann-integrable ("R-integrable") with respect to $$m$$ on $$A$$ iff $$f$$ is bounded on $$A$$ and

$R \underline{\int}_{A} f=R \overline{\int}_{A} f.$

We then set

$R \int_{A} f=R \underline{\int}_{A} f=R \overline{\int}_{A} f dm=R \int_{A} f dm$

and call it the Riemann integral ("R-integral") of $$f$$ on $$A.$$ "Classical" notation:

$R \int_{A} f(\overline{x}) dm(\overline{x}).$

If $$A=[a, b] \subset E^{1},$$ we also write

$R \int_{a}^{b} f=R \int_{a}^{b} f(x) dm(x)$

If $$m$$ is Lebesgue measure (or premeasure) in $$E^{1},$$ we write "$$dx$$" for "$$dm(x)$$."

For Lebesgue integrals, we replace "$$R$$" by "$$L$$," or we simply omit "$$R.$$"

If $$f$$ is R-integrable on $$A,$$ we also say that

$R \int_{A} f$

exists (note that this implies the boundedness of $$f);$$ note that

$R \underline{\int}_{A} f \text { and } R \overline{\int}_{A} f$

are always defined in $$E^{*}$$.

Below, we always restrict $$f$$ to a fixed $$A \in \mathcal{C}$$ (or $$A \in \mathcal{C}_{s}$$); $$\mathcal{P}, \mathcal{P}^{\prime}, \mathcal{P}^{\prime \prime}, \mathcal{P}^{*}$$ and $$\mathcal{P}_{k}$$ denote $$\mathcal{C}$$-partitions of $$A.$$

We now obtain the following result for any additive $$m : \mathcal{C} \rightarrow[0, \infty)$$.

## Corollary $$\PageIndex{1}$$

If $$\mathcal{P}$$ refines $$\mathcal{P}^{\prime}$$ (§1), then

$\underline{S}\left(f, \mathcal{P}^{\prime}\right) \leq \underline{S}(f, \mathcal{P}) \leq \overline{S}(f, \mathcal{P}) \leq \overline{S}\left(f, \mathcal{P}^{\prime}\right).$

Proof

Let $$\mathcal{P}^{\prime}=\left\{A_{i}\right\}, \mathcal{P}=\left\{B_{i k}\right\},$$ and

$(\forall i) \quad A_{i}=\bigcup_{k} B_{i k}.$

$m A_{i}=\sum_{k} m B_{i k}.$

Also, $$B_{i k} \subseteq A_{i}$$ implies

\begin{aligned} f\left[B_{i k}\right] & \subseteq f\left[A_{i}\right]; \\ \sup f\left[B_{i k}\right] & \leq \sup f\left[A_{i}\right]; \text { and } \\ \inf f\left[B_{i k}\right] & \geq \inf f\left[A_{i}\right]. \end{aligned}

So setting

$a_{i}=\inf f\left[A_{i}\right] \text { and } b_{i k}=\inf f\left[B_{i k}\right],$

we get

\begin{aligned} \underline{S}\left(f, \mathcal{P}^{\prime}\right)=\sum_{i} a_{i} m A_{i} &=\sum_{i} \sum_{k} a_{i} m B_{i k} \\ & \leq \sum_{i, k} b_{i k} m B_{i k}=\underline{S}(f, \mathcal{P}). \end{aligned}

Similarly,

$\overline{S}\left(f, \mathcal{P}^{\prime}\right) \leq \overline{S}(f, \mathcal{P}),$

and

$\underline{S}(f, \mathcal{P}) \leq \overline{S}(f, \mathcal{P})$

is obvious from (1).$$\quad \square$$

## Corollary $$\PageIndex{2}$$

For any $$\mathcal{P}^{\prime}$$ and $$\mathcal{P}^{\prime \prime}$$,

$\underline{S}\left(f, \mathcal{P}^{\prime}\right) \leq \overline{S}\left(f, \mathcal{P}^{\prime \prime}\right).$

