8.9: Riemann Integration. Stieltjes Integrals
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.
(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
\[R \int_{a}^{b} f=R \int_{a}^{b} f(x) dm(x)\]
instead.
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 \(m : \mathcal{C} \rightarrow[0, \infty)\).
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
\[(\forall i) \quad A_{i}=\bigcup_{k} B_{i k}.\]
By additivity,
\[m A_{i}=\sum_{k} m B_{i k}.\]
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).\(\quad \square\)
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.\(\quad \square\)
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.\(\quad \square\)
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.)
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.\(\quad \square\)
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 \(E^{n}.\)
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.
(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
\[\left|P_{k}\right|=\frac{d A}{2^{k}} \rightarrow 0.\]
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.\(\quad \square\)
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.
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.
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).