Hence

$R \underline{\int}_{A} f \leq R \overline{\int}_{A} f.$

Proof

Let $$\mathcal{P}=\mathcal{P}^{\prime} \cap \mathcal{P}^{\prime \prime}$$ (see §1). As $$\mathcal{P}$$ refines both $$\mathcal{P}^{\prime}$$ and $$\mathcal{P}^{\prime \prime}$$, Corollary 1 yields

$\underline{S}\left(f, \mathcal{P}^{\prime}\right) \leq \underline{S}(f, \mathcal{P}) \leq \overline{S}(f, \mathcal{P}) \leq \overline{S}\left(f, \mathcal{P}^{\prime \prime}\right).$

Thus, indeed, no lower sum $$\underline{S}\left(f, \mathcal{P}^{\prime}\right)$$ exceeds any upper sum $$\overline{S}\left(f, \mathcal{P}^{\prime \prime}\right)$$.

Hence also,

$\sup _{\mathcal{P}^{\prime}} \underline{S}\left(f, \mathcal{P}^{\prime}\right) \leq \inf _{\mathcal{P}^{\prime \prime}} \overline{S}\left(f, \mathcal{P}^{\prime \prime}\right),$

i.e.,

$R\underline{\int}_{A} f \leq R \overline{\int}_{A} f,$

as claimed.$$\quad \square$$

## Lemma $$\PageIndex{1}$$

A map $$f : A \rightarrow E^{1}$$ is $$R$$-integrable iff $$f$$ is bounded and, moreover,

$(\forall \varepsilon>0) \text{ } (\exists \mathcal{P}) \quad \overline{S}(f, \mathcal{P})-\underline{S}(f, \mathcal{P})<\varepsilon.$

Proof

By formulas (1) and (2),

$\underline{S}(f, \mathcal{P}) \leq R \underline{\int}_{A} f \leq R \overline{\int}_{A} f \leq \overline{S}(f, \mathcal{P}).$

Hence (3) implies

$\left|R \overline{\int}_{A} f-R \underline{\int}_{A} f\right|<\varepsilon.$

As $$\varepsilon$$ is arbitrary, we get

$R \overline{\int}_{A} f=R \underline{\int}_{\underline{A}} f;$

so $$f$$ is R-integrable.

Conversely, if so, definitions (b) and (c) imply the existence of $$\mathcal{P}^{\prime}$$ and $$\mathcal{P}^{\prime \prime}$$ such that

$\underline{S}\left(f, \mathcal{P}^{\prime}\right)>R \int_{A} f-\frac{1}{2} \varepsilon$

and

$\overline{S}\left(f, \mathcal{P}^{\prime \prime}\right)<R \int_{A} f+\frac{1}{2} \varepsilon.$

Let $$\mathcal{P}$$ refine both $$\mathcal{P}^{\prime}$$ and $$\mathcal{P}^{\prime \prime}.$$ Then by Corollary 1,

\begin{aligned} \overline{S}(f, \mathcal{P})-\underline{S}(f, \mathcal{P}) & \leq \overline{S}\left(f, \mathcal{P}^{\prime \prime}\right)-\underline{S}\left(f, \mathcal{P}^{\prime}\right) \\ &<\left(R \int_{A} f+\frac{1}{2} \varepsilon\right)-\left(R \int_{A} f-\frac{1}{2} \varepsilon\right)=\varepsilon, \end{aligned}

as required.$$\quad \square$$

## Lemma $$\PageIndex{2}$$

Let $$f$$ be $$\mathcal{C}$$-simple; say, $$f=a_{i}$$ on $$A_{i}$$ for some $$\mathcal{C}$$-partition $$\mathcal{P}^{*}=$$ $$\left\{A_{i}\right\}$$ of $$A$$ (we then write

$f=\sum_{i} a_{i} C_{A_{i}}$

on $$A;$$ see Note 4 of §4).

Then

$R \underline{\int}_{A} f=R \overline{\int}_{A} f=\underline{S}\left(f, \mathcal{P}^{*}\right)=\overline{S}\left(f, \mathcal{P}^{*}\right)=\sum_{i} a_{i} m A_{i}.$

Hence any finite $$\mathcal{C}$$-simple function is R-integrable, with $$R \int_{A} f$$ as in (4).

Proof

Given any $$\mathcal{C}$$-partition $$\mathcal{P}=\left\{B_{k}\right\}$$ of $$A,$$ consider

$\mathcal{P}^{*} \acdot \mathcal{P}=\left\{A_{i} \cap B_{k}\right\}.$

As $$f=a_{i}$$ on $$A_{i} \cap B_{k}$$ (even on all of $$A_{i}$$),

$a_{i}=\inf f\left[A_{i} \cap B_{k}\right]=\sup f\left[A_{i} \cap B_{k}\right].$Also,

$A=\bigcup_{i, k}\left(A_{i} \cap B_{k}\right) \text { (disjoint)}$

and

$(\forall i) \quad A_{i}=\bigcup_{k}\left(A_{i} \cap B_{k}\right);$

so

$mA_{i}=\sum_{k} m\left(A_{i} \cap B_{k}\right)$

and

$\underline{S}(f, \mathcal{P})=\sum_{i} \sum_{k} a_{i} m\left(A_{i} \cap B_{k}\right)=\sum_{i} a_{i} m A_{i}=\underline{S}\left(f, \mathcal{P}^{*}\right)$

for any such $$\mathcal{P}$$.

Hence also

$\sum_{i} a_{i} m A_{i}=\sup_{\mathcal{P}} \underline{S}(f, \mathcal{P})=R \underline{\int}_{A} f.$

Similarly for $$R \overline{\int}_{A} f.$$ This proves (4).

If, further, $$f$$ is finite, it is bounded (by max $$\left|a_{i}\right|$$) since there are only finitely many $$a_{i};$$ so $$f$$ is R-integrable on $$A,$$ and all is proved.$$\quad \square$$

Note 1. Thus $$\underline{S}$$ and $$\overline{S}$$ are integrals of $$\mathcal{C}$$-simple maps, and definition (b) can be restated:

$R \underline{\int}_{A} f=\sup_{g} R \int_{A} g \text { and } R \overline{\int}_{A} f=\inf_{h} R \int_{A} h,$

taking the sup and inf over all $$\mathcal{C}$$-simple maps $$g, h$$ with

$g \leq f \leq h \text { on } A.$

(Verify by properties of glb and lub!)

Therefore, we can now develop R-integration as in §§4-5, replacing elementary maps by $$\mathcal{C}$$-simple maps, with $$S=E^{n}.$$ In particular, Problem 5 in §5 works out as before.

Hence linearity (Theorem 1 of §6) follows, with the same proof. One also obtains additivity (limited to $$\mathcal{C}$$-partitions). Moreover, the R-integrability of $$f$$ and $$g$$ implies that of $$f g, f \vee g, f \wedge g,$$ and $$|f|.$$ (See the Problems.)

## Theorem $$\PageIndex{1}$$

If $$f_{i} \rightarrow f$$ (uniformly) on $$A$$ and if the $$f_{i}$$ are R-integrable on $$A$$, so also is $$f.$$ Moreover,

$\lim_{i \rightarrow \infty} R \int_{A}\left|f-f_{i}\right|=0 \text { and } \lim_{i \rightarrow \infty} R \int_{A} f_{i}=R \int_{A} f.$

Proof

As all $$f_{i}$$ are bounded (definition (c)), so is $$f,$$ by Problem 10 of Chapter 4, §12.

Now, given $$\varepsilon>0,$$ fix $$k$$ such that

$(\forall i \geq k) \quad\left|f-f_{i}\right|<\frac{\varepsilon}{m A} \quad \text {on } A.$

Verify that

$(\forall i \geq k) \text{ } (\forall \mathcal{P}) \quad\left|\underline{S}\left(f-f_{i}, \mathcal{P}\right)\right|<\varepsilon \text { and }\left|\overline{S}\left(f-f_{i}, \mathcal{P}\right)\right|<\varepsilon;$

fix one such $$f_{i}$$ and choose a $$\mathcal{P}$$ such that

$\overline{S}\left(f_{i}, \mathcal{P}\right)-\underline{S}\left(f_{i}, \mathcal{P}\right)<\varepsilon,$

which one can do by Lemma 1. Then for this $$\mathcal{P}$$,

$\overline{S}(f, \mathcal{P})-\underline{S}(f, \mathcal{P})<3 \varepsilon.$

(Why?) By Lemma 1, then, $$f$$ is R-integrable on $$A$$.

Finally,

\begin{aligned}\left|R \int_{A} f-R \int_{A} f_{i}\right| & \leq R \int_{A}\left|f-f_{i}\right| \\ & \leq R \int_{A}\left(\frac{\varepsilon}{m A}\right)=m A\left(\frac{\varepsilon}{m A}\right)=\varepsilon \end{aligned}

for all $$i \geq k.$$ Hence the second clause of our theorem follows, too.$$\quad \square$$

## Corollary $$\PageIndex{3}$$

If $$f : E^{1} \rightarrow E^{1}$$ is bounded and regulated (Chapter 5, §10) on $$A=[a, b],$$ then $$f$$ is R-integrable on $$A.$$

In particular, this applies if $$f$$ is monotone, or of bounded variation, or relatively continuous, or a step function, on $$A.$$

Proof

By Lemma 2, this applies to $$\mathcal{C}$$-simple maps.

Now, let $$f$$ be regulated (e.g., of the kind specified above).

Then by Lemma 2 of Chapter 5, §10,

$f=\lim _{i \rightarrow \infty} g_{i} \quad \text {(uniformly)}$

for finite $$\mathcal{C}$$-simple $$g_{i}$$.

Thus $$f$$ is R-integrable on $$A$$ by Theorem 1.$$\quad \square$$

II. Henceforth, we assume that $$m$$ is a measure on a $$\sigma$$-ring $$\mathcal{M} \supseteq \mathcal{C}$$ in $$E^{n}$$, with $$m<\infty$$ on $$\mathcal{C}$$. (For a reader who took the "limited approach," it is now time to consider §§4-6 in full.) The measure $$m$$ may, but need not, be Lebesgue measure in $$E^{n}.$$

## Theorem $$\PageIndex{2}$$

If $$f : E^{n} \rightarrow E^{1}$$ is R-integrable on $$A \in \mathcal{C},$$ it is also Lebesgue integrable (with respect to $$m$$ as above) on $$A,$$ and

$L \int_{A} f=R \int_{A} f,$

Proof

Given a $$\mathcal{C}$$-partition $$\mathcal{P}=\left\{A_{i}\right\}$$ of $$A,$$ define the $$\mathcal{C}$$-simple maps

$g=\sum_{i} a_{i} C_{A_{i}} \text { and } h=\sum_{i} b_{i} C_{A_{i}}$

with

$a_{i}=\inf f\left[A_{i}\right] \text { and } b_{i}=\sup f\left[A_{i}\right].$

Then $$g \leq f \leq h$$ on $$A$$ with

$\underline{S}(f, \mathcal{P})=\sum_{i} a_{i} m A_{i}=L \int_{A} g$

and

$\overline{S}(f, \mathcal{P})=\sum_{i} b_{i} m A_{i}=L \int_{A} h.$

By Theorem 1(c) in §5,

$\underline{S}(f, \mathcal{P})=L \int_{A} g \leq L \underline{\int}_{A} f \leq L \overline{\int}_{A} f \leq L \int_{A} h=\overline{S}(f, \mathcal{P}).$

As this holds for any $$\mathcal{P},$$ we get

$R \underline{\int}_{A} f=\sup_{\mathcal{P}} \underline{S}(f, \mathcal{P}) \leq L \underline{\int}_{A} f \leq L \overline{\int}_{A} f=\inf_{\mathcal{P}} \overline{S}(f, \mathcal{P})=R \overline{\int}_{A} f.$

But by assumption,

$R \underline{\int}_{A} f=R \overline{\int}_{A} f.$

Thus these inequalities become equations:

$R \int_{A} f=\underline{\int}_{A} f=\overline{\int}_{A} f=R \int_{A} f.$

Also, by definition (c), $$f$$ is bounded on $$A;$$ so $$|f|<K<\infty$$ on $$A.$$ Hence

$\left|\int_{A} f\right| \leq \int_{A}|f| \leq K \cdot m A<\infty.$

Thus

$\underline{\int}_{A} f=\overline{\int}_{A} f \neq \pm \infty,$

i.e., $$f$$ is Lebesgue integrable, and

$L \int_{A} f=R \int_{A} f,$

as claimed.$$\quad \square$$

Note 2. The converse fails. For example, as shown in the example in §4 , $$f=C_{R}$$ ($$R=$$ rationals) is L-integrable on $$A=[0,1].$$

Yet $$f$$ is not $$R$$-integrable.

For $$\mathcal{C}$$-partitions involve intervals containing both rationals (on which $$f=1$$) and irrationals (on which $$f=0$$). Thus for any $$\mathcal{P}$$,

$\underline{S}(f, \mathcal{P})=0 \text { and } \overline{S}(f, \mathcal{P})=1 \cdot m A=1.$

(Why?) So

$R \overline{\int}_{A} f=\inf \overline{S}(f, \mathcal{P})=1,$

while

$R \underline{\int}_{A} f=0 \neq R \overline{\int}_{A} f.$

Note 3. By Theorem 1, any $$R \int_{A} f$$ is also a Lebesgue integral. Thus the rules of §§5-6 apply to R-integrals, provided that the functions involved are R-integrable. For a deeper study, we need a few more ideas.

## Definitions (continued)

(d) The mesh $$|\mathcal{P}|$$ of a $$\mathcal{C}$$-partition $$\mathcal{P}=\left\{A_{1}, \ldots, A_{q}\right\}$$ is the largest of the diagonals $$d A_{i}:$$

$|\mathcal{P}|=\max \left\{d A_{1}, d A_{2}, \ldots, d A_{q}\right\}.$

Note 4. For any $$A \in \mathcal{C},$$ there is a sequence of $$\mathcal{C}$$-partitions $$\mathcal{P}_{k}$$ such that

(i) each $$P_{k+1}$$ refines $$P_{k}$$ and

(ii) $$\lim _{k \rightarrow \infty}\left|P_{k}\right|=0$$.

To construct such a sequence, bisect the edges of $$A$$ so as to obtain $$2^{n}$$ subintervals of diagonal $$\frac{1}{2} dA$$ (Chapter 3, §7). Repeat this with each of the subintervals, and so on. Then

$\left|P_{k}\right|=\frac{d A}{2^{k}} \rightarrow 0.$

## Lemma $$\PageIndex{3}$$

Let $$f : A \rightarrow E^{1}$$ be bounded. Let $$\left\{\mathcal{P}_{k}\right\}$$ satisfy (i) of Note 4. If $$P_{k}=\left\{A_{1}^{k}, \ldots, A_{q_{k}}^{k}\right\},$$ put

$g_{k}=\sum_{i=1}^{q_{k}} C_{A_{i}^{k}} \inf f\left[A_{i}^{k}\right]$

and

$h_{k}=\sum_{i=1}^{q_{k}} C_{A_{i}^{k} \sup } f\left[A_{i}^{k}\right].$

Then the functions

$g=\sup _{k} g_{k} \text { and } h=\inf _{k} h_{k}$

are Lebesgue integrable on $$A,$$ and

$\int_{A} g=\lim _{k \rightarrow \infty} \underline{S}\left(f, \mathcal{P}_{k}\right) \leq R \underline{\int}_{A} f \leq R \overline{\int}_{A} f \leq \lim_{k \rightarrow \infty} \overline{S}\left(f, \mathcal{P}_{k}\right)=\int_{A} h.$

Proof

As in Theorem 2, we obtain $$g_{k} \leq f \leq h_{k}$$ on $$A$$ with

$\int_{A} g_{k}=\underline{S}\left(f, \mathcal{P}_{k}\right)$

and

$\int_{A} h_{k}=\overline{S}\left(f, \mathcal{P}_{k}\right).$

Since $$\mathcal{P}_{k+1}$$ refines $$\mathcal{P}_{k},$$ it also easily follows that

$g_{k} \leq g_{k+1} \leq \sup _{k} g_{k}=g \leq f \leq h=\inf _{k} h_{k} \leq h_{k+1} \leq h_{k}.$

(Verify!)

Thus $$\left\{g_{k}\right\} \uparrow$$ and $$\left\{h_{k}\right\} \downarrow,$$ and so

$g=\sup _{k} g_{k}=\lim _{k \rightarrow \infty} g_{k} \text { and } h=\inf _{k} h_{k}=\lim _{k \rightarrow \infty} h_{k}.$

Also, as $$f$$ is bounded

$\left(\exists K \in E^{1}\right) \quad|f|<K \text { on } A.$

The definition of $$g_{k}$$ and $$h_{k}$$ then implies

$(\forall k) \quad\left|g_{k}\right| \leq K \text { and }\left|h_{k}\right| \leq K \text { (why?),}$

with

$\int_{A}(K)=K \cdot m A<\infty.$

The $$g_{k}$$ and $$h_{k}$$ are measurable (even simple) on $$A,$$ with $$g_{k} \rightarrow g$$ and $$h_{k} \rightarrow h$$.

Thus by Theorem 5 and Note 1, both from §6, $$g$$ and $$h$$ are Lebesgue integrable, with

$\int_{A} g=\lim_{k \rightarrow \infty} \int_{A} g_{k} \text { and } \int_{A} h=\lim_{k \rightarrow \infty} \int_{A} h_{k}.$

As

$\int_{A} g_{k}=\underline{S}\left(f, \mathcal{P}_{k}\right) \leq R \underline{\int}_{A} f$

and

$\int_{A} h_{k}=\overline{S}\left(f, \mathcal{P}_{k}\right) \geq R \overline{\int}_{A} f,$

passage to the limit in equalities yields (6). Thus the lemma is proved.$$\quad \square$$

## Lemma $$\PageIndex{4}$$

With all as in Lemma 3, let $$B$$ be the union of the boundaries of all intervals from all $$\mathcal{P}_{k}.$$ Let $$\left|\mathcal{P}_{k}\right| \rightarrow 0.$$ Then we have the following.

(i) If $$f$$ is continuous at $$p \in A,$$ then $$h(p)=g(p)$$.

(ii) The converse holds if $$p \in A-B$$.

Proof

For each $$k, p$$ is in one of the intervals in $$\mathcal{P}_{k};$$ call it $$A_{kp}$$.

If $$p \in A-B, p$$ is an interior point of $$A_{kp};$$ so there is a globe

$G_{p}\left(\delta_{k}\right) \subseteq A_{kp}.$

Also, by the definition of $$g_{k}$$ and $$h_{k}$$,

$g_{k}(p)=\inf f\left[A_{k p}\right] \text { and } h_{k}=\sup f\left[A_{k p}\right].$

(Why?)

Now fix $$\varepsilon>0.$$ If $$g(p)=h(p),$$ then

$0=h(p)-g(p)=\lim _{k \rightarrow \infty}\left[h_{k}(p)-g_{k}(p)\right];$

so

$(\exists k) \quad\left|h_{k}(p)-g_{k}(p)\right|=\sup f\left[A_{k p}\right]-\inf f\left[A_{k p}\right]<\varepsilon.$

As $$G_{p}\left(\delta_{k}\right) \subseteq A_{k p},$$ we get

$\left(\forall x \in G_{p}\left(\delta_{k}\right)\right) \quad|f(x)-f(p)| \leq \sup f\left[A_{k p}\right]-\inf f\left[A_{k p}\right]<\varepsilon,$

proving continuity (clause (ii)).

For (i), given $$\varepsilon>0,$$ choose $$\delta>0$$ so that

$\left(\forall x, y \in A \cap G_{p}(\delta)\right) \quad|f(x)-f(y)|<\varepsilon.$

Because

$(\forall \delta>0)\left(\exists k_{0}\right)\left(\forall k>k_{0}\right) \quad\left|\mathcal{P}_{k}\right|<\delta$

for $$k>k_{0}, A_{k p} \subseteq G_{p}(\delta).$$ Deduce that

$\left(\forall k>k_{0}\right) \quad\left|h_{k}(p)-g_{k}(p)\right| \leq \varepsilon. \quad \square$

Note 5. The Lebesgue measure of $$B$$ in Lemma 4 is zero; for $$B$$ consists of countably many "faces" (degenerate intervals), each of measure zero.

## Theorem $$\PageIndex{3}$$

A map $$f : A \rightarrow E^{1}$$ is R-integrable on $$A$$ (with $$m=$$ Lebesgue measure) iff $$f$$ is bounded on $$A$$ and continuous on $$A-Q$$ for some $$Q$$ with $$m Q=0$$.

Note that relative continuity on $$A-Q$$ is not enough-take $$f=C_{R}$$ of Note 2.

Proof

If these conditions hold, choose $$\left\{\mathcal{P}_{k}\right\}$$ as in Lemma 4.

Then by the assumed continuity, $$g=h$$ on $$A-Q, m Q=0$$.

Thus

$\int_{A} g=\int_{A} h$

(Corollary 2 in §5).

Hence by formula (6), $$f$$ is R-integrable on $$A$$.

Conversely, if so, use Lemma 1 with

$\varepsilon=1, \frac{1}{2}, \ldots, \frac{1}{k}, \ldots$

to get for each $$k$$ some $$\mathcal{P}_{k}$$ such that

$\overline{S}\left(f, \mathcal{P}_{k}\right)-\underline{S}\left(f, \mathcal{P}_{k}\right)<\frac{1}{k} \rightarrow 0.$

By Corollary 1, this will hold if we refine each $$\mathcal{P}_{k},$$ step by step, so as to achieve properties (i) and (ii) of Note 4 as well. Then Lemmas 3 and 4 apply.

As

$\overline{S}\left(f, \mathcal{P}_{k}\right)-\underline{S}\left(f, \mathcal{P}_{k}\right) \rightarrow 0,$

formula (6) show that

$\int_{A} g=\lim_{k \rightarrow \infty} \underline{S}\left(f, \mathcal{P}_{k}\right)=\lim_{k \rightarrow \infty} \overline{S}\left(f, \mathcal{P}_{k}\right)=\int_{A} h.$

As $$h$$ and $$g$$ are integrable on $$A$$,

$\int_{A}(h-g)=\int_{A} h-\int_{A} g=0.$

Also $$h-g \geq 0;$$ so by Theorem 1(h) in §5, $$h=g$$ on $$A-Q^{\prime}, m Q^{\prime}=0$$ (under Lebesgue measure). Hence by Lemma 4, $$f$$ is continuous on

$A-Q^{\prime}-B,$

with $$mB=0$$ (Note 5).

Let $$Q=Q^{\prime} \cup B.$$ Then $$m Q=0$$ and

$A-Q=A-Q^{\prime}-B;$

so $$f$$ is continuous on $$A-Q.$$ This completes the proof.$$\quad \square$$

Note 6. The first part of the proof does not involve $$B$$ and thus works even if $$m$$ is not the Lebesgue measure. The second part requires that $$mB=0$$.

Theorem 3 shows that R-integrals are limited to a.e. continuous functions and hence are less flexible than L-integrals: Fewer functions are R-integrable, and convergence theorems (§6, Theorems 4 and 5) fail unless $$R \int_{A} f$$ exists.

III. Functions $$f : E^{n} \rightarrow E^{s}\left(C^{s}\right).$$ For such functions, R-integrals are defined componentwise (see §7). Thus $$f=\left(f_{1}, \ldots, f_{s}\right)$$ is R-integrable on $$A$$ iff all $$f_{k}$$ $$(k \leq s)$$ are, and then

$R \int_{A} f=\sum_{k=1}^{s} \overline{e}_{k} R \int_{A} f_{k}.$

A complex function $$f$$ is R-integrable iff $$f_{re}$$ and $$f_{im}$$ are, and then

$R \int_{A} f=R \int_{A} f_{re}+i R \int_{A} f_{im}.$

Via components, Theorems 1 to 3, Corollaries 3 and 4, additivity, linearity, etc., apply.

IV. Stieltjes Integrals. Riemann used Lebesgue premeasure $$v$$ only. But as we saw, his method admits other premeasures, too.

Thus in $$E^{1},$$ we may let $$m$$ be the $$LS$$ premeasure $$s_{\alpha}$$ or the $$LS$$ measure $$m_{\alpha}$$ where $$\alpha \uparrow$$ (Chapter 7, §5, Example (b), and Chapter 7, §9).

Then

$R \int_{A} f dm$

is called the Riemann-Stieltjes (RS) integral of $$f$$ with respect to $$\alpha,$$ also written

$R \int_{A} f d \alpha \quad \text {or} \quad R \int_{a}^{b} f(x) d \alpha(x)$

(the latter if $$A=[a, b]$$); $$f$$ and $$\alpha$$ are called the integrand and integrator, respectively.

If $$\alpha(x)=x, m_{\alpha}$$ becomes the Lebesgue measure, and

$R \int f(x) d \alpha(x)$

turns into

$R \int f(x) dx.$

Our theory still remains valid; only Theorem 3 now reads as follows.

## Corollary $$\PageIndex{4}$$

If $$f$$ is bounded and a.e. continuous on $$A=[a, b]$$ (under an LS measure $$m_{\alpha}$$) then

$R \int_{a}^{b} f d \alpha$

exists. The converse holds if $$\alpha$$ is continuous on $$A$$.

For by Notes 5 and 6, the "only if" in Theorem 3 holds if $$m_{\alpha} B=0.$$ Here consists of countably many endpoints of partition subintervals. But (see Chapter §9) $$m_{\alpha}\{p\}=0$$ if $$\alpha$$ is continuous at $$p.$$ Thus the later implies $$m_{\alpha} B=0$$.

RS-integration has been used in many fields (e.g., probability theory, physics, etc.), but it is superseded by LS-integration, i.e., Lebesgue integration with respect to $$m_{\alpha},$$ which is fully covered by the general theory of §§1-8.

Actually, Stieltjes himself used somewhat different definitions (see Problems 10-13), which amount to applying the set function $$\sigma_{\alpha}$$ of Problem 9 in Chapter 7, §4, instead of $$s_{\alpha}$$ or $$m_{\alpha}.$$ We reserve the name "Stieltjes integrals," denoted

$S \int_{a}^{b} f d \alpha,$

for such integrals, and "RS-integrals" for those based on $$m_{\alpha}$$ or $$s_{\alpha}$$ (this terminology is not standard).

Observe that $$\sigma_{\alpha}$$ need not be $$\geq 0.$$ Thus for the first time, we encounter integration with respect to sign-changing set functions. A much more general theory is presented in §10 (see Problem 10 there).

8.9: Riemann Integration. Stieltjes Integrals is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.